Student’s BookNumerical Functional Analysis

Editor: Stefan Engblom

Fall 2014

Preface

This little book collects some of the student’s output during the course Numerical Func-tional Analysis, which was given for the first time in the fall 2014 at the Department ofInformation technology, Uppsala university.

The course consisted of a total of five lectures, the first three of which went over Metricspaces, Normed spaces, and Inner product spaces, thus following closely the first threechapters of Kreyszig’s book Introductory functional analysis with applications. Two morelectures were devoted to the five ‘Big’ theorems of functional analysis as presented by thestudents themselves: the Hahn-Banach theorem, the Uniform boundedness theorem, theOpen mapping theorem, the Closed graph theorem, and the Banach fixed point theorem.

The final part of the course consisted of short essays on various topics, typically con-necting Numerical analysis and Functional analysis in one way or the other. The essayswere improved by double open review among the students themselves after which the finalversion entered this collection. A selection of student’s solution to book exercises has alsobeen included: the selection was very loosely based on the difficulty of the exercise itselfand on the solution quality.

Stefan EngblomUppsala, January 2015

Contents

I Student’s essays 3

Dual-consistent approximations of PDEs and super-convergent functionaloutputMartin Almquist 4

Lax equivalence theoremSiyang Wang 15

The Schauder fixed point theoremHannes Frenander 20

Existence of a weak solution to the Stokes equationsHanna Holmgren 24

The Lax-Milgram theoremAndrea Alessandro Ruggiu 31

A time dependent mapping from Cartesian coordinate systems into curvi-linear coordinate systems and the geometric conservation lawSamira Nikkar 37

The Babuska inf-sup conditionSimon Sticko 42

Uniformly best wavenumber approximations by spatial central differenceoperatorsViktor Linders 47

The Ekeland variational principle with applicationsMarkus Wahlsten 54

Construction of Sobolev spacesCristina La Cognata 59

1

The Arzela -Ascoli theoremTomas Lundquist 65

A completeness theorem for non-self-adjoint eigenvalue problemsSaleh RezaeiRavesh 69

The metric Lax and applicationsCheng Gong 77

Fixed-point proof of Lax-Milgram theoremFatemeh Ghasemi 82

II Solutions to exercises 87

1 Metric spaces 88

2 Normed spaces 96

3 Inner product spaces 112

2

Part I

Student’s essays

3

Dual-consistent approximations of PDEs andsuper-convergent functional output

Martin Almquist∗

December 22, 2014

Abstract

Finite-difference operators satisfying the summation-by-parts (SBP)rule can be used to obtain high-order accurate time-stable schemes forpartial differential equations (PDEs), when the boundary and inter-face conditions are imposed weakly by the simultaneous approximationterm method (SAT). The most common SBP-SAT discretizations areaccurate of order 2p in the interior and order p close to the boundaries,which yields a global accuracy of p+ 1 for first order PDEs. However,Berg and Nordstrom [2012] and Hicken and Zingg [2014] have shownthat any linear functional computed from the time-dependent numer-ical solution will be accurate of order 2p if the spatial discretization isdual-consistent.

This text discusses the concepts of dual problem and dual consis-tency. We walk through some of the definitions and results in Bergand Nordstrom [2012] and Hicken and Zingg [2014] while emphasizingthe underlying functional analysis considerations.

1 Dual problems and dual consistency

We introduce the concepts of dual problem and dual consistency by consid-ering linear partial differential equations with homogeneous boundary condi-tions. Let L be a linear differential operator of order m on a domain Ω andconsider the problem

Lu− f = 0, x ∈ Ω, (1.1)

∗Division of Scientific Computing, Department of Information Technology, Uppsalauniversity, SE-751 05 Uppsala, Sweden. martin.almquist@it.uu.se

1

subject to homogeneous boundary conditions. f ∈ L2(Ω) is independent ofthe solution u ∈ Hk(Ω) ⊂ L2(Ω). Here Hk(Ω) = W k,2(Ω) is the Sobolevspace consisting of functions whose weak derivatives up to order k (for someinteger k) exist and are square-integrable on Ω. We assume that m ≤ k.

For v, w ∈ L2(Ω) we use the notation

(v, w) :=

∫

Ω

vw dx. (1.2)

for the standard L2 inner product. We now introduce the formal adjoint ofthe operator L.

Definition 1.1. Let u ∈ Hk(Ω). The formal adjoint of a linear differentialoperator L is the operator L† such that

(φ,Lu) = (L†φ, u)

The word formal in the definition emphasizes that we are not yet specify-ing the domain D(L†) of the adjoint operator. The reason for this definitionis that L is not bounded on L2(Ω), in general. Hence we need to distinguishbetween the formal adjoint and the Hilbert-adjoint (which concerns boundedoperators) even though L2(Ω) is a Hilbert space. Note also that the defi-nition of L† is abstract; an explicit expression for L†φ can be obtained via(repeated) integration by parts on (φ,Lu).

We now consider a bounded linear functional of the solution u,

J (u) = (z, u) (1.3)

for some z ∈ L2(Ω). Note that since L2(Ω) is a Hilbert space, the representa-tion (1.3) includes all linear functionals, by Riesz’s representation theorem.To derive the adjoint equation of (1.1) with respect to the functional J , wetake the inner product of (1.1) with a function φ ∈ D(L†) and express J as

J (u) = (z, u) = (z, u)− (φ,Lu− f) = (φ, f)− (L†φ− z, u). (1.4)

We now obtain the adjoint or dual equation by seeking φ such that J becomesindependent of u, which requires

L†φ− z = 0, (1.5)

yielding J = (φ, f).

2

Remark. To simplify the analysis we have disregarded the boundary con-ditions. In general, the dual problem depends on the type of boundaryconditions in the primal problem, but not on the particular boundary data.Boundary conditions for the dual problem can be constructed as the minimalset of homogeneous conditions which, together with the homogeneous primalboundary conditions, make all boundary terms resulting from the integrationby parts procedure vanish. We outline the procedure for a model problem inSection 2.3.

Definition 1.2. The continuous dual problem associated with the primalproblem (1.1) and the functional J is

L†φ− z = 0

subject to the dual boundary conditions. The solution φ is called the dualvariable.

Note that the forcing function f and functional representation z in theprimal problem have switched roles in the dual problem.

We now turn to the discrete case. Let

Lhuh − f = 0 (1.6)

be a consistent discretization of (1.1) including boundary conditions, whereuh ∈ Rn may hold basis function coefficients and/or collocation values. Lh isa bounded linear operator which can be represented by matrix multiplication,Lh : Rn −→ Rn. Let (·, ·)h denote a discrete inner product on Rn × Rn suchthat (Rn, (·, ·)h) is a Hilbert space. We can then use the Hilbert-adjoint:

Definition 1.3. The Hilbert-adjoint of Lh is the operator L†h : Rn −→ Rn

such that(φh,Lhuh)h = (L†hφh, uh)h

for all φh, uh ∈ Rn.

Let Jh(uh) = (z, uh)h be an approximation of the functional J . Wenow derive the discrete dual problem by taking the inner product of thediscretization (1.6) with φh ∈ Rn and expressing Jh as

Jh(uh) = (z, uh)h − (φh,Lhuh − fh)h = (φh, fh)h − (L†hφh − z, uh)h . (1.7)

We obtain the discrete dual problem by seeking φh such that Jh is indepen-dent of uh, which requires

L†hφh − z = 0 . (1.8)

3

Definition 1.4. The discrete dual problem associated with (1.6) and thefunctional Jh is

L†hφh − z = 0

Definition 1.5. A discretization (1.6) of a continuous problem (1.1) is calleddual consistent if the discrete dual problem (1.8) is a consistent discretizationof the continuous dual problem (1.5).

1.1 Time-dependent problems

We now extend the concepts of dual problems and dual consistency to time-dependent problems. Consider the problem

ut + Lu− f = 0, x ∈ Ω, t > 0 (1.9)

subject to homogeneous boundary and initial conditions. Let

ddtuh + Lhuh − f = 0, t > 0 (1.10)

be a semi-descretization of (1.9), including the boundary conditions.

Definition 1.6. The semi-discretization (1.10) is called spatially dual-consistentif

Lhuh − f = 0

is a dual-consistent discretization of the steady adjoint problem, i.e., the dualproblem corresponding to (1.9) with ut = 0.

2 SBP-SAT discretizations

We begin this section by introducing notation and some basic concepts re-lated to SBP-SAT discretizations, and then proceed to consider stable anddual-consistent SBP-SAT discretizations.

2.1 Definitions

Consider as an example the advection equation on 0 ≤ x ≤ 1,

ut + ux = f

u(0, t) = g(t).(2.1)

4

We introduce the N + 1 equidistant grid points

xj = jh, j = 0, 1, . . . , N, h =1

N(2.2)

and denote the semi-discrete solution vector corresponding to grid size hby uh, where uh = [uh,0, uh,1, . . . , uh,N ]T ∈ RN+1 and uh,j(t) approximatesu(xj, t). The restriction [f(x0), f(x1), . . . , f(xN)]T of f onto the grid willalso be denoted by f , with no risk for confusion. Spatial derivatives will beapproximated by first-derivative SBP operators:

Definition 2.1. A matrix D is called a first-derivative SBP operator if Dcan be written as

D = H−1Q

where H = HT is positive definite, and thus defines a norm, and Q satisfies

Q+QT = diag[−1, 0, . . . , 0, 1].

In this text we will restrict ourselves to diagonal norm matrices H. Inthat case, D consists of a 2pth-order accurate centered difference stencil inthe interior and pth-order accurate one-sided stencils near the boundaries.The global accuracy for first order problems can then be shown to be p + 1Svard and Nordstrom [2006].

Let u, v ∈ RN+1. We define the discrete inner product and norm as

(u, v)h = uTHv, ‖u‖2h = (u, u)h. (2.3)

The inner product space (RN+1, (·, ·)h) is complete and thus a Hilbert space.Now consider a discretization matrix Lh. We can compute the Hilbert-adjointL†h with respect to the inner product (2.3) explicitly:

Lemma 2.1. Let Lh be an (N + 1)× (N + 1) matrix with real entries. Thenthe Hilbert adjoint is

L†h = H−1LThH (2.4)

Proof. Starting from Definition 1.3, we have

(φh, Lhuh)h = (L†hφh, uh)h ∀φh, uh ∈ Rn ⇐⇒φThHLhuh = (L†hφh)THuh ∀φh, uh ∈ Rn ⇐⇒φThHLhuh = φT

h (L†h)THuh ∀φh, uh ∈ Rn ⇐⇒HLh = (L†h)TH ⇐⇒

(L†h)T = HLhH−1 ⇐⇒

L†h = H−1LThH

5

2.2 Stable SBP-SAT discretizations

The SBP-SAT semi-discretization of (2.1) can be written

d

dtuh +Duh = f + τH−1e0(eT0 uh − g) (2.5)

where e0 = [1, 0, . . . , 0]T and τ is a penalty parameter, which will be chosensuch that the scheme (2.5) is stable. Multiplying (2.5) by uThH from the leftand setting f = 0, g = 0 yields

d

dt‖uh‖2

h = (1 + 2τ)u2h,0 − u2

h,N , (2.6)

from which it is clear that τ ≤ −12

leads to a stable scheme. Thus, the penaltyparameter τ is allowed to vary in a semi-infinite range. This flexibility createsa possibility to impose additional conditions on the scheme, such as dualconsistency.

2.3 Dual-consistent SBP-SAT discretizations

Consider again the advection equation (2.1) on 0 ≤ x ≤ 1,

ut + ux = f

u(0, t) = 0.(2.7)

with some initial condition and an associated linear functional

J (u) =

∫ 1

0

zu dx. (2.8)

Note that J is a time-dependent functional. The dual problem is obtainedby setting ut = 0 and finding φ such that J = (φ, f). Using integration byparts and inserting the boundary condition in (2.7), we obtain

J (u) = J (u)−∫ 1

0

φ(ux − f) dx = −φ(1, t)u(1, t) +

∫ 1

0

(z + φx)u dx+ (φ, f),

and hence the dual problem is

−φx = z

φ(1, t) = 0 .(2.9)

Note that the sign has changed and the boundary condition is located at theopposite boundary, compared to the primal problem. We are now ready tostate the main result in this text:

6

Theorem 2.2. Letd

dtuh + Lhuh = f (2.10)

be a stable and spatially dual consistent SBP-SAT approximation of the con-tinuous problem

ut + Lu = f. (2.11)

A linear functionalJ (u) = (z, u) (2.12)

approximated byJh(uh) = (z, uh)h (2.13)

satisfiesJh(uh) = J (u) +O(h2p) (2.14)

Proof. The proof can be found in Berg and Nordstrom [2012].

We now investigate whether the scheme (2.5) can be made spatially dual-consistent. Assuming homogeneous boundary conditions (g = 0) and ne-glecting the time-derivative, the scheme can be written as

Lhuh = f (2.15)

whereLh = D − τH−1e0e

T0 . (2.16)

By definition the scheme is spatially dual-consistent if the discrete dual prob-lem

L†hφh − z = 0 (2.17)

is a consistent discretization of (2.9). Using Lemma 2.1 and the SBP prop-erty, we have

L†hφh−z = H−1LThHφh−z = H−1

(−Q+ eNe

TN − (τ + 1)e0e

T0

)φh−z (2.18)

which is a consistent discretization of (2.9) if and only if τ = −1. For τ 6= −1the scheme imposes a boundary condition at x = 0 which does not exist in thecontinuous equation. Note that the requirement τ = −1 does not contradictthe stability requirement τ ≤ −1

2. Thus, we expect convergence of order 2p

for any linear functional of the solution if we choose τ = −1.

7

3 Numerical experiments

For the interested reader, we include a set of numerical experiments whichvalidate the theoretical results. We solve the advection equation using thescheme (2.5) and consider the functional

J (u) =

∫ 1

0

cos(4πx)u dx. (3.1)

Using the technique of manufactured solutions, we choose the exact solution

u = cos(4πx) sin(2πt), (3.2)

which corresponds to the forcing function

f = 2π cos(4πx) cos(2πt)− 4π sin(4πx) sin(2πt). (3.3)

The analytic solution is used as initial and boundary data. With the solution(3.2), the exact time-dependent functional becomes

J =1

2sin 2πt. (3.4)

We use an SBP-SAT discretization of internal order 6 (2p = 6) and henceexpect 4th order convergence in the l2-norm for any τ ≤ −1

2. We expect

the functional error to converge as (at least) h4 for τ ≤ −12. For the dual-

consistent choice τ = −1, we expect 6th order convergence in the functional.This is verified by Figures 3.1-3.3, which show the convergence rates forτ = −0.5, τ = −1 and τ = −1.5, respectively. The l2-error converges as h4

regardless of τ , whereas the functional error converges as h6 for τ = −1 andas h4 otherwise.

Figure 3.4 shows both the l2-error and the functional error as functionsof τ , using N = 256. As expected, the functional error is minimized by thedual-consistent choice τ = −1. It is interesting to note that also the l2-erroris minimized by τ ≈ −1 in this case, although none of the theory in thistext leads us to expect that. However, the l2-error is much less sensitive tothe choice of τ than the functional error, which decreases by approximately4 magnitudes in the vicinity of τ = −1.

References

J. Berg and J. Nordstrom. Superconvergent functional output fortime-dependent problems using finite differences on summation-by-parts

8

N

101

102

103

10-10

10-8

10-6

10-4

10-2

100

l2-error

4th order reference

functional error

6th order reference

Figure 3.1: Convergence in ‖·‖l2 and functional for τ = −0.5; dual-inconsistent discretization.

N

101

102

103

10-15

10-10

10-5

100

l2-error

4th order reference

functional error

6th order reference

Figure 3.2: Convergence in ‖·‖l2 and functional for τ = −1; dual-consistentdiscretization.

9

N

101

102

103

10-10

10-8

10-6

10-4

10-2

100

l2-error

4th order reference

functional error

6th order reference

Figure 3.3: Convergence in ‖·‖l2 and functional for τ = −1.5; dual-inconsistent discretization.

τ

-2 -1.5 -1 -0.5

l2-e

rror

10-6

10-5

10-4

(a)

τ

-2 -1.5 -1 -0.5

functional err

or

10-12

10-10

10-8

10-6

(b)

Figure 3.4: The (a) l2-error ‖u−uh‖l2 and (b) functional error, as functions ofthe penalty parameter τ . Only τ = −1 yields a dual-consistent discretization.

10

form. Journal of Computational Physics, 231(20):6846–6860, Aug. 2012.ISSN 00219991. doi: 10.1016/j.jcp.2012.06.032. URL http://www.

sciencedirect.com/science/article/pii/S0021999112003476.

J. Hicken and D. Zingg. Dual consistency and functional accuracy: a finite-difference perspective. Journal of Computational Physics, 256:161–182,Jan. 2014. ISSN 00219991. doi: 10.1016/j.jcp.2013.08.014. URL http://

www.sciencedirect.com/science/article/pii/S0021999113005524.

M. Svard and J. Nordstrom. On the order of accuracy for difference approx-imations of initial-boundary value problems. J. Comput. Physics, 218:333–352, October 2006.

11

Lax Equivalence Theorem

Siyang Wang∗

December 26, 2014

Abstract

Convergence of numerical methods for solving differential equa-tions is of great importance, but difficult to prove in a direct way. TheLax equivalence theorem links consistency, stability and convergence.This enables us to prove convergence by consistency and stability anal-ysis. In this essay, the Lax equivalence theorem is proved from thefunctional analysis point of view. A differential equation example isalso presented.

1 Introduction

The Lax equivalence theorem is originally presented in Lax and Richtmyer[1956]. It applies to an initial value problem

ut = f, 0 ≤ t ≤ tf , (1.1)

u = u0, t = 0. (1.2)

The theorem readsGiven a properly posed initial value problem 1.1, 1.2 and a finite dif-

ference approximation C(∆t) to it that satisfies the consistency condition,stability is a necessary and sufficient condition that C(∆t) be a convergentapproximation.1

Remark. The term properly posed is equivalent to wellposed, which is usedmore often nowadays. That is, a unique solution U exists, and a smallperturbation of the data leads to a small perturbation of the solution.

∗Division of Scientific Computing, Department of Information Technology, UppsalaUniversity, SE-751 05 Uppsala, Sweden. siyang.wang@it.uu.se

1This theorem is copied entirely from Lax and Richtmyer [1956].

1

2 The theorem

We prove the Lax equivalence theorem from the functional analysis point ofview. We reformulate the mathematical problem to a more general one.

Find x ∈ X such thatTx = y, (2.1)

where y ∈ Y , T : X → Y is a linear operator from a normed space X ontoa normed space Y .

We assume that the above problem is wellposed, that is, T−1 : Y → D(T )is continuous. This ensures that a small perturbation in the data y leads toa small perturbation in the solution x. Since T is linear, T−1 is also linear.The spaces X and Y are equipped with appropriate norms.

The mathematical problem 2.1 is solved numerically as

Tnxn = yn, (2.2)

where Tn : Xn → Yn, xn ∈ Xn and yn ∈ Yn. The spaces Xn and Ynare equipped with appropriate norms. Here, n can be considered as theresolution of 2.2. As n →∞, we hope for xn → x. In practice (think of 2.1as a differential equation), the true solution x is a continuous function, whilethe numerical solution xn is discrete. In order to compare xn with x to checkconvergence, we define the following two operators

RXn : X → Xn,

RYn : Y → Yn.

These two operators project x and y (in X and Y ) to RXn x and RY

n y (in Xn

and Yn). We make the following definition

• The numerical method 2.2 is consistent with 2.1 if

TnRXn z → RY

n Tz,

for any z ∈ D(T ).

• The numerical method 2.2 is stable if

‖T−1n ‖ ≤ K,

for all n, where K is a constant independent of n.

• The numerical method is convergent if

T−1n RYn y → RX

n T−1y,

for every y ∈ Y .

2

The Lax equivalence theorem from the functional analysis point of view reads:If the numerical method is consistent, it is convergent if and only if it is

stable.

Proof. 1. consistent + stable ⇒ convergent:

‖T−1n RYn y −RX

n T−1y‖

=‖T−1n RYn Tx−RX

n x‖=‖T−1n (RY

n Tx− TnRXn x)‖

≤‖T−1n ‖‖RYn Tx− TnRX

n x‖≤K ‖RY

n Tx− TnRXn x‖︸ ︷︷ ︸

→0 by consistency

Therefore, T−1n RYn y → RX

n T−1y. The method is convergent. Note that

this proof does not require deep functional analysis knowledge.

2. consistent + convergent ⇒ stable:From convergence, we know that the sequence (T−1n RY

n y) is bounded[Kreyszig, 1978, Lemma 1.4–2], that is

‖T−1n RYn y‖ ≤ cy,

for every y ∈ Y and n = 1, 2, .... By uniform boundedness theorem[Kreyszig, 1978, Lemma 4.7–3], the sequence of the norms ‖T−1n RY

n ‖ isbounded independent of n. That is, there exists a constant β such that

‖T−1n RYn ‖ ≤ β.

In order to proceed, we have to make the following assumption (*):For every wn ∈ Yn, ‖wn‖ ≤ 1, there exists w ∈ Y, ‖w‖ ≤ α, such thatRY

nw = wn, where α is a constant independent of n. This is a plausibleassumption in most cases in practice, since RY

n is just a restrictionoperator bringing a continuous function to its corresponding discreteone. It then follows

‖T−1n RYn ‖ = sup

‖w‖6=0

‖T−1n RYnw‖

‖w‖ = sup‖w‖6=0

‖T−1n wn‖‖w‖ ≤ β

The norm of T−1n can be written as ‖T−1n ‖ = sup‖z‖=1 ‖T−1n z‖. Choosewn such that ‖wn‖ = 1 and ‖T−1n ‖ = ‖T−1n wn‖. With this wn, we stillhave ‖w‖ ≤ α. Therefore,

‖T−1n ‖ ≤ αβ.

This proves that the numerical method is stable.

3

Remark. There are interesting examples Sanz-Serna and Palencia [1985] wherethe assumption (*) does not hold, so that a consistent and convergent methodis not stable. Also, the uniform boundedness theorem requires that Yn is com-plete, i.e. a Banach space. This is not assumed or guaranteed, and might be astrong assumption in practice. Rosinger even claims that the Lax equivalencetheorem is wrong Rosinger [2005].

3 Application

We choose the operator T in 2.1 as a differential operator d/dt. Then wehave a differential equation of the type

xt = y, 0 ≤ t ≤ tf , (3.1)

with the initial condition x(0) = c. Comparing with the mathematical prob-lem 2.1, the space X and Y can be chosen as x ∈ C2[0, tf ] | x(0) = cequipped with the maximum norm in [0, tf ].

We introduce the time discretization

tj = jk, j = 0, 1, · · · , N,

where the step size k = tf/N . The corresponding discrete solutions arex0, x1, · · · , xN . We equip the discrete space with the maximum norm. Thefinite difference scheme can be written as

xn+1 = P (k)xn, n = 0, 1, · · · , N − 1,

where P (k) is the finite difference operator taking the discrete solution fromthe current time step to the next time step. The recursion relation can bewritten as

1−P (k) 1

−P (k) 1. . . . . .

−P (k) 1

︸ ︷︷ ︸Q(k)

x1x2x3...xN

=

P (k)x000...0

.

Stability requires that ‖Q−1(k)‖ ≤ cQ, where cQ is independent of k. Due to

4

the special structure of Q(k), its inverse can be computed exactly, yielding

Q−1(k) =

1P (k) 1P 2(k) P (k) 1

.... . . . . . . . .

PN−1(k) · · · P 2(k) P (k) 1

.

Therefore, ‖Q−1(k)‖ =∑N−1

i=0 |P i(k)| = 1−|P (k)|N1−|P (k)| . Stability is ensured if

|P (k)| < 1.

Remark. By taking y = −λx, we get the familiar test equation, which isoften used to analyze the stability properties of numerical methods for timeintegration. For forward Euler method, P (k) = 1−kλ, and stability requires|1− kλ| < 1. This is the same result as if we would do the standard stabilityanalysis.

References

E. Kreyszig. Introductory functional analysis with applications. John Wiley& Sons, 1978.

P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite differ-ence equations. Comm. Pure Appl. Math., IX:267–293, 1956.

E. E. Rosinger. What is wrong with the Lax-Richtmyer fundamental theoremof linear numerical analysis? Technical report, 2005.

J. M. Sanz-Serna and C. Palencia. A general equivalence theorem in thetheory of discretization methods. Math. Comp., 45(171):143–152, 1985.

5

The Schauder fixed point theorem

Hannes Frenander∗

January 5, 2015

Abstract

The Brouwer and Schauder fixed point theorems are presented; thelatter with a complete proof. An application of the Schauder fixedpoint theorem regarding solutions to partial differential equations hasalso been investigated.

1 Brouwer fixed point theorem

For completeness, we start by introducing the Brouwer fixed point theorem(BFPT) since it is essential for understanding the proof of our main the-orem: the Schauder fixed point theorem (SFPT). However, we neglect theproof since it would obscure the main points of this essay. The theorem istypically used in numerical analysis to prove existence of initial boundaryvalue problems, which will be demonstrated in the end of this essay. In gen-eral, the BFPT states that every continous mapping that maps a convex setinto itself has at least one fixed point. That is:

Theorem 1.1 (Brouwer fixed point theorem). Let X be a compact and con-vex set and f : X → X a continuous map. Then there is a x0 ∈ X s.tf(x0) = x0.

To illustrate, lets take a simple example by considering the domain X =[0, 1] ⊂ R, i.e. the set of all real numbers between zero and one. This setis compact and convex. BFPT now states that every mapping f(x) thatremains in this domain has at least one fixed point. For example, f(x) = x2

has a fixed point at x0 = 0, 1, f(x) = sin(x) has a fixed at x0 = 0 and so on.

∗Division of Computational Mathematics, Department of Mathematics, Linkoping uni-versity, SE-58183 Linkoping, Sweden. hannes.frenander@liu.se

1

2 Shauder fixed point theorem

The SFPT was proved 1930 by Juliusz Schauder, but only for special cases.It remained for Robert Cauty to prove it in its final form in 2001. Thetheorem is a generalization of BFPT into vector spaces and claims that everycontinuous mapping from a convex vector space into itself has at least onefixed point. That is:

Theorem 2.1 (Shauder fixed point theorem). Let X be a normed vectorspace and K ⊂ X a convex and compact subset of X. Then every continuousmap f : K → K has a fixed point.

Proof. Since K is compact, it has a finite subcover. That is: there is a finitesequence xi such that the open balls Bϵ(xi) cover the set K, where ϵ > 0 isthe radius of the balls. Lets define the continuous functions gi(x) as

gi(x) =

ϵ − ||x − xi|| if ||x − xi|| ≤ ϵ

0 if ||x − xi|| ≥ ϵ.

Since gi(x) ≥ 0 and∑

i gi(x) > 0 for all x ∈ K, we can define anothercontinuous function in K:

g(x) =

∑i gi(x)xi∑i gi(x)

.

We can easily verify that g is a map from K into the convex hull 1 K0 of K.That is, g =

∑i αixi where αi ≥ 0 and

∑i αi = 1. One can also observe that

||g(x) − x|| ≤ ϵ

for all x ∈ K. The continuous map H = g f maps K0 into itself. Since K0

is compact, the BFPT tells us that there is a z ∈ K0 such that

H(z) = g(f(z)) = z

and therefore||f(z) − z|| = ||f(z) − g(f(z))|| ≤ ϵ.

Since ϵ can be arbitrarily small, there is for all m ∈ N a zm ∈ K such that

||f(zm) − zm|| ≤ 1

m.

1The convex hull K0 of K is the smallest convex set that contains K. In this case, theset ∑

i αixi|αi > 0,∑

i αi = 1 is a convex hull of K.

2

Remember that the theorem demands K to be compact, so there is a subsequence (zm)k such that f((zm)k) → x0 for some x0 ∈ K. Hence, we have

||(zm)k − x0|| = ||(zm)k − f((zm)k) + f((zm)k) − x0||≤ ||(zm)k − f((zm)k)|| + ||f((zm)k) − x0||

≤ 1

mk

+ ||f((zm)k) − x0|| → 0

which means (zm)k → x0. Since f is continuous, we also have f((zm)k) →f(x0). This means f(x0) = x0, i.e. x0 ∈ K is a fixed point.

3 An application

The Schauder fixed point theorem is commonly used to prove existence ofsolutions to partial differential equations (Pouso [2012],Cao et al. [2014],Agar-wal et al. [2013]), and we will in this section consider an example given byPouso [2012]. For clarity, we skip some details of the proof and the reader isreferred to Pouso [2012] for the full version.

Consider the problem

xtt = f(t, x)

x(0, t) = x(1, t) = 0

t ∈ I = [0, 1]

(3.1)

under the Caratheodory’s conditions Pouso [2012]. Hence, in (3.1), the func-tion f(t, x) is allowed to be discontinuous. We now prove that (3.1) has asolution.

First, we demand that there is a M , such that |f(t, x)| ≤ M(t), andconsider the domain

K = x ∈ C1(I), x(0) = x(1) = 0, |xt(t) − xt(s)| ≤∫ t

s

M(r)dr

which can be proved to be compact and convex subset of the Banach spaceC1(I) if we define the norm

||x|| = maxt∈I |x(t)| + maxt∈I |xt(t)|.

Consider now the operator T , defined by

Tx(t) =

∫ t

0

G(t, s)f(s, x(s))ds

3

where G(t, s) is the Green’s function to (3.1). T maps K into itself if all theabove conditions are fulfilled. Therefore, according the SFPT, T has a fixedpoint:

Tx(t) = x(t)

which is then the solution to (3.1).

References

R. Agarwal, S. Arshad, and D. O’regan. A Schauder fixed point theorem insemilinear spaces and applications. Fixed Point Theory and Applications,306, 2013.

Z. Cao, C. Yuan, and X. Li. Applications of Schauders fixed point theo-rem to semipositone singular differential equations. Journal of AppliedMathematics, 2014, 2014.

H. Pouso. Schauder’s fixed-point theorem: new applications and a new ver-sion for discontinous operators. Boundary Value Problems, 92, 2012.

4

Existence of a weak solution to the Stokesequations

Application of the Babuška-Brezzi inf-sup theoremto the Stokes equations

Hanna Holmgren∗

January 5, 2015

Abstract

The dynamics of incompressible viscous fluids are described by the Stokesequations if stationary laminar flow conditions with low Reynolds numbersare assumed, see Figure 0.1 for an example. The existence of a uniquesolution to the weak formulation of the Stokes equations is proved in thisessay. First the Two Hilbert Spaces H1

0(Ω) and L20(Ω) are introduced and

the weak formulation of the Stokes equations is derived. Then, the Babuška-Brezzi inf-sup theorem is applied to the corresponding variational problemto show existence and uniqueness of a weak solution.

Figure 0.1: Laminar flow conditions on the River Derwent in North Yorkshire. 1

∗Division of Scientific Computing, Department of Information Technology, Uppsala university,SE-751 05 Uppsala, Sweden. hanna.holmgren@it.uu.se

1

1 The Stokes EquationsThe dynamics of incompressible fluids are generally described by the Navier-Stokesequations. These equations are derived by considering the conservation of mass,momentum and energy of a fluid flow. If the flow is “slow and calm” it is calleda laminar flow and an example is illustrated in Figure 0.1. For steady laminarflows which have a small Reynolds number (intertial forces are small compared withviscous forces) the Navier-Stokes equations are reduced to the Stokes equations.In the Stokes equations all time dependent terms as well as the term describingadvection have been neglected. This essay aims at proving the existence of aunique solution to the weak formulation of the Stokes equations.

Assume an incompressible viscous fluid with kinematic viscosity ν > 0 is filling upa domain Ω ⊂ R3 with boundary Γ, then the Stokes system of equations is givenby

−ν∆u +∇p = f, in Ω (1.1)∇ · u = 0, in Ω (1.2)

u = 0, on Γ (1.3)

where f is a given volume force, u is the unknown vector describing the velocityof the fluid and p is the unknown pressure.

2 The Two Hilbert Spaces H10(Ω) and L2

0(Ω)Before deriving the weak formulation of the Stokes system (1.1)−(1.3) we firstneed to introduce two Hilbert spaces related to the two unknowns u and p. Weassume the velocity u to be in the Hilbert space H1

0(Ω) (here Ω ⊂ R3), which isthe space of all vector fields v = (v1, v2, v3) with components vi in the space H1

0defined by

H10 (Ω) = v ∈ H1(Ω) : v|Γ = 0. (2.1)

Remember that the space H1(Ω) is equipped with the norm defined by

‖v‖2H1(Ω) = ‖v‖2

L2(Ω) + ‖∇v‖2L2(Ω).

The Hilbert space(s) H10(Ω) is equipped with the inner product and seminorm

defined by

(u,v)H10(Ω) =

∫

Ω∇u : ∇v dx (2.2)

|v|2H10(Ω) = (v,v)H1

0(Ω) (2.3)1Picture from http://evidence.environment-agency.gov.uk

2

where∫

Ω∇u : ∇v dx =

∫

Ω∇u1 · ∇v1 dx+

∫

Ω∇u2 · ∇v2 dx+

∫

Ω∇u3 · ∇v3 dx.

Further, we assume the pressure p to be in the Hilbert space L20(Ω) defined by

L20(Ω) = q ∈ L2(Ω) :

∫

Ωq dx = 0 (2.4)

and equipped with the usual inner product and norm of the space L2(Ω).

3 The Weak Form of Stokes EquationsNow, the weak formulation is derived by first multiplying the momentum equation(1.1) by a test vector v ∈ H1

0(Ω) and then integrating by parts which gives thefollowing weak formulation of the momentum equation

ν∫

Ω∇u : ∇v dx−

∫

Ω(∇ · v)p dx =

∫

Ωf · v dx,

where all boundary terms from the integration by parts vanishes since v|Γ = 0.Further, multiplying the incompressibility constraint (1.2) by a test functionq ∈ L2

0(Ω) and integrating yields∫

Ω(∇ · u)q dx = 0.

Summarizing, the variational formulation derived from the Stokes system of equa-tions reads:

Find the solution pair (u,p) ∈ H10(Ω)× L2

0(Ω) such that

a(u,v) + b(v,p) = l(v) ∀v ∈ H10(Ω) (3.1)

b(u,q) = 0 ∀q ∈ L20(Ω), (3.2)

where the following linear forms have been introduced:

a(u,v) = ν∫

Ω∇u : ∇v dx, (3.3)

b(u,q) = −∫

Ω(∇ · u)q dx, (3.4)

l(v) =∫

Ωf · v dx. (3.5)

3

4 The Babuška-Brezzi inf-sup theoremFor the abstract variational problem:

Find u ∈ V , such thata(u,v) = l(v), ∀v ∈ V,

where V is a Hilbert space with inner product (·,·), a(u,v) is a coercive continuousbilinear form on V and l(v) is a continuous linear form on V , the Lax-Milgramlemma can be used to prove the existence and uniqueness of a solution [1] . Re-member that a bilinear form a(·,·) is said to be V -coercive if there is a positiveconstant m such that

m‖v‖2V ≤ a(v,v) ∀v ∈ V.

However, in the case of the weak formulation of the Stokes equations (3.1)-(3.2),the existence of a unique solution does not follow from the Lax-Milgram lemmaalone, since it is impossible to establish coercivity for the bilinear form b(·, ·) in(3.4) [2]. In this case we need the Babuška-Brezzi inf-sup theorem [1]:

Theorem 1: The Babuška-Brezzi inf-sup theorem. Let V and M be twoHilbert spaces, and let a(·,·) : V × V −→ R and b(·,·) : V × M −→ R be twocontinuous bilinear forms with the following properties: There exists a positiveconstant α such that

a(v,v) ≥ α‖v‖2V for all v ∈ U0 = v ∈ V : b(v,q) = 0 ∀q ∈M,

i.e. a(·,·) is U0-coercive, and there exists a positive constant β such that

infq∈M,q 6=0

supv∈V,v 6=0

|b(v,q)|‖v‖V ‖q‖M

≥ β,

where the latter is called the Babuška-Brezzi inf-sup condition.Finally, let l : V −→ R and g : M −→ R be two continuous linear forms. Thenthe variational problem: Find (u,p) ∈ V ×M such that

a(u,v) + b(v,p) = l(v) ∀v ∈ V,b(u,q) = g(q) ∀q ∈M,

has one and only one solution.

4

5 Existence and uniqueness of a weak solutionWe use the Babuška-Brezzi inf-sup theorem to prove existence and uniqueness of asolution to the variational problem (3.1)-(3.2) derived from the Stokes equations.

LetV = H1

0(Ω) and M = L20(Ω),

where V and M are the two spaces from the Babuška-Brezzi inf-sup theorem.

First, we prove that a(·,·) is U0-coercive:

1. a(·,·) is H10(Ω)-coercive since

a(v,v) = ν|v|2H10(Ω) ∀v ∈ H1

0(Ω).

2. Then a(·,·) is also coercive on U0 = v ∈ H10(Ω) : b(v,q) = 0 ∀q ∈ L2

0(Ω)since the above equality holds for all v ∈ H1

0(Ω).

Then, for proving the second part of the Babuška-Brezzi inf-sup theorem (i.e. thatb(·,·) fulfills the Babuška-Brezzi inf-sup condition) we need the following theorem[1]:

Theorem 2. Let Ω be a domain in RN . Then the injective continuous linearoperator

div : (ker div)⊥ −→ L20(Ω)

is surjective and has a continuous inverse.

Here “div” is another notation used for the divergence, i.e. divv = ∇ · v for anyv. The space (ker div) is defined by

(ker div) = v ∈ H10(Ω) : divv = 0 in L2

0(Ω).

Now, we prove that b(·,·) fulfills the Babuška-Brezzi inf-sup condition:

1. Since div : (ker div)⊥ −→ L20(Ω) is surjective (by Theorem 2), there exists

a unique vector field w ∈(ker div)⊥ ⊂ H10(Ω) such that

divw = q in L20(Ω),

and besides, since div : (ker div)⊥ −→ L20(Ω) has a continuous inverse (also

by Theorem 2), there exists a constant C such that

|w|H10(Ω) ≤ C‖q‖L2

0(Ω) ∀q ∈ L20(Ω).

5

2. Now we can prove the Babuška-Brezzi inf-sup condition holds for b(·,·), foreach non-zero q ∈ L2

0(Ω):

supv∈H1

0(Ω),v6=0

|b(v,q)||v|H1

0(Ω)‖q‖L20(Ω)

= supv∈H1

0(Ω),v6=0

| ∫Ω(∇ · v)q dx||v|H1

0(Ω)‖q‖L20(Ω)

[∇ · v = divv

]

= supv∈H1

0(Ω),v6=0

| ∫Ω(divv)q dx||v|H1

0(Ω)‖q‖L20(Ω)

[since w ∈ (ker div)⊥ ⊂ H1

0(Ω) we can choose v = w]

≥ |∫

Ω(divw)q dx||w|H1

0(Ω)‖q‖L20(Ω)

[divw = q

]

= | ∫Ω q2 dx|

|w|H10(Ω)‖q‖L2

0(Ω)

[ ∫Ω q

2 dx = ‖q‖2L2

0(Ω)

]

=‖q‖L2

0(Ω)

|w|H10(Ω)

[|w|H1

0(Ω) ≤ C‖q‖L20(Ω)

]

≥ 1C

= β

So, by the Babuška-Brezzi inf-sup theorem, there exists a unique solution pair(u,p) ∈ H1

0(Ω) × L20(Ω) to the variational problem (3.1)-(3.2) derived from the

Stokes equations.

6

References

[1] Philippe G. Ciarlet, Linear and Nonlinear Functional Analysis with Applica-tions. Society for Industrial and Applied Mathematics, Philadelphia, 2013.

[2] Mats G. Larson, Fredrik Bengtzon, The Finite Element Method - Theory, Im-plementation and Applications. Springer, Umeå, Sweden, 2012.

7

The Lax–Milgram Theorem

Andrea Alessandro Ruggiu∗

January 7, 2015

Abstract

In this essay we state and prove the Lax–Milgram Theorem, aresult which claims well–posedness of variational problems. This the-orem is here proven through the classical approach, that is to say inthis work the Banach Fixed Point Theorem is not taken into account.

1 Introduction

Various problems in science can be formulated through PDEs, which neces-sarly don’t have a classic solution. Solving a problem weakly means find asolution for the problem, called weak solution, whose derivatives could alsonot exist in a classical sense. Not rarely they are the only solutions thatis possible to find. In particular, it’s possible to formulate any differentialproblem as a variational problem of the following kind

given F ∈ H∗, with H∗ dual space of H, find u ∈ H such that

a (u, v) = F (v) , ∀v ∈ H. (1)

A common example is the Poisson problem

−∆u = f, x ∈ Ωu = 0, x ∈ ∂Ω

(2)

that can be formulated as a variational problem through the integrationby parts.

∗Computational Mathematics, Department of Mathematics, Linkoping university, SE-581 83 Linkoping, Sweden. andrea.ruggiu@liu.se

1

Indeed, multiplying the equation by v ∈ H10 (Ω) and integrating by parts

leads to

∫

Ω

∇u · ∇v dx =

∫

Ω

fv dx, ∀v ∈ H10 (Ω) . (3)

The solution of (3), u ∈ H10 (Ω), is said weak solution for (2).

2 Background theorems

In this section we recall two important theorems that we want to use inorder to prove the Lax–Milgram Theorem. Both of them were studied andanalysed in the Numerical Functional Analysis course. The first one is theHilbert Projection Theorem.

Theorem 2.1. Let us consider a Hilbert space H, a non–empty subspaceV ⊂ H closed in H and x ∈ H. Then ∃!v = PV (x) ∈ V such that ‖x− v‖ =infv∈V ‖x− v‖ = d ≥ 0. Moreover

• PV (x) = x ⇐⇒ x ∈ V ,

• QV (x) = x− PV (x) ∈ V ⊥.

The second theorem that we want to use is the Riesz RepresentationTheorem

Theorem 2.2. Let H be a Hilbert space. Then

∀L ∈ H∗ ∃!u ∈ H s.t. L (v) = (u, v) ∀v ∈ H

and moreover ‖u‖H = ‖L‖H∗.

3 Lax–Milgram Theorem

In this section we prove the following statement

Theorem 3.1. Let H be a Hilbert space. If a (·, ·) is a bilinear form suchthat

• ∃M > 0 such that |a (u, v)| ≤M ‖u‖ ‖v‖,

• ∃α > 0 such that a (u, u) ≥ α ‖u‖2,

2

and F ∈ H∗ is linear and bounded, then ∃!u solution for (1) and

‖u‖ ≤ 1

α‖F‖∗ , (4)

where ‖·‖∗ is the norm associated to H∗.

Proof. By Theorem 2.2 ∃!z ∈ H s.t. F (v) = (z, v) , ∀v ∈ H and ‖F‖∗ =‖z‖. But a (u, ·) : V 7→ a (u, v) is a linear operator in H∗, then, again byTheorem 2.2 we know that

∃!h ∈ H s.t. a (u, v) = (h, v) , ∀v ∈ H

with u fixed. Thus we can call h = A (u). Notice that if u is fixed, then hexists and it is unique.Therefore (1) can be rewritten as

(A (u) , v) = (z, v) , ∀v ∈ H,

which implies A (u) = z, A : H → H. In order to show the uniqueness of uwe have to prove that A is invertible. This proof requires 5 steps

• A is linear. As stated before, by Theorem 2.2

(A (u) , v) = a (u, v) , ∀v ∈ H.

Then we can use the bilinearity of a (·, ·) in order to prove the linearityof A

(A (λu1 + µu2) , v) = a (λu1 + µu2, v) = λa (u1, v) + µa (u2, v) =

= λ (A (u1) , v) + µ (A (u2) , v) =

= (λA (u1) + µA (u2) , v) ,

∀v ∈ H. This implies that A (λu1 + µu2) = λA (u1) + µA (u2), that isto say A is linear. For this reason from now on we write Au instead ofA (u).

• A is bounded. We can write

‖Au‖2 = (Au,Au) = a (u,Au) ≤M ‖u‖ ‖Au‖ . (5)

If ‖Au‖ = 0, then the boundedness follows trivially. Therefore weassume ‖Au‖ 6= 0: dividing by ‖Au‖ both the left and the right–handside of (5), the boundedness of A follows.

3

• A is injective. This statement is equivalent to nul (A) = 0, wherenul is the null space of A. Using the hypotheses and Cauchy–Schwarzinequality in a (u, u) = (Au, u) leads to

α ‖u‖2 ≤ a (u, u) = (Au, u) ≤ ‖Au‖ ‖u‖ .

This implies that if u 6= 0 (trivially an element of nul (A)) then‖Au‖ ≥ α ‖u‖, that is to say

‖u‖ ≤ 1

α‖Au‖ , ∀u ∈ H. (6)

Therefore if we consider u ∈ nul (A) (‖Au‖ = 0), then u = 0. Thismeans that the only element in the nullspace of A is 0, then A isinjective.

• The image of A is closed. Let yn ∈ Im (A) an arbitrary convergentsequence in H, let’s say yn → y. The statement is proven if, and onlyif, y ∈ Im (A). But since

yn ∈ Im (A)⇐⇒ ∃xn s.t. yn = Axn

we can also prove the statement by showing the existence of x s.t.y = Ax.

As a first step we can prove that xn is convergent. Using (6) and thelinearity of A we can write

0 ≤ ‖xn − xm‖ ≤1

α‖A (xn − xm)‖ =

1

α‖Axn − Axm‖ =

1

α‖yn − ym‖ .

But since yn is convergent in H, then it is also Cauchy. In particular,this implies by the above equation that xn is Cauchy in H. The com-pleteness of H leads to the convergence of xn to a certain x ∈ H. Byhypothesis it is known that

Axn = yn → y. (7)

But, on the other hand, we know by the continuity of A (it is linearand bounded) that

Axn → Ax. (8)

By (7), (8) and the uniqueness of the limit we have y = Ax, whichprove the closedness of Im (A) since x is such that y = Ax ∈ Im (A).

4

• A is surjective, that is to say Im (A) = H. We want to prove thisstatement by contradiction: let us assume that Im (A) ( H. Then∃q0 /∈ Im (A) and applying Theorem 2.1 (notice that Im (A) is closedby the previous point) leads to

QIm(A) (q0) = q0 − PIm(A) (q0) ∈ Im (A)⊥ .

If we define

q =q0 − PIm(A) (q0)∥∥q0 − PIm(A) (q0)

∥∥ ,

then q ∈ Im (A)⊥ and ‖q‖ = 1. Then we can write

0 = (Aq, q) = a (q, q) ≥ α ‖q‖2 = α > 0.

Since we have a contradiction, then is impossible to have Im (A) ( Hand therefore A is surjective.

Finally, we have proven that A is invertible by proving that A is injectiveand surjective. In particular Au = z has the unique solution u = A−1z andthis means that there exists a unique solution for (1).

In order to prove the remaining part of the statement we use (6) andTheorem 2.2

‖u‖ ≤ 1

α‖Au‖ =

1

α‖z‖ =

1

α‖F‖∗ .

4 Example of application

Let’s take into account the problem (2) and its weak formulation given by(3). We can write

a (u, v)

∫

Ω

∇u · ∇v dx, F (v) =

∫

Ω

fv dx. (9)

With this notation we have written (3) as a problem of the same kind of(1) with H ≡ H1

0 (Ω). It can be shown that a suitable norm for H10 (Ω) is

given by‖·‖H1

0 (Ω) = ‖∇·‖L2(Ω)

and, equivalently, we can use as inner product in H10 (Ω)

(u, v)H10 (Ω) = (∇u,∇v)L2(Ω) .

5

We can easily check that

|a (u, v)| ≤ ‖∇u‖L2(Ω) ‖∇v‖L2(Ω) = ‖u‖H10 (Ω) ‖v‖H1

0 (Ω)

anda (u, u) = ‖∇u‖2

L2(Ω) = ‖u‖2H1

0 (Ω) .

Moreover F is linear and it is bounded, indeed

F (v) =

∫

Ω

fv dx ≤ ‖f‖L2(Ω) ‖v‖L2(Ω) ≤ cp ‖f‖L2(Ω) ‖v‖H10 (Ω) ,

where we have used the Cauchy–Schwarz and the Poincare inequalities withconstant cp.

Then the Lax–Milgram theorem holds with α = M = 1 and therefore (2)admits a unique solution u ∈ H1

0 (Ω) that is such that

‖u‖ ≤ cp ‖f‖L2(Ω) .

References

[1] Evans L.C., Partial Differential Equations, American Mathematical So-ciety, PP.297-299, 1997.

[2] Salsa S., Equazioni a derivate parziali. Metodi, modelli e applicazioni,Springer, Chapter 6, 2nd edition, 2010.

6

A time dependent mapping from Cartesiancoordinate systems into curvilinear coordinatesystems and the geometric conservation law

Samira Nikkar∗

January 9, 2015

Abstract

A time-dependent transformation that maps a moving domainin the physical space (in terms of the Cartesian coordinates) intoa fixed domain in the computational space (in terms of the curvi-linear coordinates) is considered. The continuous transformation isusually non-singular, which means that the Geometric ConservationLaw (GCL) holds exactly. In the numerical set up, the differential op-erators and the metric formulations must satisfy some certain criteriain order for the Numerical GCL (NGCL) to be fulfilled. With the useof the Summation-by-Parts (SBP) operators for the space and timediscretization, we can prove that even for the discrete mapping, theNGCL is also guaranteed.

1 The continuous problem

Consider a time-dependent transformation from the Cartesian coordinatesinto curvilinear coordinates, which results in a fixed spatial domain, as

x = x(τ, ξ, η), y = y(τ, ξ, η), t = τ,ξ = ξ(t, x, y), η = η(t, x, y), τ = t.

(1.1)

such that 0 ≤ ξ ≤ 1, 0 ≤ η ≤ 1, 0 ≤ τ ≤ T , see Figure 1. The Jacobian

∗Division of scientific computing, Department of Mathematics, Linkoping university,Sweden. samira.nikkar@liu.se

1

n

η

ξ

y

x

d' c'

b' a'

c

b

a

d

Ω

ab a'b' : South (s) bc b'c' : East (e) cd c'd' : North (n) da d'a' : West (w)

Figure 1.1: A schematic of the moving and fixed domains and boundarydefinitions.

matrix of the transformation is

[J ] =

xξ yξ 0xη yη 0xτ yτ 1

, (1.2)

The relation between [J ], and its inverse, which transforms the derivativesback to the Cartesian coordinates leads to the metric relations

Jξt = xηyτ − xτyη, Jξx = yη, Jξy = −xηJηt = yξxτ − xξyτ , Jηx = −yξ, Jηy = xξ,

(1.3)

in which J=xξyη−xηyξ>0 is the determinant of [J ].If we want to preserve any conservation property while transforming the

physical domain into the computational domain, the transformation and themetric relations must satisfy the GCL Farhat et al. (2001); Sjogreen et al.(2014), summarized as the following

Jτ + (Jξt)ξ + (Jηt)η = 0,(Jξx)ξ + (Jηx)η = 0,(Jξy)ξ + (Jηy)η = 0.

(1.4)

In practice (1.4) holds for any non-singular transformation. Now let us in-vestigate the requirements for the discrete transformation such that the nu-merical version of (4) also holds.

2

2 The discrete problem

The spatial computational domain Ω is a square in ξ, η coordinates, seeFigure 1.1, and discretized using N and M nodes in the direction of ξ andη respectively. In time we use L time levels from 0 to T. The fully-discretegrid vector is a column vector of size LMN organized as follows

Z=

Z0

...

[Zk]

...

ZL

; [Zk]=

Z0...

[Zi]...

ZN

k

; [Zi]k=

Z0...

[Zj ]...

ZM

ki

, (2.1)

where Zkij = Z(τk, ξi, ηj) and Z ∈ x,y.The first derivative xξ is approximated by Dξx, where Dξ is a so-called

SBP operator of the formDξ = P−1

ξ Qξ, (2.2)

and x=[x0, x1, · · · , xN ]T is the x coordinate of the grid points. Pξ is a sym-metric positive definite matrix, and Q is an almost skew-symmetric matrixthat satisfies

Qξ +QTξ = E1−E0 =B = diag(−1, 0, ..., 0, 1). (2.3)

In (2.3), E0 =diag(1, 0, ..., 0) and E1 =diag(0, ..., 0, 1). The η and τ directionsare discretized in the same way.

A finite difference approximation including the time discretization Nord-strom and Lundquist (2013), on SBP-SAT form, is constructed by extendingthe one-dimensional SBP operators in a tensor product fashion as

Dτ =P−1τ Qτ ⊗ Iξ ⊗ Iη,

Dξ = Iτ ⊗ P−1ξ Qξ ⊗ Iη,

Dη = Iτ ⊗ Iξ ⊗ P−1η Qη

(2.4)

where ⊗ represents the Kronecker product Loan (2000). Note that in (2.4) wehave used the same names for the differential operators in multi dimensionalcases as the operators in the one dimensional case. All matrices in the firstposition are of size L×L, the second position N×N , the third position M×M .I denotes the identity matrix with a size consistent with its position in theKronecker product.

3

The Kronecker product is bilinear and associative. For square matricesthe following rules exist Nordstrom and Berg (2013),

(A⊗B)(C⊗D)=(AC⊗BD), (A⊗B)−1=A−1⊗B−1, (A⊗B)T =AT⊗BT . (2.5)

For later reference we need

Lemma 2.1. The difference operators in (2.4) commute.

Proof. The properties (2.5) of the Kronecker product lead to

DτDξ = (P−1τ Qτ ⊗ Iξ ⊗ Iη)(Iτ ⊗ P−1

ξ Qξ ⊗ Iη)= P−1

τ Qτ ⊗ P−1ξ Qξ ⊗ Iη

= (Iτ ⊗ P−1ξ Qξ ⊗ Iη)(P−1

τ Qτ ⊗ Iξ ⊗ Iη) = DξDτ .The proof is analogous for the other coordinate combinations.

We show

Lemma 2.2. The Numerical Geometric Conservation Law (NGCL) holds:

Jτ + (Jξt)ξ + (Jηt)η = 0,

(Jξx)ξ + (Jηx)η = 0,

(Jξy)ξ + (Jηy)η = 0.

(2.6)

Proof. Consider the following definitions,

Jτ = diag[Dτ (DηM(1) −DξM

(2))]

(Jξt)ξ = diag[Dξ(DτM(2) −DηM

(3))]

(Jηt)η = diag[Dη(DξM(3) −DτM

(1))]

(Jξx)ξ = diag[Dξ(Dηy)]

(Jξy)ξ = diag[−Dξ(Dηx)]

(Jηx)η = diag[−Dη(Dξy)]

(Jηy)η = diag[Dη(Dξx)]

(2.7)

in which x and y are the discrete Cartesian coordinates defined in (2.1). AlsoM (1) = diag(y)(Dξx), M (2) = diag(y)(Dηx) and M (3) = diag(y)(Dτx). ByLemma 2.1 we find that the NGCL holds exactly.

3 Summary and conclusions

We have considered a time-dependent curvilinear coordinates. By using SBPoperators in space and time and a special definitions for the metric coeffi-cients the Geometric Conservation Law is proven to hold numerically.

4

References

C. Farhat, P. Geuzaine, C. Grandmont, The discrete geometric conservationlaw and the nonlinear stability of ALE schemes for the solution of flowproblems on moving grids, Journal of Computational Physics 174 (2001)669–694.

B. Sjogreen, H. C. Yee, M. Vinokur, On high order finite-difference metricdiscretizations satisfying GCL on moving and deforming grids, Journal ofComputational Physics 265 (2014) 211–220.

J. Nordstrom, T. Lundquist, Summation-by-parts in time, Journal of Com-putational Physics 251 (2013) 487–499.

C. F. V. Loan, The ubiquitous Kronecker product, Journal of Computationaland Applied Mathematics 123 (2000) 85–100.

J. Nordstrom, J. Berg, Conjugate heat transfer for the unsteady compressibleNavier-Stokes equations using a multi-block coupling, Computers & Fluids72 (2013) 20–29.

5

The Babuska Inf-Sup Condition

Simon Sticko∗

January 12, 2015

Abstract

A general weak formulation, typically arising from a linear time-independent partial differential equation, is discussed. The Babuskainf-sup condition is shown to guarantee that the weak formulation isa well-posed problem.

1 Introduction

The finite element method applied to a linear time-independent partial dif-ferential equation typically leads to a weak formulation of the following form:find u ∈ U such that

a(u, v) = l(v) ∀v ∈ V. (1.1)

Here U and V are Hilbert spaces, a(·, ·) : U × V → R a bounded bilinearform:

|a(u, v)| ≤ αC ‖u‖U ‖v‖V , αC ∈ (0,∞)

and l(v) : V → R is a bounded linear functional. Our goal is to find acondition for when (1.1) is a well-posed problem. That is, the problem shouldfulfill the following requirements:

(E) There exist a solution u.

(U) The solution is unique.

(C) The solution depends continuously on initial data.

∗Division of Scientific Computing, Department of Information Technology, Uppsalauniversity, SE-751 05 Uppsala, Sweden. simon.sticko@it.uu.se

1

2 Theory

In order to understand if the problem in (1.1) is well posed or not we wouldlike to reformulate it into something which is a bit more familiar. In thefollowing we denote the scalar product in V and U by 〈·, ·〉V and 〈·, ·〉U re-spectively. Since U and V are Hilbert spaces a(·, ·) has a Ritz-representationA : U → V such that:

a(u, v) = 〈Au, v〉V ,where A is a bounded linear operator with the same norm as a(·, ·). Also l(·)has a Ritz-representations f ∈ V :

l(v) = 〈f, v〉V .

Using this we can now express (1.1) as

〈Au− f, v〉V = 0 ∀v ∈ V.

Since this should hold for all v ∈ V it holds in particular for v = Au− f . Sothis leads to the equation

Au = f. (2.1)

This formulation is equivalent to (1.1), but it’s easier to understand what isrequired for the problem to be well posed. In order for a solution of (2.1)to exist for an arbitrary f ∈ V we must require that there is at least oneu ∈ U such that Au = f . Thus A must be a surjective mapping. If wewant uniqueness we can not have two different elements u1, u2 ∈ U suchthat Au1 = f and Au2 = f . Thus we must also require A to be injective.Finally, consider two different right-hand sides f1, f2, such that Au1 = f1and Au2 = f2. Since A is a linear operator we obtain

A(u1 − u2) = f1 − f2,

which leads to‖u1 − u2‖U ≤

∥∥A−1∥∥V‖f1 − f2‖V .

Thus if we want a small change in right hand side to generate a small changein the solution we must require that A−1 is a bounded operator. So insummary we have the following requirements for our problem to be wellposed:

2

(E) A is surjective.

(U) A is injective.

(C) A−1 is a bounded operator.

Our goal is to show that A fulfills these three requirements. This was shownby Babuska [1971] and can be summarized as:

Theorem 2.1 (Babuska-Lax-Milgram). The weak formulation in (1.1) iswell posed if there exists αV , αU > 0 such that a(·, ·) fulfills:

αV ‖u‖U ≤ supv∈V

|a(u, v)|‖v‖V

∀u ∈ U (2.2)

αU ‖v‖V ≤ supu∈U

|a(u, v)|‖u‖U

∀v ∈ V. (2.3)

Proof. The three requirements (E), (U) and (C) are proved separately inLemma 2.2, 2.3 and 2.4.

Remark. The conditions (2.2) and (2.3) are frequently referred to as inf-supconditions. The reason for this is that they can be formulated as: there existαU , αV > 0 such that

infu∈U

supv∈V

|a(u, v)|‖u‖U ‖v‖V

≥ αV

infv∈V

supu∈U

|a(u, v)|‖u‖U ‖v‖V

≥ αU .

Lemma 2.2 ((U) Injectivity). If a(·, ·) satisfies (2.2) its Ritz-representationis injective.

Proof. By the definition of the Ritz-representation and the Cauchy-Schwartzinequality we have

|a(u, v)|‖v‖V

=| 〈Au, v〉V |‖v‖V

≤ ‖Au‖V ‖v‖V‖v‖V= ‖Au‖V .

By taking the supremum and utilizing (2.2) we obtain:

αV ‖u‖U ≤ supv∈V

|a(u, v)|‖v‖V

≤ ‖Au‖V . (2.4)

3

This gives us thatAu = 0 =⇒ u = 0

which implies injectivity since

Au1 − Au2 = 0 =⇒ A(u1 − u2) = 0 =⇒ u1 − u2 = 0.

Lemma 2.3 ((E) Surjectivity). If a(·, ·) satisfies (2.2) and (2.3) its Ritz-representation is surjective.

Proof. Consider the range of A: R(A), this is a subspace of V . In order toshow that this is a complete subspace let yj∞j=1 be a Cauchy sequence inR(A). For every ε > 0 there exists N such that

‖yi − yj‖V ≤ ε ∀i, j ≥ N.

But since yi ∈ R(A) we have yi = Axi with xi ∈ U so we get

‖Axi − Axj‖V ≤ ε ∀i, j ≥ N,

which by linearity and (2.4) gives us:

αV ‖xi − xj‖U ≤ ‖A (xi − xj)‖V ≤ ε ∀i, j ≥ N.

This shows us that xj∞j=1 is a Cauchy sequence, which converges since U isa Hilbert space. Denote the limit by x and let y = Ax. Since we originallyassumed that A is a bounded operator we now see that yj converges to y:

‖y − yj‖V = ‖Ax− Axj‖V ≤ αC ‖x− xj‖U → 0 j →∞.

Thus the conclusion is that R(A) is a complete subspace of V . This can onlyhappen if R(A) is also closed. Since R(A) is a closed subspace of V we candecomposed V as a direct product space:

V = R(A)⊕R(A)⊥. (2.5)

But consider now a point v ∈ R(A)⊥. From (2.3) we get

αU ‖v‖V ≤ supu∈U

| 〈Au, v〉 |‖u‖U

= 0,

since Au ∈ R(A) and v ∈ R(A)⊥. So this implies v = 0, which means thatR(A)⊥ = 0 since v was arbitrary. By (2.5) we must then have R(A) = V ,so that A is surjective.

4

Lemma 2.4 ((C) Bounded Ritz-Inverse). Given that a(·, ·) fulfills (2.2), theinverse of the Ritz-projection fulfills

∥∥A−1∥∥U≤ 1

αV

,

and thus is a bounded operator.

Proof. By rewriting (2.4) we obtain

‖u‖U ≤1

αV

‖Au‖V .

Since we now know that A is a bijection we knot that for each u ∈ U thereexist an unique f ∈ V such that u = A−1f . Inserting this gives us

∥∥A−1f∥∥U≤ 1

αV

‖f‖V .

Since this holds for all f ∈ V we obtain

∥∥A−1∥∥V

= supf∈V

‖A−1f‖U‖f‖V

≤ 1

αV

.

References

I. Babuska. Error-bounds for finite element method. Numerische Mathe-matik, 16:322–333, 1971. ISSN 0029599X. doi: 10.1007/BF02165003.

5

Uniformly best wavenumber approximationsby spatial central difference operators

Viktor Linders∗

January 12, 2015

Abstract

We show that the problem of finding uniformly best wavenumber ap-proximations by central difference schemes is equivalent to approxi-mating a continuous function from a finite dimensional subspace. Acharacterisation theorem for best approximations is proven.

1 Introduction and motivation

Consider a central difference approximation of a first derivative ux of somefunction u(x, t) at the point x = xi,

(∂u

∂x

)

i

+O(∆x2p) =1

∆x

p∑

k=1

c(p)k (ui+k − ui−k). (1.1)

Here ∆x is the spatial step size of the discretisation and c(p)k are the coeffi-

cients of the 2pth order classical central difference scheme. It is well knownthat the numerical dispersion relation of this scheme is

ξc = 2

p∑

k=1

c(p)k sin (kξ). (1.2)

where ξ = κ∆x is the normalised wavenumber. The subscript c serves as a re-minder that (1.2) corresponds to a classical stencil. Here we have introducedan overbar to signify that the wavenumber is a numerical approximation.

∗Division of Computational Mathematics, Department of Mathematics, Linkoping Uni-versity, SE-581 83 Linkoping, Sweden. viktor.linders@liu.se

1

Noting that the smallest wavelength the scheme can resolve is λmin = 2∆x,the largest wavenumber is κmax = 2π/λmin = π/∆x so that we have |ξ| ≤ π.

Typically, waves with high frequencies and wavenumbers require small spa-tial increments, ∆x, in order to be properly resolved. Over sizeable intervals,the so called dispersion error, |ξ− ξ|, may come to dominate the error in theapproximation, which gravely restricts the ∆x that may be used. Prob-lems of this type are commonly encountered in computational fluid dynam-ics, aeroacoustics, electromagnetism, elasticity, seismology, and other fieldswhere energy is propagated. Wave properties encoded within the dispersionrelation include phase velocity, group velocity and dissipation. It is thereforeof interest to develop difference schemes that preserve the analytic disper-sion relation of the governing equations for a wide range of spatial increments.

2 Problem formulation

Let us perturb the stencil (1.1) by adding n new coefficients, without increas-ing the accuracy. Thus, the scheme we consider has the form

(∂u

∂x

)

i

+O(∆x2p) =1

∆x

p+n∑

k=1

ak(ui+k − ui−k). (2.1)

Such a scheme uses (p + n) points on either side of xi to approximate thederivative, however the formal order of accuracy remains O(∆x2p).

In view of (1.2) the numerical wavenumber corresponding to this scheme is

ξ = 2

p+n∑

k=1

ak sin (kξ). (2.2)

Our goal may thus be formulated in terms of the following problem:

Choose ak, k = 1, . . . , p+ n, such that the error function

‖E‖∞ := ‖ξ − ξ‖∞is minimised, ensuring that the stencil has formal

accuracy O(∆x2p).

(2.3)

Note that if ap+j = 0, ∀j ≥ 1, we have ak = c(p)k . By linearity we may thus

write

ak = c(p)k +

n∑

j=1

α(j)k ap+j, k = 1, . . . , p (2.4)

2

where the parameters α(j)k are independent of ap+j. We summarise in the

following proposition:

Proposition 2.1. The numerical wavenumber of the 2pth order central dif-ference stencil (2.1) can be written as

ξ = ξc +n∑

j=1

ap+jφj(ξ)

where ξc is the wavenumber apprpximation of the classical central differencestencil and

φj(ξ) = 2 sin ((p+ j)ξ) + 2

p∑

k=1

α(j)k sin (kξ).

The coefficients α(j)k are independent of ap+j for all k = 1, . . . , p and j =

1, . . . , n.

Proof. Inserting (2.4) into (2.2) and collecting terms multiplying ap+j imme-diately gives the desired result.

The set φj is linearly independent and thus span some n-dimensional vec-tor space, which we will denote by Ξn. We may thus consider each φj as abasis function of Ξn.

Let ψ(ξ, a) =∑n

j=1 ap+jφj(ξ). We see from Proposition 2.1 that the problem(2.3) may equivalently be written

find a ∈ Rn that minimises ‖E(ξ, a)‖∞ = ‖Ec(ξ)− ψ(ξ, a)‖∞, (2.5)

where Ec(ξ) = ξ − ξc is the error of the classical stencil and ψ(ξ, a) ∈ Ξn.

3 Approximation theory

For convenience we use the following definition:

Definition 3.1. If a vector a solves problem (2.5), then ψ(ξ, a) is called abest approximation.

Finding and characterising best approximations are central problems in thefield of approximation theory. It is well-known that a solution to problem(2.5) exists Riesz [1918]. It is in place to introduce some relevant concepts:

3

Definition 3.2. A set of functions, θi, i = 1, . . . , n forms a Chebyshevset on the interval [0, ξmax] if any non-trivial linear combination has at most(n− 1) zeros in [0, ξmax].

We will also need the following definition:

Definition 3.3. Let M(a) denote the set of points ξ1, ξ2, . . . at which|E(ξ, a)| = ‖E(ξ, a)‖∞. Then E(ξ, a) is said to alternate n times on [0, ξmax]if M(a) contains (n+ 1) points 0 ≤ ξ1 < ξ2 < · · · < ξn+1 ≤ ξmax such that

E(ξi, a) = −E(ξi+1, a), i = 1, 2, . . . , n.

The set ξi is referred to as an alternating set.

The usefulness of the above definitions becomes apparent when we considerthe following well-known extension of the Chebyshev alternation theorem:

Theorem 3.1. Let θi(ξ), i = 1, . . . , n be a Chebyshev set on [0, ξmax].Then a solves problem (2.5) if and only if there is an alternating set of(n+ 1) points in [0, ξmax]. Such a solution exists and is unique.

For a proof, see e.g. Watson [1980].

Theorem 3.1 identifies best approximations as those whose error oscillatesbetween a positive and a negative fixed value. The main task left is thus toshow that the set φj constitutes a Chebyshev set.

4 Characterisation of best approximations

The following observation is useful. The proof is omitted for brevity.

Lemma 4.1. Consider the function

f (p)(x, a) = 1− 2

p+n∑

k=1

akkTk(x)

where ak is defined as before and Tk(x) is the kth order Chebyshev polynomialof the first kind, uniquely defined through the relation Tk(cos (φ)) = cos (kφ).Then f (p) has a root at x = 1 of multiplicity p.

We are in position to prove our main result:

Theorem 4.2. The set φj, j = 1, . . . , n defined in Proposition 2.1 formsa Chebyshev set on the semi-open interval (0, ξmax].

4

Proof. Recall the form of the basis functions,

φj(ξ) = 2 sin ((p+ j)ξ) + 2

p∑

k=1

α(j)k sin (kξ).

Note first that φj(ξ) is a trigonometric polynomial. Thus the number of zerosof any non-trivial combination, ψ(ξ, a) is finite in a bounded interval.

The function f (p)(x, a) is a polynomial in x of degree p + n with a root ofmultiplicity p at x = 1, i.e.

f (p)(x, a) = (1− x)pPn(x, a),

where Pn(x, a) is some polynomial of degree n that depends linearly on a.

Let x = cos (ξ). We have

f (p)(cos (ξ), a) = 1− 2

p+n∑

k=1

akkTk(cos (ξ)) = 1− 2

p+n∑

k=1

akk cos (kξ) =dE

dξ.

Thus, E(ξ, a) has an extreme point at ξ = 0 and at most n other extremepoints ξ1 ≤ ξ2 ≤ · · · ≤ ξn in [0, ξmax]. If ξ1 = 0 the stencil would be of higherorder than O(∆x2p) so we may assume that 0 < ξ1.

Consider now another error function, E(ξ,b). Linearity gives

∂ (E(ξ, a)− E(ξ,b))

∂ξ= (1− cos (ξ))p [Pn(cos (ξ), a)− Pn(cos (ξ),b)]

= (1− cos (ξ))pPn(cos (ξ), a− b)

but also

∂ (E(ξ, a)− E(ξ,b))

∂ξ=

∂

∂ξ[(Ec(ξ)− ψ(ξ, a))− (Ec(ξ)− ψ(ξ,b))]

=∂

∂ξψ(ξ,b− a),

where, as before, ψ is a linear combination of the basis functions, φj. Thevectors a and b are general so we may write c = b−a for any vector c ∈ Rn.Thus ψ(ξ, c) has at most n extrema in the open interval (0, ξmax]. Obviouslythere is at most one zero between any two consecutive extreme points. Thusψ(ξ, c) has at most n− 1 zeros in (0, ξmax]. This completes the proof.

5

Corollary 4.3. Let φj(ξ), j = 1, . . . , n be defined as in Proposition 2.1.Then a solves problem (2.5) if and only if there exists an alternating set of(n+ 1) points in (0, ξmax]. Such a solution exists and is unique.

Proof. Theorem 4.2 shows that for any δ satisfying 0 < δ < ξmax, φj is aChebyshev set on [δ, ξmax]. Thus Theorem 3.1 applies. Letting δ → 0 andnoting that at ξ = 0 the approximation is exact yields the desired result.

5 Example

We conclude with a brief example: Figure 5.1 shows the error of a stenciloptimised with respect to the max-norm. The stencil is second order accurateand is 31 points wide. Here we have set ξmax = π/2. In this region thedispersion error is bounded from above by ‖E‖∞ ∼ O(10−12). Clearly thealternation property holds as expected from Theorem 3.1.

ξ0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

E

×10-12

-1.5

-1

-0.5

0

0.5

1

1.5

‖E‖∞ = 1.337520e− 12

Figure 5.1: Optimised stencil for the case p = 1 and n = 14, i.e. the stencilhas formal accuracy O(∆x2) and is 31 points wide.

6

References

F. Riesz. Uber lineare functionalgleichungen. Acta Mathematica, 41:71–98,1918.

G. A. Watson. Approximation Theory and Numerical Methods. John Wiley& Sons Ltd., 1980.

7

The Ekeland Variational Principle WithApplications

Markus Wahlsten ∗

January 12, 2015

Abstract

We state and prove the Ekeland Variational Principle and exem-plify its usefulness with some applications.

1 Introduction

The Ekeland Variational Principle is an important tool in areas such as non-linear analysis and optimization theory. The principle is often used to estab-lish existence of solutions for optimization problems where the Weierstrass’theorem does not apply. Further, it’s proven to be a valuable tool whenstudying partial differential equations Ekeland [1974].

2 The Ekeland Variational Principle

Before we state The Ekeland Variational Principle we recall the Cantor’sintersection theorem and some definitions.

Theorem 2.1. (Cantor’s intersection theorem) Let Sk be a sequence ofnon-empty, closed and bounded sets satisfying

S0 ⊇ S1 ⊇ · · ·Sk ⊇ Sk+1 · · · , (2.1)

then (∞⋂

k

Sk

)6= ∅. (2.2)

∗Department of Mathematics, Computational Mathematics, Linkoping University, SE-581 83 Linkoping, Sweden markus.wahlsten@liu.se

1

Definition 2.1. (Upper and lower semicontinuity) Let f be a function andx0 a point such that f : R → R and x0 ∈ R. The functiom f is said to beupper semicontinuous at the point x0 if

f(x0) ≥ lim supx→x0

f(x) (2.3)

or lower semicontinuous if

f(x0) ≤ lim infx→x0

f(x) (2.4)

Theorem 2.2. (The Ekeland Variational Principle) Let X be a completemetric space and φ : X → R an upper semi-continuous map that is boundedfrom above. Then, for any ε > 0 there exists a y ∈ X such that,

φ(y) > φ(x)− εd(x, y), ∀x ∈ X\y (2.5)

Remark. The theorem can be modified by replacing the words ”upper” with”lower” when also ”above” is replaced by ”below” and finally in (2.5) the”−” and ”>” is replaced by ”+” and ”<” respectively.

Proof. By the fact that supφ(X) <∞, ∃z0 ∈ X such that,

φ(z0) > supφ(X)− ε. (2.6)

If we now construct two sequences zi∞i=0 and Si∞i=0 inductively in thefollowing manner

1. Assume zi is given2. Define Si := x ∈ X : φ(zi) ≤ φ(x)− εd(x, zi)3. Pick any element zi+1 ∈ Si such thatφ(zi+1) ≥ supφ(Si)− 1

i+1

(2.7)

Since φ is upper semi-continuous, the map φ(·)− εd(·, zi) is also upper semi-continuous, which leads to the fact that Si is closed.

Next, we need to prove that Si+1 ⊆ Si for i = 0, 1, . . . . First, we pick anarbitrary i ∈ Z+ and let x ∈ Si+1. Then, by using the triangle inequality,and the facts that x ∈ Si+1 and zi+1 ∈ Si, we obtain the relation

φ(x)− εd(x, zi) ≥ φ(x)− εd(x, zi+1)− εd(zi+1, zi)≥ φ(zi+1)− εd(zi+1, zi)≥ φ(zi)

(2.8)

which means x ∈ Si and since x was arbitrarily chosen Si+1 ⊆ Si.

2

Further, in order to apply the Cantor Frechet Intersection Theorem wealso need to prove that

δ(Si) = supx,y∈Si

d(x, y) ≤ 2

εi. (2.9)

By picking an i ∈ Z+, and an x ∈ Si+1, which also is x ∈ Si which we knowfrom the proof above. By using this fact and the inequality in (2.7) we get,

εd(x, zi) ≤ φ(x)− φ(zi)≤ supφ(Si−1)− φ(zi)≤ 1

i

(2.10)

which proves the desired result. By applying the Cantor Frechet IntersectionTheorem we conclude that the intersection

⋂+∞i=0 Si = y for some y ∈ X.

To conclude our proof we need to show that the element y satisfies (2.7).This is done by assuming the contrary, let x ∈ X\y, hence we have

φ(x)− εd(x, y) ≥ φ(y), (2.11)

thenφ(x)− εd(x, y) ≥ φ(y) ≥ φ(zi) + εd(x, zi) (2.12)

since (2.12) is satisfied for all i = 0, 1, . . . we have

φ(x) ≥ φ(zi) + ε(d(x, y) + d(y, zi))≥ φ(zi) + εd(x, zi)

(2.13)

(2.13) is true for all i = 0, 1, . . . , hence x ∈ ⋂+∞i=0 Si, which means x = y,

hence a contradiction.

3 Applications

In this section several useful applications of the Ekeland Variational Principleis presented.

Theorem 3.1. Caristi’s Fixed Point Theorem Let ψ be a self-map on acomplete metric space X. If

d(x, ψ(x)) ≤ φ(x)− φ(ψ(x)), ∀x ∈ X (3.1)

for some lower semi-continuous φ ∈ RX that is bounded from below, then ψhas a fixed point in X.

3

Proof. From the Ekeland Variational Principle, we know that there existssome y ∈ X such that,

φ(y) < φ(x) + d(x, y), ∀x ∈ X\y. (3.2)

However, by (3.1) we have that

φ(y) ≥ φ(ψ(y)) + d(y, ψ(y)). (3.3)

So ψ(y) /∈ X\y, that is ψ(y) = y.

Definition 3.1. (Palais-Smale condition) We say that a C1 functional φ :X → R satisfies the Palais-Smale condition if every sequence un∞n=1 ∈ Xwhich satisfies

‖ψ(un)‖ ≤ const, and ψ(un)→ 0 in X (3.4)

possesses a convergent subsequence.

Theorem 3.2. Let ψ : X → R be a C1 functional where X is a Hilbertspace. Let S be a closed and convex subset of X. Suppose that

• K ≡ I − ψ′ which maps S into S

• ψ is bounded from below in S

• ψ satisfies definition 3.1 in S

Then there exists a u0 ∈ S such that ψ′(u0) = 0 and infu∈S

ψ(u) = ψ(u0)

Proof. We start by applying the Ekeland Variational Principle to ψ. Givenε there exists a uε ∈ C such that ψ(uε) ≤ inf

u∈Sψ(u) + ε and

ψ(uε) ≤ ψ(u) + ε ‖u− uε‖ , ∀u ∈ S (3.5)

If we now let u = (1 + t)uε + tKuε, where 0 < t < 1. Next, we use Taylorsformula to expand ψ(uε + t(Kuε − uε)) about uε and obtain,

t ‖ψ′(uε)‖2 ≤ ε ‖ψ′(uε)‖+O(t), (3.6)

which implies ‖ψ′(uε)‖ < ε. Finally we use the property (3.1).

4 Summary

The Ekeland Variational Principle has been stated and proved in section 2followed by some useful applications of the theorem in section 3.

4

References

I. Ekeland. On the variational principle. Journal of Mathematical Analysisand applications, 47:324–353, 1974.

5

Construction of Sobolev spaces

Cristina La Cognata∗

January 15, 2015

Abstract

Sobolev spaces are Banach spaces that are characterized by the defini-tion of ”weak” derivatives. The functions in Sobolev spaces and theirweak derivatives lie in Lp spaces.

Introduction

This essay is an introduction to the theory of “Sobolev” spaces. They arepowerful tools from Functional Analysis for defining the solutions of PartialDifferential Equations (PDE). In particular, they are used in those classes ofproblems in which smoothness and continuity are too strong requirements.In a certain sense, functions that belong to Sobolev spaces represent a goodcompromise as they have some, but not all, smoothness properties [Evans,2002]. We start with the definition of the concept of weak derivatives andthen we will apply it to the construction of the Sobolev spaces. We concludewith an overview of some of the most interesting properties concerning thesespaces.

1 Weak derivatives

A kth-order PDE can be symbolically represented by the following expression

F (Dku(x), Dk−1u(x), ..., Du(x), u(x), x) = 0, x ∈ U, (1.1)

where k ≥ 1 is a fixed integer, U is an open set in Rn and u : U → R isthe unknown. The equation (1.1) involves an unknown function u : U → R,

∗Division of Computational Mathematics, MAI, Linkoping University, SE-581 83Linkoping, Sweden. cristina.la.cognata@liu.se

1

of variable x ∈ U , and some of its partial derivatives. By solving a PDEof type (1.1) we mean (ideally) finding a simple (in the best case), explicitexpression for such a u, or at least guaranteeing its existence together withsome properties. Let us ignore for the moment the strict concept of a well-posed PDE problem. What do we require for u to be a “solution” of (1.1)?It is natural to desire having a solution to be at least k times continuouslydifferentiable (u ∈ Ck). A solution of this type is called “classical” solution.Unfortunately there exist certain PDEs (or most of them) which cannot besolved in a classical sense. For instance, in general the conservation law ofthe form

ut + F (u)x = 0, x ∈ U, (1.2)

has no classical solutions Evans [2002],R.LeVeque [1992]. Then we need toallow a more general or weaker class of functions to be our solutions.

Let us denote by C∞c (U) the spaces of infinitely many times differentiablefunctions φ : U → R, with compact support on U . We will refer to a functionin C∞c (U) as a test function. Furthermore, we will widely use Lp spaces andtheir norms. For a a reader which is not familiar with these type of spaceswe remand to Evans [2002], R.LeVeque [1992].

Definition. Suppose that u, v ∈ L1loc(U)1 and α is a multi-index. We say

that v is the αth-weak derivative of u if∫

U

uDαφ dx = (−1)|α|∫

U

vφ dx (1.3)

for all test functions φ ∈ C∞c (U).

In other words, a weak derivative is a generalization of the concept of thederivative of a function (strong derivative) for functions not assumed differ-entiable, but only integrable in the Lebesgue sense. Note that, since u is notnecessarily a differentiable function, the locally summability assumption isneeded to ensure the existence of the right-hand side of (1.3).

The notion of weak derivatives originated with the works of J. Leray (1906-1998) and later from S. Sobolev (1908-1989) when he came to deal withnon-continuous functions with almost everywhere existing derivatives.

Lemma 1.1. The αth-weak derivative of u is uniquely defined almost every-where2

1see Appendix2see Appendix

2

Proof. Let us assume that v, w ∈ L1loc(U) both satisfy

∫

U

uDαφ dx = (−1)|α|∫

U

vφ dx = (−1)|α|∫

U

wφ dx,

for all test functions φ ∈ C∞c (U). Then∫

U

(v − w)φ dx = 0,

for all test functions φ ∈ C∞c (U). Hence, (v − w) = 0 a.e.

Example. Let n = 1, U = (a, b) and consider u, v defined as follows

u(x) =

x a < x ≤ (a+ b)/21 (a+ b)/2 ≤ x < b

v(x) =

1 a < x ≤ (a+ b)/20 (a+ b)/2 ≤ x < b.

We can show that v is the weak derivative of u by choosing any test functionφ ∈ C∞c (U) and calculate

∫ bauφ′ dx =

∫ (a+b)/2

axφ′ dx+

∫ b(a+b)/2

φ′ dx

= −∫ (a+b)/2

aφ dx+ [φ((a+ b)/2)− φ((a+ b)/2)] ((a+ b)/2) =

= −∫ bavφ dx.

as required.

2 Sobolev spaces

We now define a function space which contains elements with weak derivativesof various orders from Lp spaces. Let us fix 1 ≤ p ≤ ∞ and let k be a non-negative integer.

Definition (Sobolev spaces). The Sobolev space W k,p(U) consists of all lo-cally summable functions u : U → R such that for each multi-index α with|α| ≤ k, the weak derivative Dαu exists and lie in Lp(U).

Definition (Sobolev norms). W k,p(U) is a normed space equipped with thenorm defined as

‖u‖Wk,p(U) =

(∑|α|≤k

∫U|Dαu|p dx

)1/p(1 ≤ p ≤ ∞)

∑|α|≤k ess supU |Dαu| (p =∞)

, (2.1)

where u ∈ W k,p(U). For the definition of ess sup f see 3

3

We finally introduce the following notation for convergence:

Definition (Sobolev convergence). We say that um∞m=1 , um ∈ W k,p(U)converge to u ∈ W k,p(U) provided that

limm→∞

‖um − u‖Wk,p(U) = 0.

Next we list some elementary properties Sobolev functions. The proofs canbe found in Evans [2002]. Note that the following properties are triviallytrue for functions in Ck(U) spaces. For Sobolev functions, instead, we mustthink about these rules as in the weak definition.

Theorem 2.1. Assume u, v ∈ W k,p(U), |α| ≤ k. Then

1. Dβ(Dαu) = Dα(Dβu) = Dα+βu for all |α|+ |β| ≤ k.

2. For each λ, µ ∈ R, λu+ µv ∈ W k,p(U) and λDαu+ µDα, |α| ≤ k.

3. if V is an open set in U , then U ∈ W k,p(V ).

Some of these properties are needed to prove that the Sobolev spaces areBanach spaces.

Theorem 2.2 (Sobolev spaces as a function spaces). For each k = 1, ..., and1 ≤ p ≤ ∞, the Sobolev space W k,p(U) is a Banach space.

Proof. First we check that (2.1) is a norm. It is easy to check from thedefinition that

‖λu‖Wk,p(U) = |λ|‖u‖Wk,p(U) and ‖u‖Wk,p(U) = 0 iff u = 0 a.e.(2.2)

Next we assume that u, v ∈ W k,p(U) with 1 ≤ p <∞. Then

‖u+ v‖Wk,p(U) =(∑

|α|≤k‖Dαu+Dαv‖pLp(U)

)1/p

≤(∑

|α|≤k(‖Dαu‖Lp(U) + ‖Dαv‖Lp(U)

)p)1/p

≤(∑

|α|≤k‖Dαu‖pLp(U)

)1/p+(∑

|α|≤k‖Dαv‖pLp(U)

)1/p

= ‖u‖Wk,p(U) + ‖v‖Wk,p(U).

Note that in the second inequality we have used Minkowski’s relation definedin 3. Next we prove that W k,p(U) is complete by showing that any Cauchysequence um ∈ W k,p(U) converges.

4

1. For any |α| ≤ k,Dαum is a Cauchy sequence in Lp(U). Since Lp(U) iscomplete, the existence of u and uα,∀α, such that

um → u and Dαum → uα in Lp(U), (2.3)

is guaranteed.

2. Consider any test function φ ∈ C∞c (U). Then we have

∫UuDαφ dx = limm→∞

∫UumD

αφ dx= limm→∞(−1)|α|

∫UDαumφ dx

= (−1)|α|∫Uuαφ dx.

Thus, uα = Dαu, |α| ≤ k.

3. From the two previous steps we find that um → u in Lp(U) and the uαare the weak derivatives of u. Therefore um → u in W k,p(U).

Thus W k,p(U) is a Banach space.

We conclude by noting that in the special case p = 2, the Sobolev space hasa more interesting structure.

Definition (Hilbert-Sobolev spaces). We define the Hilbert-Sobolev spaceas Hk(U) = W k,2(U). The space Hk(U) is a Hilbert space endowed with theinner product

(u, v)Hk(U) =∑

|α|≤k

∫

U

DαuDαv dx.

Note that H0(U) = L2(U).

For the curious reader, more details and results can be found in Brezis[1983].

3 Appendix

Definition (Almost everywhere). If a property holds everywhere in Rn, ex-cept for a measurable set of Lebesgue measure zero, we say that the propertyis true almost everywhere (a.e.).

Definition (ess sup). If a real function f is measurable, the essential supre-mum of f is defined as

ess sup f = inf µ ∈ R s.t. |f > µ| = 0 .

5

Definition (L1loc functions).

L1loc = u : U → R s.t. u ∈ Lp(V ) for each V ⊂⊂ U .

Definition (Minkowski’s inequality). Assume 1 ≤ p ≤ ∞ and u, v ∈ Lp(U).Then

‖u+ v‖Lp(U) ≤ ‖u‖Lp(U) + ‖v‖Lp(U).

The Minkowski inequality is the triangle inequality in Lp(U).

References

H. Brezis. Analyse Fonctionnelle. Masson, 1983.

L. Evans. Partial Differential Equations. AMS, 2002.

R.LeVeque. Numerical Methods for Conservation Laws. Birkhauser, 1992.

6

The Arzela-Ascoli theorem

Tomas Lundquist∗

January 15, 2015

Abstract

We discuss the Arzela-Ascola theorem from a functional analysispoint of view, and give examples of applications within the field, inparticular relating to the theory of compact embeddings in functionspaces.

1 Introduction

The Arzela-Ascola theorem is of fundamental importance in both real analy-sis and functional analysis, as well as in topology. In functional analysis, it isparticularly useful in function space theory, providing a constructive criterionfor compactness of subsets in C(X), the set of complex valued continuousfunctions on a metric space X .

In its original form, the theorem was formulated for sequences of contin-uous functions on compact intervals of the real line, giving a criterion forthe existence of uniformly bounded subsequences. Later it was extended tocompact metric spaces then further to compact Hausdorff topological spaces.Here we will focus on the functional analysis version of the theorem. In sec-tion 2 we present the theorem and discuss some of its implications. Section3 is a brief introduction to the theory of compact embeddings in continuousfunction spaces, to which the theorem is closely related.

2 Statement of the theorem

The Arzela-Ascoli theorem gives a necessary and sufficient condition for com-pactness of continuous function spaces. Before we proceed with stating thetheorem, we need the following definition.

∗Division of Computational Mathematics, Department of Mathematics, Linkoping uni-versity, SE-581 83 Linkoping, Sweden. tomas.lunqduist@liu.se

1

Definition 2.1. A set F of functions between two metric spaces X andY is said to be equicontinuous if for each ϵ > 0, ∃ a δ > 0 such thatd(f(x0), f(x1)) < ϵ for all f ∈ F , and x0, x1 ∈ X with d(x0, x1) < δ. Theparameter δ may depend on x0 but not on f .

Note the contrast between equicontinuity and uniform continuity. In theformer case, ϵ is not allowed to depend on f , whereas in the latter case, ϵ isnot allowed to depend on x0.

The metric space version of the Arzela-Ascoli theorem now follows below.

Theorem 2.1. Let X be a compact metric space, and denote with C(X)the space of real (or complex) valued continuous functions on X with respectto the supremum norm. Then a subset M of C(X) is relatively compact inC(X) if and only if M is equicontinuous and bounded.

Proof. A proof can be found e.g. in Nagy [2006].

We recall that relatively compact in this case means that the closure M ofM is compact in C(X); that is, each sequence in M has a convergent subse-quence. The Arzela-Ascola theorem is used in the proof of many importanttheoretical results. One example is Peano’s theorem of local existence forcontinuous ordinary differential equations. Another is to prove that an op-erator between Banach spaces is compact if and only if the adjoint operatoris compact (see e.g. Garrett [2012]).

3 Application: compact embeddings of con-

tinuous functions

The concept of compact embeddings between function spaces is importantnot the least in the study of numerical methods for partial differential equa-tions. For example, the famous Sobolev embedding theorems are examplesof results relating to compact embeddings. These results are important whenapplying Sobolev spaces to partial differential equations. For a discussion ofthis, see e.g. Dlotko [2014].

An embedding E : X ← Y from a subset X ⊂ Y into Y is sim-ply the identity operator. As an example, consider the space C1(Ω) ofC1−differentiable functions on the domain Ω ∈ Cn, equipped with the norm∥f∥C1 = sup|f | + sup|∇f |. With this definition, the embedding E(C1(Ω))into C(Ω) is the space formed by the set C1(Ω) and standard supremumnorm. As we shall later see, this embedding is indeed compact.

2

Definition 3.1. Let (X, ∥·∥X) and (Y, ∥·∥Y ) be two normed spaces, where Xis a subset of Y . We say that X is compactly embedded in Y if the followingholds:

1. ∃ c > 0 : ∥x∥Y ≤ c∥x∥X ∀x ∈ X. (uniform continuity)2. The embedding of X into Y is a compact operator. (compactness)

We recall that an operator is compact if it maps bounded sets to relativelycompact sets.

The Arzela-Ascoli theorem provides a constructive way to show compact-ness of embeddings. Too see this, we continue with the example above. Thus,consider a bounded subset M of the space C1(Ω). The boundedness of sup|f |and sup|∇f | then implies equicontinuity of the set (this part is left as an ex-ercise). From the Arzela-Ascoli theorem, it also follows that M is relativelycompact in C(Ω). i.e. the embedding E(M) into C(Ω) is compact. More-over, uniform continuity follows directly from the definitions of the normsinvolved:

∥f∥C(Ω) = sup|f | ≤ sup|f | + sup|∇f | = ∥f∥C1(Ω).

Thus the space C1(Ω) is compactly embedded in C(Ω) with respect to thenorm given above.

Analogous results to the one given above can be derived using the Arzola-Ascoli theorem for a variety of continuous function spaces. In particularthe Sobolev embedding theorems, of fundamental importance for Sobolevspace theory, can be proven using similar arguments. The most famous ofthese result states that the embedding Hm+1(Ω) in Hm(Ω) is compact, whereHm(Ω) is the Sobolev space with square integrable weak derivatives up toorder m. Note that H0(Ω) = L2(Ω), so that all Sobolev spaces are compactlyembedded in the space of square integrable functions. A full proof of this,generalized to Lp(Ω), is given e.g. in Adams [1975].

4 Summary

We have reviewed the metric space formulation of the Arzola-Ascoli theorem,and shown its usefulness as a constructive criterion for compactness of con-tinuous function spaces. An example of application is given by the famousSobolev embedding theorems.

References

R. Adams. Sobolev spaces. Academic Press, New York, 1975.

3

T. Dlotko. Sobolev spaces and embedding theorems. Technical report,www.icmc.usp.br/~andcarva/sobolew.pdf, 2014.

P. Garrett. Compact operators on banachspaces: Fredholm-riesz. Technical report,http://www.math.umn.edu/~garrett/m/fun/fredholm-riesz.pdf,2012.

G. Nagy. A functional analysis point of view of the arzela-ascoli theorem.Real Analysis Exchange, 32:583–586, 2006.

4

A Completeness Theorem for Non-Self-AdjointEigenvalue Problems

Saleh RezaeiRavesh∗

January 16, 2015

Abstract

An important completeness theorem for a general class of non-self-adjoint eigenvalue problems is stated. The corresponding proof isreproduced in connection with basic material in spectral theory andfunctional analysis.

1 Introduction

In many applications in hydrodynamic stability analysis we are confrontedwith eigenvalue problems in the form

(L− λM)ϕ = 0 (1.1)

where L and M are differential operators and λ ∈ σ(L,M). The associatedboundary conditions for (1.1) are ϕ(0) = Dϕ(0) = ϕ(1) = Dϕ(1) = 0, where,D := d/dy is the differentiation operator. One of the famous examples is theOrr-Sommerfeld equation which is derived in linearized stability analysis ofparallel flows (e.g., cf. Drazin and Reid [1981]). The general idea for studyingthe spectrum of (1.1) is to write L := Ls+B where Ls and M are self-adjointoperators and B is called perturbation.In Section 2, some of the definitions and theorems required in this documentare expressed. Then, theorem 3.1 shows how we can find a self-adjoint ex-tension operator G for M−1Ls in such a way that two problems Gϕ = λϕand Lsϕ = λMϕ become equivalent. Subsequently, theorem 3.2 proves thecompleteness of the generalized eigenvectors of (1.1). Finally, an applicationof the completeness theorem is briefly explained.

∗Division of Scientific Computing, Department of Information Technology, Uppsalauniversity, SE-751 05 Uppsala, Sweden. saleh.rezaeiravesh@it.uu.se

1

2 Spectral Theory

Definition 2.1 (Resolvent Set & Spectrum of T ) Let X 6= 0 be acomplex normed space and T : D(T )→ X1 a linear operator with D(T ) ⊂ X.Then, (a) The resolvent set ρ(T ) of T is the set of all regular values λ ofT . If T : X → X is a bounded linear operator and X is a Banach space,then, ρ(T ) = λ ∈ C|(T − λI) is one-to-one and onto Evans [2010] (Anyλ ∈ σ(T ) is called a spectral value of T (cf. 7.2-1 Kreyszig [1989])), (b) Thespectrum of T is the complement of ρ(T ); i.e., σ(T ) = C − ρ(T ) (cf. 7.2-1Kreyszig [1989]).

Definition 2.2 (Resolvent Operator) If Tλ := (T −λI) has an inverse,denoted by Rλ(T ) := T−1λ = (T − λI)−1, we call it the resolvent operator ofT (cf. 7.2 Kreyszig [1989]).

Definition 2.3 (Point, Continuous, Residual Spectra) The point ordiscrete spectrum σp(T ) is the set such that Rλ(T ) does not exist. λ ∈ σp(T )is called an eigenvalue of T .The continuous spectrum σc(T ) is the set such that Rλ(T ), exists and has adomain dense in X, but is unbounded.The residual spectrum σr(T ) is the set such that Rλ(T ), (a) exists (may bebounded or not) but its domain is not dense in X (cf. 7.2-1 Kreyszig [1989]).

Theorem 2.1. (Distance from σ(T )) If T ∈ B(X,X), where X is com-plex Banach space and λ ∈ ρ(T ), then (cf. 7.5-3 Kreyszig [1989])

‖Rλ(T )‖ ≥ 1

d(λ, σ(T )), d(λ, σ(T )) := inf

s∈σ(T )|λ− s| (2.1)

Definition 2.4 (Eigenvalue & Eigenvector) We say that λ ∈ σ(T ) isan eigenvalue of T provided N (T − λI) 6= 0. We write σp(T ) to denotethe collection of eigenvalues of T ; clearly, σp(T ) is a point spectrum. If λis an eigenvalue and ϕ 6= 0 satisfies Tϕ = λϕ, we say ϕ is an associatedeigenvector (cf. D.3 Evans [2010]).

Definition 2.5 (Generalized Eigenvector of (L,M)) Let L and Mbe linear operators with D(L) ⊂ D(M) dense in a Hilbert space H. We saythat ψ ∈ H is a generalized eigenvector of (L,M) if and only if for some λand some integer p > 1, there exist p non-zero vectors ψ1, ψ2, · · · , ψp = ψsuch that Lψi = λMψi + Mψi−1, i = 1, · · · , p, where ψ0 = 0. By allowingp = 1, eigenvectors become included in generalized eigenvectors (DiPrimaand Habetler [1969]).

1In this manuscript, D(T ), R(T ), and N (T ) respectively denote domain, range andnull space of operator T .

2

Definition 2.6 (Completeness) The generalized eigenvectors of a linearoperator with domain dense in H are said to span H if the following is true.Let ϕk1, ϕk2, · · · , ϕknk

be the eigenvectors (and possibly generalized eigenvec-tors) corresponding to the eigenvalues in rk−1 < |λ| < rk for k > 1 and|λ| < r1 for k = 1; then each ϕ ∈ H has a representation

ϕ =∞∑

i=1

ni∑

j=1

aijϕij (2.2)

with convergence in the norm of H (DiPrima and Habetler [1969]).

Theorem 2.2. (Friedrichs Extension) A positive-bounded below (hencepositive definite) operator with domain dense in a Hilbert space can be ex-tended to a self-adjoint operator which possesses an inverse defined on theentire space (cf. p 11 Mikhlin [1965]).

Theorem 2.3. (Naimark’s Theorem) If T is a closed linear operatoron a dense subspace of a Hilbert space H with norm ‖.‖ whose resolvent(T − λI)−1 is compact for some λ = λ0, and if there exists a sequence ofcircles Ck, |λ| = rk such that (a) there are no eigenvalues of T on any Ck,(b) limk→∞ rk = ∞, and (c) limk→∞ supλ∈Ck

‖(T − λI)−1‖ = 0, then thegeneralized eigenvectors of T span H in the sense of (2.2) (e.g., see DiPrimaand Habetler [1969]).

3 Completeness Theorem

Theorem 3.1. (Equivalent EVP’s) Let Ls and M be linear operatorswith D(Ls) ⊂ D(M) dense in a Hilbert space H and such that,

1. M and Ls are positive-bounded below with R(M) = R(Ls),

2. G−1 is compact and R(G−1) ⊂ D(M) where G−1 is the inverse of theself-adjoint extension of M−1Ls in HM .2

Then, the two problems Gϕ = λϕ and Lsϕ = λMϕ are equivalent (DiPrimaand Habetler [1969]).

Proof: I Step 1. (Construct G−1): M is psoitive-bounded below, i.e.,

∀x ∈ D(M) ∃α > 0 : 〈Mx, x〉 ≥ α‖x‖2

2A new Hilbert space HM which is embedded in H can be constructed by completingthe pre-Hilbert space HM (D(M), [., .]), where [f, g] = 〈f,Mg〉 for f, g ∈ D(M) is the

corresponding inner product and the norm is ‖f‖HM= [f, f ]

12 (cf. 3 Mikhlin [1965]).

3

Combining this with the Schwarz inequality results in,

α‖x‖2 ≤ 〈Mx, x〉 ≤ ‖Mx‖‖x‖

Note that y = Mx for x and y in D(M) and R(M) of M , respectively; hence,

‖M−1y‖ ≤ 1

α‖y‖

Therefore, M−1 exists and is bounded. Furthermore, M−1Ls is positive-bounded below in HM ; since ∀x ∈ D(Ls) ⊂ D(M),

[x,M−1Lsx] = 〈x,MM−1Lsx〉 = 〈x, Lsx〉 ≥ β‖x‖2, β > 0

where the last inequality holds since Ls is positive-bounded below in H.According to Theorem 2.2, M−1Ls has a self-adjoint extension (call it G)that has an inverse G−1 as mentioned in hypothesis 2.

I Step 2. (Find the Spectrum of G): Since G−1 is compact, then it has apoint spectrum, σp(G) (cf. 8.3-1, 8.3-3 Kreyszig [1989]). Furthermore, theset of eigenvalues G is countable (may be finite and even empty) and λ = 0 isthe only possible point of accumulation of that set. In general, on D(G) ⊂ Xwhich is a complex normed space two cases can be considered; if X is finitedimensional, then G has a matrix representation and clearly 0 may or maynot belong to σ(G) = σp(G); that is, if dimX < ∞ we may have 0 /∈ σ(G);then 0 ∈ ρ(G). However, if dimX = ∞, then we must have 0 ∈ σ(G) (cf.pp. 420 and 433 Kreyszig [1989]).

I Step 3. (Equivalent Spectra): We show that σ(G) = σp(G) = σp(Ls,M).3.a) Suppose λ ∈ σ(G); then, there exists ϕ ∈ D(G) such that Gϕ = λϕ.Since from hypothesis 2, D(G) = R(G−1) ⊂ D(M), then we have ϕ ∈ D(M)and hence Gϕ ∈ D(M). But, since R(M) = R(Ls), M

−1Ls is onto D(M).This means that there exists ψ ∈ D(Ls) such that M−1Lsψ = Gϕ, and sinceG = M−1Ls on D(Ls) and G−1 exists, we have ψ = ϕ. As a result,

M−1Lsψ = M−1Lsϕ = λϕ⇒ Lsϕ = λMϕ⇒ λ ∈ σp(Ls,M)

3.b) Conversely, consider λ ∈ σp(Ls,M) and correspondingly, Lsϕ = λMϕ,ϕ 6= 0. Then immediately we have M−1Lsϕ = λϕ an hence Gϕ = λϕ andλ ∈ σ(G). Consequently, σp(Ls,M) ∈ σ(G) and an eigenvector of G isan eigenvector of (Ls,M). This completes the proof of equivalency of theGϕ = λϕ and Lsϕ = λMϕ .

Theorem 3.2. (Completeness) Let L = Ls + B, and let Ls, B, and Mbe linear operators with D(Ls) = D(B) ⊂ D(M) dense in a Hilbert spaceH, such that,

4

1. Ls and M satisfy conditions of theorem 3.1,

2. M−1B is bounded in HM and D(Ls) is dense in HM ,

3. for some λ, R(Ls + B + λM) = R(M) and (G + M−1B + λI) has aninverse,

4. there exists a sequence of concentric circles Ck with the radii rksuch that, (a) limk→∞ rk = ∞, (b) d(Ck, σp(Ls,M)) > 0 for all k, (c)limk→∞ d(Ck, σp(Ls,M)) =∞.

Then,

1. the eigenvalues of (L,M) lie within circles of radii ‖M−1B‖HMabout

those of (Ls,M), and,

2. the generalized eigenvectors of (L,M) span HM in the sense of (2.2)(DiPrima and Habetler [1969]).

Proof: I Step 1. (Define T and Show it is Closed) Consider T :=G+M−1B where M−1B is a unique bounded extension from D(B) to D(G).Now, we show that T satisfies the condition of Naimark’s theorem 2.3. Ac-cording to the proof of theorem 3.1, G is self-adjoint and therefore closed.In addition by hypothesis 2., M−1B is bounded and when we add it to theclosed operator G, we end up with closed operator T .

I Step 2. (Show that ∃λ0 : Rλ(T ) = (T − λ0I)−1 is Compact): Let’s chooseλ0 such that d(λ0, σ(G)) > ‖M−1B‖HM

. This is always possible; since ac-cording to hypothesis 4.(c), it is enough to choose λ0 on Ck for sufficientlylarge k. G−1 is compact and G is closed, so (G − λI)−1 is compact for allλ ∈ ρ(G) (cf. 5.7 Engel and Nagel [2006]), and hence for λ = λ0. Now, wecan show that (T − λ0I)−1 is compact. we have,

(T − λ0I)−1 = (G+M−1B − λ0I)−1 = [I + (G− λ0I)−1M−1B]−1(G− λ0I)−1

By using theorem 2.1, we can show that [I+(G−λ0I)−1M−1B]−1 is bounded:

‖I+(G−λ0I)−1M−1B‖HM≥ 1−‖(G−λ0I)−1‖HM

‖M−1B‖HM≥ 1−‖M

−1B‖HM

d(λ0, σ(G))> 0

Therefore, (T − λ0I)−1 which is the product of a bounded operator and acompact operator, is compact (cf. 8.3-2 Kreyszig [1989]).

I Step 3. (Obtain Conclusion 1.): 3.a) Since (T − λ0I)−1 is compact andT is closed, then (T − λI)−1 is compact for all λ ∈ ρ(T ) (see 5.7 Engeland Nagel [2006]). Therefore, σ(T ) is a point spectrum such that each λ ∈σ(T ) = σp(T ) has finite multiplicity (cf. 8.3-1 Kreyszig [1989]). On the otherhand, if λ is such that d(λ, σ(G)) > ‖M−1B‖HM

then necessarily λ ∈ ρ(T ).Consequently, for λ ∈ σ(T ), we have d(λ, σ(G)) < ‖M−1B‖HM

. In other

5

words, σ(T ) lies within the union of a set of circles of radii ‖M−1B‖HMand

center σ(G) = σp(Ls,M).3.b) By a similar discussion made in Step 3. of the proof of theorem 3.1,it can be shown that σ(T ) = σp(M

−1L). For this, replace Ls by L + λM ;similarly, existence of G−1 is replaced by existence of (G+M−1B+λI)−1 forλ mentioned in hypothesis 3. Since it is possible that G + M−1B does nothave an inverse, it is required that such λ exists. Hence, σp(M

−1L) lies withinthe union of a set of circles of radii ‖M−1B‖HM

centered about σp(M−1Ls).

I Step 4. Obtain Conclusion 2.: According to hypothesis 4.(c) we can choosea k0 such that for all integers k ≥ k0, d(Ck, σp(Ls,M)) = d(Ck, σp(G)) >‖M−1B‖HM

. Then for such a k and λ ∈ Ck, we have

‖(T − λI)ϕ‖HM= ‖(G− λI)ϕ+M−1Bϕ‖HM

≥ ‖(G− λI)ϕ‖HM− ‖M−1Bϕ‖HM

≥ d(λ, σp(Ls,M))− ‖M−1B‖HM‖ϕ‖HM

Furthermore, ‖(T − λI)ϕ‖HM≤ ‖(T − λI)‖HM

‖ϕ‖HM. Combining this with

the latest inequality and then taking supremum of both sides results in,

supλ∈Ck

‖(T − λI)−1‖HM≤ 1

d(λ, σp(Ls,M))− ‖M−1B‖HM

Clearly, by hypothesis 4.(c), limk→∞ supλ∈Ck‖(T −λI)−1‖HM

= 0. Thereforeall the conditions of Naimark’s theorem are satisfied by T for circles Ck withk ≥ k0. Consequently, the generalized eigenvectors of T , and hence those of(L,M) span HM in the sense of (2.2) .

4 Application

To show an application of theorem 3.2, consider the Orr-Sommerfeld equationrewritten in the form of equation (1.1); then

λ = iαcRe (4.1)

Ls = (D2 − α2)2, B = −iαRe[U(D2 − α2)− U ′′] (4.2)

M = −(D2 − α2) (4.3)

Here, U(y) is the base flow profile and U ′′ = d2U/dy2. Moreover, α andRe arewavenumber and Reynolds number, respectively. In hydrodynamic temporalstability analysis we are interested in finding the phase speed c appearing inλ for a set of input parameters. In association with the completeness theoremmentioned above, consider

H = L2(Ω) = H0(Ω), HM = H10 (Ω), where Ω = [0, 1]

6

Then, through a detailed process, one can show that differential operatorsin the Orr-Sommerfeld equation satisfy all hypotheses of theorems 3.1 and3.2. Here as a step of that process, the construction of concentric circles ispresented. Consider σp(Ls,M) = µn ∪ ϑn where the µn are the orderedpositive eigenvalues of the problem

Lsϕ1 = µMϕ1, ϕ1(0) = ϕ′1(0) = ϕ1(1) = ϕ′1(1) = 0

(repeated eigenvalues, if any occurs, are listed once) and the numbers νn arethe ordered positive eigenvalues of the problem

Mϕ2 = ϑϕ2, ϕ2(0) = ϕ2(1) = 0

For the operators in the Orr-Sommerfeld equation, it can be shown that asn→∞, µn+1 − µn →∞ and ϑn+1 − ϑn →∞. Thus, the sequence Ck∞k=1

is a set of concentric circles with center at the origin and with radii rk asfollows. For n ≥ 1 consider the interval (µn, µn+1). Then,(a) if there is no ϑl ∈ (µn, µn+1), take rn = 1

2(µn + µn+1);

(b) if there is only one ϑl ∈ (µn, µn+1), then when ϑl ∈ (µn,12(µn+µn+1)) take

rn = 14µn+ 3

4µn+1 and when ϑl ∈ (1

2(µn+µn+1), µn+1) take rn = 3

4µn+ 1

4µn+1;

(c) if there are several (say m) ϑl in (µn, µn+1), take rn = 1m

∑mi=1 ϑi (DiPrima

and Habetler [1969]).

As a result of theorem 3.2, for a bounded flow, the spectrum of the Orr-Sommerfeld equation consists of an infinite number of discrete eigenvaluesand the associated eigenvectors span HM = H1

0 (Ω) in the sense of (2.2). Bythis completeness, an arbitrary initial disturbance which satisfies the phys-ical boundary conditions can be expanded in terms of these eigenfunctions.This technique is very useful in studying different hydrodynamic instabilitymechanisms.

References

R. C. DiPrima and G. J. Habetler. The completeness theorem for nonselfad-joint eigenvalue problems in hydrodynamic stability. Archieve for RationalMechanics and Analysis, 34(3):218–227, 1969.

P. G. Drazin and W. H. Reid. Hydrodynamic Stability. Cambridge Univ.Press, 1981.

K.-J. Engel and R. Nagel. A Short Course on Operator Semigroups. Springer,2006.

7

L. C. Evans. Partial Differential Equations. American Mathematical Society,2nd edition, 2010.

E. Kreyszig. Indtroductory Functional Analysis with Applications. John Wi-ley & Sons, 1989.

S. G. Mikhlin. The Problem of the Minimum of a Quadrature Functional.Holden-Day INC., 1965.

8

The Metric Lax and Applications∗

Cheng Gong†

January 16, 2015

Abstract

We define consistency, convergence, and stability of numerical meth-ods in the setting of metric spaces. The Lax equivalence theorem isformulated and a proof is given in this setting. We also indicate itsuse in some applications.

1 Introduction

The Lax (or Lax-Richtmyer) equivalent theorem Lax and Richtmyer [1956] isthe fundamental theorem in numerical analysis. It was established by PeterLax and Robert D. Richtmyer in 1956. The theorem states that “a consis-tent finite difference method for a well-posed linear initial value problem isconvergent if and only if it is stable”.

The convergence of a numerical method is a measurement of the dif-ferences between the numerical solution and the analytic solution from themathematical problem. In general, the analytic solutions are usually not ac-cessible and the convergence is difficult to be observed directly. However, theconsistency and stability can be evaluated from the properties of the numer-ical method, e.g. error estimate, Fourier analysis and Von Neumann analysisGustafsson et al. [2013]. Thus, the Lax equivalent theorem provides a practi-cal way to prove the convergence of a numerical method from stability. Thisprinciple can be formulated and proved in metric spaces.

∗This essay is based on the notes from Stefan Engblom. stefan.engblom@it.uu.se†Division of Scientific Computing, Department of Information Technology, Uppsala

university, SE-751 05 Uppsala, Sweden. cheng.gong@it.uu.se

1

2 Definition

Let X = (X, d) and Y = (Y, d) be two metric spaces and we define themathematical problem as: find x ∈ X, such that

Tx = y, (2.1)

where T : X −→ Y is an operator defined by the problem and y ∈ Y isgiven.

Definition 2.1. The mathematical problem in (2.1) is well-posed if T−1

exists and is continuous in some neighborhood containing y.

Then, the solution of a well-posed problem with the same form as in (2.1)is given by the inverse operator of T as

x = T−1y. (2.2)

The problem in (2.1) is approximated by a sequence of numerical problems:find xn ∈ X such that

Tnxn = y, (2.3)

where Tn : X −→ Y is an operator of the corresponding numerical methodand y ∈ Y is the same right-hand-side as in (2.1). The family of equations in(2.3) is called the numerical approximations for the problem in (2.1) [Hunterand Nachtergaele, 2001] and n refers to the indices of the sequence such thatTn −→ T as n −→∞.

Definition 2.2. Let D(T ) be the domain of T . The numerical method in(2.3) is consistent, if ∀x ∈ D(T ), where D(T ) ⊂ X is the domain of T inX, then Tnx −→ Tx as n −→∞.

Definition 2.3. The numerical method in (2.3) is stable if ∀n ∈ N, T−1n

exists and is continuous in some neighborhood containing y.

Definition 2.4. The numerical method in (2.3) is convergent if xn −→ xas n −→∞.

For a stable numerical method, the solution is given as xn = T−1n y by the

inverse operator of T−1n . If the numerical method is convergent, the numerical

solution will converge to the analytic solution of the mathematical problem.

2

3 The Lax Equivalent Theorem

A general form of Lax equivalent theorem is given in Theorem 3.1.

Theorem 3.1. A consistent method applied to a well-posed problem is con-vergent if and only if it is stable.

By introducing the definitions from Section 2, the Lax equivalent theoremis formulated in Theorem 3.2.

Theorem 3.2. Define a mathematical problem as (2.1) and the numericalapproximation as (2.3). Suppose T−1 is continuous and Tnx −→ Tx for anyx in the domain of T . Then, T−1

n y −→ x as n −→ ∞ if and only if T−1n is

continuous for all n.

Proof. [Engblom, 2014]‘=⇒’By the consistency of Tn, for any given δ, ∃N ∈ N such that, for all n > N

d(Tnx, Tx) < δ. (3.1)

From the assumption in (2.1) and (2.3),

Tx = y = Tnxn, (3.2)

together with the well-posedness of the problem,

d(Tnx, Tnxn) < δ. (3.3)

By the stability of the numerical method, T−1n is continuous at y for all n.

Denote y1 = Tnx, then ∀ε > 0 and for all y1 satisfying (3.3), ∃δ > 0, suchthat

d(T−1n y1, T

−1n y) < ε, (3.4)

which is equivalent tod(x, xn) < ε. (3.5)

Therefore, xn −→ x as n −→∞ and the numerical method is convergent.‘⇐=’If the numerical method in (2.3) is not stable, then ∃ε > 0, for ∀δ > 0,

∀y1 ∈ Y satisfying d(y1, y) < δ, we have

d(T−1n y1, T

−1n y) > ε. (3.6)

From the consistency, there exists an N ∈ N, ∀n > N ,

d(Tnx, Tx) < δ. (3.7)

3

If we take y1 = Tnx, then (3.6) is written as

d(T−1n Tnx, T

−1n Tnxn) > ε, (3.8)

by (3.2)d(x, xn) > ε, (3.9)

and consider the well-posedness of the problem that T−1 exists in some neigh-borhood of y,

d(T−1y, T−1n y) > ε, (3.10)

which indicates that the numerical method is not convergent. Therefore, aconsistent and convergent numerical method is stable.

4 Applications

In a metric space, the metric can be used to define convergence, continuityand compactness. In section 3, the metric form of Lax equivalent theoremis shown in Theorem 3.2. Usually, this theorem is applied to the problemsdefined in normed spaces, or even in Banach spaces, to prove the numericalmethod is convergent by using the stability. There are many linear andnonlinear examples for this forward direction of Lax equivalent theorem, e.g.[Sanz-Serna and Palencia, 1985, Palencia and Sanz-Serna, 1984].

On the other hand, to get stability from convergence is not very practi-cal in some finite difference method since the convergence always relates tothe analytic solution of the problem which is not accessible in many cases.However, this direction of the Lax equivalent theorem can be use to showa consistent numerical method does not converge from the condition that itis unstable. An interesting example is to show that Newton-Cotes formu-las for numerical integration do not converge [Engels, 1980] by knowing thatNewton-Cotes formulas are numerically unstable [Krommer and Ueberhuber,1998].

The importance of the Lax equivalent theorem is that the convergence of anumerical method is difficult to be established while it is usually desired. Theconsistency is straightforward to be verified and the stability which concernsthe ‘local’ properties of numerical method is much easier to show than theconvergence. Moreover, the metric Lax equivalent theorem only requires theoperator for the numerical method to be defined on a metric space which mayprovide a wider use than the classical one for numerical analysis. However,the operators Tn and T are defined on the same metric space which couldcause some troubles for the applications in the future.

4

References

S. Engblom. Notes of lax equivalent theorem, 2014.

H. Engels. Numerical quadrature and cubature. 1980.

B. Gustafsson, H.-O. Kreiss, and J. Oliger. Time-dependent problems anddifference methods, volume 121. John Wiley & Sons, 2013.

J. K. Hunter and B. Nachtergaele. Applied analysis. World Scientific, 2001.

A. R. Krommer and C. W. Ueberhuber. Computational integration, vol-ume 53. Siam, 1998.

P. D. Lax and R. D. Richtmyer. Survey of the stability of linear finite differ-ence equations. Communications on Pure and Applied Mathematics, 9(2):267–293, 1956.

C. Palencia and J. Sanz-Serna. An extension of the lax-richtmyer theory.Numerische Mathematik, 44(2):279–283, 1984.

J. Sanz-Serna and C. Palencia. A general equivalence theorem in the theoryof discretization methods. Mathematics of computation, 45(171):143–152,1985.

5

Fixed-point proof of Lax-Milgram theorem

Fatemeh Ghasemi∗

January 16, 2015

Abstract

Lax-Milgram theorem gives conditions under which a bilinear formcan be inverted to show existence and uniqueness of weak solution of agiven partial differential equation (PDE). There are different ways forproving this theorem. Here we use the well-known Banach fixed-pointtheorem.

1 Introduction

Riesz representation theorem is of the fundamental theorem about Hilbertspace that is the basis for the existence and uniqueness for (symmetric) vari-ational problems. For using this theorem, the variational problems needs tobe symmetric. But a wide variety of variational problems are not symmet-ric and this means that this theorem is not applicable. Fortunately, thereis an analogous result, the Lax-Milgram theorem, that does apply to non-symmetric problems.

The Lax-Milgram lemma has played a critical role over the last half cen-tury by establishing existence and uniqueness of weak solutions of operatorequations Au = f where A is a continuous and coercive linear operator from aHilbert space H into its topological dual H ′. The result has had far-reachingimpact on the analytic and numerical study of elliptic and parabolic partialdifferential equations; boundedness and coercivity of the weak operators areall that need be verified.

While in 1954 the paper (1) of Lax and Milgram provided a nonconstruc-tive proof of the result, constructive proofs appeared over the next several

∗Division of Computational Mathematics, Department of Mathematics, Linkoping Uni-versity, SE-581 83 Linkoping, Sweden . fatemeh.ghasemi@liu.se

1

years. In 1960, Zaran- tanello provided a generalization of Lax-Milgram tostrongly monotone nonlinear operators via Banach fixed points. A similarfixed point proof in the linear case was given by Lions and Stampacchia ina 1967 paper (2) on variational inequalities. In both cases, the mappingshown to be contractive involves applying the Riesz map for H to a residualAv − f ∈ H for some v ∈ H. Also in 1967, Petryshyn gave a constructiveproof of the result based on so-called upper and lower semi-orthogonal basesfor subspaces of H.

In this essay, after introducing some preliminaries in the first section, weprove Lax-Milgram theorem by using Banach fixed-point theorem.

2 Preliminaries

Let H denote a real Hilbert space equipped with inner product (., .)H andassociated norm ‖.‖. H ′

denotes the topological dual with 〈., .〉 the H′ ×H

duality pairing. In this essay we use Banach fixed-point theorem and RieszRepresentation theorem in order to prove Lax-Milgram Theorem. In thefollowing we review these two theorems.

Theorem 2.1. (Banach fixed-point theorem)Let (X, d) be a non-empty complete metric space, and T : X → X a mapsuch that d(Tx, Ty) ≤ kd(x, y) for some real positive k < 1. Then T hasa unique fixed point, i.e. there exists a unique x ∈ X such that Tx = x;furthermore for any initial x0 , the iterates T nx0 converge to the fixed pointx.

Theorem 2.2. (Riesz Representation theorem) For every bounded linearfunctional f ∈ H there exists a unique element u ∈ H such that

f(v) = (u, v), ∀v ∈ H.

3 Lax-Milgram Theorem

Theorem 3.1. Suppose that B : H ×H → R is a bilinear form that is

1. bounded, i.e. there exists an α > 0 such that

|B(u, v)| ≤ α‖u‖‖v‖, ∀u, v ∈ H,

and

2

2. coercive, i.e. there exists a β such that

B(u, u) ≥ β‖u‖2, ∀u ∈ H.

Then for any f ∈ H ′there exists a unique uf ∈ H such that

B(uf , v) = f(v), ∀v ∈ H. (3.1)

Furthermore‖uf‖ ≤ β−1‖f‖H′ , (3.2)

where‖f‖H′ = sup

v∈H,‖v‖=1

|〈f, v〉|.

in particular uf depends continuously on f , i.e.

‖uf − ug‖ ≤ β−1‖f − g‖.

Proof. The uniqueness is straightforward. Suppose there exist two solu-tions u, u then

B(u, v) = B(u, v) = f(v), ∀v ∈ H.Since B is bilinear, we have B(u − u, v) = 0 for all v ∈ H. Let v = u − u.Using the coercivity of B leads to

β‖u− u‖2 ≤ B(u− u, u− u) = 0,

i.e. u = u. Let v = uf in (3.1) to get

B(uf , uf ) = f(uf )

By definition of ‖.‖H′ we know that f(uf ) ≤ ‖f‖H′ . By using this inequalityand coercivity of B, the bound (3.2) is derived. The continuity result followsby considering

B(uf , v)−B(ug, v) = B(uf − ug, v) = (f − g, v),

and setting v = uf − ug. So only existence requires any work.Fix u ∈ H and consider the map v → B(u, v). We claim that this defines abounded linear functional on H. Since B is bilinear and bounded, it is clearlylinear and bounded. It follows from the Riesz Representation Theorem thatthere exists a w ∈ H such that

(w, u) = B(u, v), ∀v ∈ H.

3

We define Au = w. By this definition we have

(Au, u) = B(u, v), ∀v ∈ H.and claim that this definition yields a bounded linear operator from H intoitself.Indeed, for every v ∈ H

(A(α1u1 + α2u2), v) =B(α1u1 + α2u2, v)

=α1B(u1, v) + α2B(u2, v)

=α1(Au1, v) + α2(Au2, v)

=(α1Au1 + α2Au2, v)

Since this holds for every v ∈ H, it follows that

A(α1u1 + α2u2) = α1Au1 + α2Au2,

and this implies that A is linear. To show that A is bounded, note that

‖Au‖2 = (Au,Au) = B(u,Au) ≤ α‖u‖‖Au‖,hence ‖Au‖ ≤ α‖u‖ and A is bounded.Using the Riesz Representation, we know that there exists a ϕ ∈ H such that

(ϕ, v) = f(v), ∀v ∈ H.We can rewrite our equation in the following way

(Au, v) = (ϕ, v), ∀v ∈ H.This means that u satisfies (3.1) if and only if Au = ϕ. Now we have toshow that this equation has a solution. In this step of proof we use Banachfixed-point theorem.Clearly, for any λ > 0

Au = ϕ⇔ u = u− (Au− ϕ).

Let Tu = u − λ(Au − ϕ). So we are investigating a fixed point for T. SinceH is a Hilbert space, it is enough to show that T is a contraction. We have

‖Tu− Tv‖2 =‖(u− v)− λA(u− v)‖2=‖u− v‖2 − 2λ(A(u− v), u− v) + λ2‖A(u− v)‖2=‖u− v‖2 − 2λB(u− v, u− v) + λ2‖A(u− v)‖2≤‖u− v‖2 − 2λβ‖u− v‖2 + λ2α2‖u− v‖2=(1− 2λβ + λ2α2)‖u− v‖2,

where we have used the coercivity of B and the fact that A is bounded. Itfollows that if we choose sufficiently small λ, T is a contraction. It thereforehas a unique fixed point, which provides our solution.

4

References

[1] P. D. Lax and A. N. Milgram, Parabolic equations , in Contributionsto the theory of partial differential equations, Annals of MathematicsStudies, no. 33, Princeton University Press, Princeton, N. J., 1954, pp.167-190.

[2] J.-L. Lions and G. Stampacchia, Variational inequalities ,Comm. PureAppl. Math. Comm. Pure Appl. Math., 20 (1967), pp. 493-519.

5

Part II

Solutions to exercises

87

Chapter 1

Metric spaces

88

Ex. 1.2.4

Consider the sequence

ξn =1

log (n), n ≥ 2.

Clearly this sequence converges to zero. However, log (n) grows slower than any polynomialsequence so for each 1 ≤ p <∞ there exists an Np ∈ N such that ξpn ≥ 1/n for all n ≥ Np.Therefore, since the harmonic series diverges, the sequence ξn is not in any space lp.

Ex. 1.3.8

Show that the closure B(x0; r) of an open all B(x0; r) in a metric space can differ from theclosed ball B(x0; r).

Solution: Let x0 = (0, 0) and r = 1. Define a metric space (X, d), where X = (x1, x2) |x2

1 + x22 ≤ 1, x1 and x2 are real and nonnegative ∪ (x1, x2) = (cos(1.25π), sin(1.25π)),

and d is the Euclidean distance. The points in set X are shown in Figure 1.1. According

−1.5 −1 −0.5 0 0.5 1 1.5−1.5

−1

−0.5

0

0.5

1

1.5

Figure 1.1: Points in X

to the definitions,

B(x0; r) = (x1, x2) ∈ X | x21 + x2

2 < 1,B(x0; r) = x ∈ X | d(x, x0) ≤ 1 = X,

B(x0; r) = (x1, x2) ∈ X | (x1, x2) 6= (cos(1.25π), sin(1.25π)).

89

Therefore, B(x0; r) can differ from B(x0; r).

Ex. 1.3.8

Let X be a set with at least two elements and let the metric space (X, d) have the discretemetric

d(x, y) =

0 x = y

1 x 6= y.

Let B(x0, 1) be the closed ball around x0 ∈ X

B(x0, 1) = x ∈ X : d(x, x0) ≤ 1.

By the definition of the metric this means

B(x0, 1) = X.

Let now B(x0, 1) = x ∈ X : d(x, x0) < 1 be the open ball around x0. Which means

B(x0, 1) = x0.

Let now x ∈ X be an accumulation point of B(x0, 1). This means that

d(x, x0) < ε, ∀ε > 0

which implies x = x0 thus the closure of the open ball is

B(x0, 1) = x0,

which is not equal to the closed ball.

Ex. 1.3.8

Consider the discrete metric on R2, X = (R2, d) where B(x0; 1) = X and B(x0; 1) =x0, clearly differing.

Ex. 1.3.12

Let us consider the family of characteristic functions

X =x (t) = χ(a,c) (t) , c ∈ (a, b)

90

that is non–countable because (a, b) is a non–countable set. Notice that

d (x, y) = 1, x 6= y, x, y ∈ X.

Now, let us assume that there exists a dense set in B [a, b], call it G: then there exist asmany non–empty balls B

(x0,

12

), x0 ∈ X as the elements of X. On the other hand, we

know that

B

(x0,

1

2

)∩B

(x

′0,

1

2

)= ∅, x0 6= x

′0

and therefore G is non–countable as well. Thus, it does not exist a countable dense set inB [a, b], since we have proved that every dense set in X is non–countable.

Ex. 1.3.12

Consider the subset M ⊂ B[a, b] consisting of the functions δx = 1 at x and zero everywhereelse on [a, b]. Then M is uncountable, and

sup |δx − δy| = 1

whenever x 6= y. Consider now all open balls B(δx; 1/2). These are disjoint, and so ifa subset S ⊂ B[a, b] is dense, then it must intersect each ball in a different point. Thisrenders S uncountable. Since S is general this shows that B[a, b] is not separable.

Ex. 1.4.2

If xnk→ x, then for every ε > 0 there is an N1 = N1 (ε) such that

d (xnk, x) <

ε

2∀nk > N1

and, since xn is Cauchy we know that there is an N2 = N2 (ε) such that

d (xn, xm) <ε

2∀n,m > N2.

Hence by the triangular inequality we obtain for n, nk > max N1, N2

d (xn, x) ≤ d (xn, xnk) + d (xnk

, x) <ε

2+ε

2= ε

and the claim follows.

91

Ex. 1.4.2

For the subsequence we haved(xnk

, x) −→ 0,

and since (xn) is Cauchy,d(xn, xnk

) −→ 0.

Using the triangle inequality we get

d(xn, x) ≤ d(xn, xnk) + d(xnk

, x) −→ 0

which proves that (xn) is convergent with the limit x.

Ex. 1.4.8

If d1 and d2 are metrics on the same set X and there are positive numbers a and b suchthat for all x, y ∈ X,

ad1(x, y) ≤ d2(x, y) ≤ bd1(x, y), (1.0.1)

show that the Cauchy sequences in (X, d1) and (X, d2) are the same.

Solution: We use the relation (1.0.1) to show

(xn) is Cauchy in (X, d1)⇐⇒ (xn) is Cauchy in (X, d2) .

First assume that (xn) is Cauchy in (X, d1). That is, for any ε > 0 there is an N such that

d1(xn, xm) <ε

b∀m,n > N.

Using (1.0.1) we obtain

d2(xn, xm) ≤ bd1(xn, xm) < ε ∀m,n > N.

which proves that (xn) is Cauchy also in (X, d2).

Now assume that (xn) is Cauchy in (X, d2). That is, for any ε > 0 there is an N such that

d2(xn, xm) < aε ∀m,n > N.

Using (1.0.1) we obtain

d1(xn, xm) ≤ 1

ad2(xn, xm) < ε ∀m,n > N.

which proves that (xn) is Cauchy also in (X, d1). Thus, (xn) is Cauchy in (X, d1) iff (xn)is Cauchy in (X, d2), i.e., the Cauchy sequences in (X, d1) and (X, d2) are the same.

92

Ex. 1.5.6

Show that the set of all real numbers constitutes an incomplete metric space if we choosed(x, y) = |arctanx− arctan y|.

Solution: Let X = R. We show that (X, d) is incomplete by constructing a Cauchysequence in X which does not converge.

Consider the sequence (xn), where xn = n. Clearly, (xn) does not converge. Since

limx→∞

arctanx =π

2,

for any ε > 0 we can find N = N(ε) such that∣∣∣arctann− π

2

∣∣∣ < ε

2∀n > N.

Hence,

d(xn, xm) = |arctanxn − arctanxm| =∣∣∣arctann− π

2− arctanm+

π

2

∣∣∣

≤∣∣∣arctann− π

2

∣∣∣+∣∣∣arctanm− π

2

∣∣∣ < ε

2+ε

2= ε ∀n,m > N

which shows that (xn) is Cauchy.

Ex. 1.5.8

Let y1, y2 ∈ Y . Since Y is a subspace it is enough to show that Y is closed. Let yn∞n=1

be a sequence in Y with yn → y ∈ C[a, b], this means

∀ε > 0∃N : supt∈[a,b]

|y(t)− yn(t)| < ε|.

But by definition|y(a)− yn(a)| ≤ sup

t∈[a,b]

|y(t)− yn(t)|

so we getyn(a)→ y(a),

and by the same argument alsoyn(b)→ y(b).

But since each yn ∈ Y we have yn(a) = yn(b) so we get

|y(a)− y(b)| ≤ |y(a)− yn(a)|+ |y(b)− yn(b)| → 0

which shows that y(a) = y(b). This means that y ∈ Y and thus is a closed subspace andcomplete.

93

Ex. 1.5.9

Assume: that (xm) is the sequence of continuous functions on [a, b] that uniformly con-verges on [a, b];Prove: that the limit function x is continuous on [a, b].

Solution: Since (xm(t)) is converging to x,

∀ε > 0 ∃N(ε), ∀t ∈ [a, b], |xm(t)− x(t)| < ε

3(1.0.2)

On the other hand, (xm(t)) is continuous at any to ∈ [a, b]; so, we have,

∀ε > 0 ∃δ > 0, |t− to| < δ ⇒ |xm(t)− xm(to)| <ε

3(1.0.3)

Now, we want to show that x(t) is continuous on [a, b]; therefore,

∀ε > 0 ∃δ1 > 0, |t− to| < δ1 ⇒ |x(t)− x(to)| < ε (1.0.4)

For this purpose, we can find the following relation, using (1) and (2):

|x(t)− x(to)| = |x(t)± xm(t)± xm(to)| == |(x(t)− xm(t)) + (xm(t)− xm(to)) + (xm(to)− x(to))|≤ |x(t)− xm(t)|+ |xm(t)− xm(to)|+ |xm(to)− x(to)|≤ ε

3+ε

3+ε

3= ε

which is equivalent to (3).

Ex. 1.6.14

The only question here is whether d(x, y) = 0⇒ x = y or not.i) Let d(x, y) = 0 and let h(t) = x(t)− y(t) which implies

∫ b

a

|h(t)|dt = 0.

Assume that there exist t0 ∈ [a, b] such that |h(t0)| 6= 0. Since x and y are continuous h iscontinuous. So let 0 < ε = |h(t0)| and let δ be so small that if

|t− t0| < δ

we have|h(t)− h(t0)| < ε.

94

The reverse triangle inequality gives us:

|h(t0)| − |h(t)| ≤ |h(t)− h(t0)| < ε.

Since ε = |h(t0)| this now implies

|h(t)| > 0, ∀t ∈ [t0 − δ, t0 + δ].

But since: ∫ b

a

|h(t)|dt ≥∫ t0+δ

t0−δ|h(t)|dt

this implies ∫ b

a

|h(t)|dt > 0

which is a contradiction. So no t0 with |h(t0)| 6= 0 can exist, thus x(t) = y(t).ii) For Riemann-integrable functions d(x, y) = 0 does not imply x = y. An example is:

let tm = (a+ b)/2 and

x(t) =

0 t < tm

1 t ≥ tm

y(t) =

0 t ≤ tm

1 t > tm

We have x(tm) 6= y(tm) but d(x, y) = 0.

95

Chapter 2

Normed spaces

96

Ex. 2.3.10

Suppose that the Banach space (X, ‖ · ‖) has a Schauder basis en∞n=1 so that for anyx ∈ X there is a unique sequence of scalars, α1, α2, . . . such that

x =∞∑

k=1

αkek.

We will for simplicity assume that αk ∈ R, though generalisation to complex numbers isstraight forward.

Given any ε > 0 there is an N such that

‖x−N∑

k=1

αkek‖ <ε

2.

Now, recall that Q is dense in R, meaning that for any αk ∈ R we can find a βk ∈ Q suchthat |αk − βk| < ε/2k+1. By the triangle inequality we have

‖x−N∑

k=1

βkek‖ ≤ ‖x−N∑

k=1

αkek‖+ ‖N∑

k=1

αkek −N∑

k=1

βkek‖ <ε

2+

N∑

k=1

ε

2k+1< ε.

Thus the Schauder basis constitutes a basis for a dense subset of X, which is countablesince βk ∈ Q for all k. Consequently, X is separable.

Ex. 2.3.10

Let Y be the set,

Y = N∑

j=1

αjej : αj ∈ R, N ∈ N+.

Let x ∈ X, since we have a Schauder-basis there exist αj∞j=1 such that

‖x− xn‖ → 0

where

xn :=n∑

j=0

αjej.

Thus for every x ∈ X we have a sequence in Y converging to x. This shows that Y = X.

97

Let now instead xn =∑n

j=1 αjej ∈ Y be an arbitrary element in Y and let M be theset

M = N∑

j=1

βjej : bj ∈ Q, N ∈ N+,

which is countable. Since Q is dense in R there exist a sequence βkj ∞k=1 with βkj ∈ Q suchthat

|αj − βkj | ≤ε

Cn, ∀k ≥ K,

whereC = max

j∈1,...,n]‖ej‖ .

Let xkn =∑n

j=1 βkj ej ∈M , we get:

∥∥xn − xkn∥∥ =

∥∥∥∥∥n∑

j=1

(αj − βkj )ej

∥∥∥∥∥ ≤n∑

j=1

|αj − βkj | ‖ej‖ ,

so for k ≥ K we get∥∥xn − xkn

∥∥ ≤n∑

j=1

ε

Cn‖ej‖ ≤

ε

n

n∑

j=1

1 ≤ ε.

Showing that xkn → xn when k →∞. Thus for every xn ∈ Y we can find a sequence in Mconverging to xn. So M = Y and in total

X = Y = (M) = M

showing that M is dense in X so that X is separable.

Ex. 2.4.4

Show that equivalent norms on a vector space X induce the same topology for X.Solution: Assume that ‖·‖1 and ‖·‖2 are equivalent norms on X. Let T1 be the topology

for X induced by ‖·‖1 and let T2 be the topology for X induced by ‖·‖2. In other words, T1

and T2 are the collections of all open subsets of X in the sense of ‖·‖1 and ‖·‖2, respectively.We show that T1 = T2 by showing that any subset of X is open in the sense of ‖·‖1 iff itis open in the sense of ‖·‖2.

Since ‖·‖1 and ‖·‖2 are equivalent, we have

a ‖·‖1 ≤ ‖·‖2 ≤ b ‖·‖1 (2.0.1)

for some a, b > 0. Let Bi(x; r) denote the open ball of radius r around x in the sense of‖·‖i. Now assume that M is an open subset of X in the sense of ‖·‖1. By the definition ofan open set, M contains an open ball around each of its points, i.e., for any x ∈M ∃ε > 0

98

such that B1(x; ε) ⊂ M . Because of (2.0.1) we have B2(x; aε) ⊂ B1(x; ε), which meansthat M contains an open ball in the sense of ‖·‖2 around each of its points. This provesthat any M ⊂ X is open in the sense of ‖·‖2 if it is open in the sense of ‖·‖1. By reversingthe above argument, we can show that any M ⊂ X is open in the sense of ‖·‖1 if it is openin the sense of ‖·‖2.

Ex. 2.4.4

Suppose that the norms ‖‖0 and ‖‖1 are equivalent. Then there are constants c1 and c2

with 0 ≤ c1 ≤ c2 such that c1 ‖ x ‖0≤‖ x ‖1≤ c2 ‖ x ‖0 for all x ∈ X. Let U be a subset ofX which is open with respect to the topology on X induced by the norm ‖‖1. For u ∈ Uthere is an ε such that x ∈ X ‖ u − x ‖1< ε ⊂ U . But since the norms are equivalentx ∈ X ‖ u − x ‖1< ε ⊂ x ∈ X ‖ u − x ‖0< ε/c2. This means that U is open withrespect to the topology induced by norm ‖‖0. By the same argument, we can show thatany open subset of X with respect to the topology induced by the norm ‖‖0 is open withrespect to the topology induced by ‖‖1 .

Ex. 2.5.4

Show that for an infinite subset M in the space s to be compact, it is necessary that thereare numbers γ1, γ2, . . . such that for all x = (ξk(x)) ∈M we have |ξk(x)| ≤ γk.

Solution: The metric on the space s is

d(x, y) =∞∑

k=1

1

2k|ξk − ηk|

1 + |ξk − ηk|

where x = (ξj), y = (ηj). Assume that not all (ξk) in M are bounded by |ξk| ≤ γk. Thenthere is at least one k0 such that |ξk0| is unbounded. Now consider a sequence of elements

in M , x1, x2, . . ., where xm =(ξ

(m)k

). We choose the sequence such that ξ

(m)k0

= m. We

then have, for m 6= n,

|xm − xn| =∞∑

k=1

1

2k|ξ(m)k − ξ(n)

k |1 + |ξ(m)

k − ξ(n)k |≥ 1

2k0|m− n|

1 + |m− n| ≥1

2k01

2.

Clearly (xm) can not have a convergent subsequence. Thus, M is not compact.

Ex. 2.5.10

Let X and Y be metric spaces, X compact, and T : X → Y bijective and continuous.Show that T is a homeomorphism.

99

In the proof, I use the result from Prob. 14 in Section 1.3: a mapping T : X → Y iscontinuous if and only if the inverse image of any closed set M ⊂ Y is a closed set in X.This result is equivalent to a mapping T−1 : Y → X is continuous if and only if for anyclosed set M ⊂ X, T (M) is a closed set in Y . (*)

Let M be a closed subset in X. Since X is a compact metric space, M is compact(Prob. 2.5-9). Since T is continuous, according to Theorem 2.5-6 we have that T (M) iscompact. By using Lemma 2.5-2, we have that T (M) is closed and bounded. Since Mis arbitrary and T is bijective, by (*) we have that T−1 is continuous. Therefore, T is ahomeomorphism.

Ex. 2.5.10

Let X and Y be metric spaces, X be compact, T : X → Y bijective and continuous andF = T−1.

• Let V ⊆ X and assume that V is closed and bounded so that it is compact.

• Since T is continuous, it follows that T (V ) is compact.

• This implies that T (V ) is closed and bounded.

• But T (V ) = F−1(V ) so F−1(V ) is closed and bounded.

Hence F is continuous, and thus T is a homeomorphism.

Ex. 2.5.10

We study the continuous and bijective operator T : X → Y where X is compact. Bytheorem 2.5-6, Y is compact. Lets assume that T−1 is discontinuous, that is there is ay ∈ Y and x ∈ X such that

dy(yn, y)→ 0 (2.0.2)

dx(xn, x) > δ > 0 (2.0.3)

where the sequences xn, yn satisfy Txn = yn and Tx = y. However, X is compact, so thereis a subsequence of xn such that xk → x0 6= x. We have

dy(Tx0, Tx) ≤ dy(Txk, Tx) + dy(Tx0, Txk) = dy(yk, y) + dy(y0, yk).

Since T is continuous, we have dy(y0, yk) = dy(Tx0, Txk) → 0 as k → ∞. Therefore, wehave

dy(Tx0, Tx) ≤ dy(yk, y) + dy(y0, yk)→ 0

100

as k →∞. However, according to (8) we should have for bijective T dy(Tx, Tx0) > ε > 0,which is a contradiction. T−1 must therefore be continuous and T is then a homeomor-phism.

Ex. 2.5.10

A Homeomorphism is a continuous bijective mapping whose inverse is continuous. Sincewe have a continuous and bijective mapping T : X −→ Y (X and Y metric spaces, Xcompact), we need to show that the inverse T−1 is continuous.

Generally, a mapping T : X −→ Y is continuous if and only if, for any closed setK ⊂ Y , the inverse image T−1(K) is a closed set in X (cf. Prob. 14, Sec. 1.3).So here, to show that T−1 : Y −→ X is continuous, we want to show that T (M) is aclosed set in Y , for any closed set M ⊂ X.

1. Let M be a closed subset in X-compact, then from Prob. 9, Sec. 2.5, M is alsocompact.

2. Since T is continuous, and M is compact ⇒ T (M) is compact(Theorem 2.5-6)

3. Since T (M) is a compact subspace of the metric space Y , T (M) is closed and bounded(Lemma 2.5-2)

Ex. 2.7.2

• Consider T bounded. If there exists M > 0 such that ‖x‖ ≤M , then

‖Tx‖ ≤ c ‖x‖ = cM

and therefore T maps bounded sets in X into bounded sets in Y .

• Let us consider T which maps bounded sets in X into bounded sets in Y . If x = 0,then T is trivially bounded since

T (0) = T (0Rz) = 0RT (z) = 0 ∀z ∈ X ⇒ ‖T0‖ = 0.

Let us consider x 6= 0. We know that exists C > 0 such that ‖Tx‖ ≤ C, ∀x 6= 0 suchthat ‖x‖ = 1. Therefore

∥∥∥∥T(

x

‖x‖

)∥∥∥∥ ≤ C

and for the homogeneity of the norm we have ‖Tx‖ ≤ C ‖x‖.

101

Ex. 2.7.2

Let X and Y be normed spaces. A linear operator T : X → Y is bounded iff T mapsbounded sets in X into bounded sets in Y .

If T is bounded:We want to show that for any bounded set M ⊂ D(T ) s.t. for any x, y ∈ M we havethat ‖x − y‖ ≤ r < ∞ than also ‖Tx − Ty‖ < ∞. Since D(T ) is a vector space thenx, y ∈ M ⊂ D(T ) → x − y = z ∈ D(T ). Furthermore ‖Tz‖ ≤ c‖z‖ since T is bounded.Hence ‖Tx− Ty‖ = ‖T (x− y)‖ = ‖Tz‖ ≤ c‖z‖ <∞.

If T maps bounded sets in X into bounded sets in Y :We know that for any x, y ∈M bounded than ‖Tx− Ty‖ <∞. Then we call

δ(M) = supx,y∈M

‖x− y‖ and δ(T (M)) = supTx,Ty∈T (M)

‖Tx− Ty‖.

For any zD(T ), z 6= 0 we define

x = y +δ(M)

‖z‖ z ⇒ ‖x− y‖ = δ(M).

Then

‖Tx− Ty‖ = ‖T (x− y)‖ = ‖T(δ(M)

‖z‖ z)‖ =

δ(M)

‖z‖ ‖Tz‖

which means that

‖Tz‖ = ‖Tx− Ty‖ ‖z‖δ(M)

≤ δ(T (M))

δ(M)‖z‖ = c‖z‖.

where c = δ(T (M))δ(M)

.

Ex. 2.7.2

A linear operator T : D(T ) −→ Y , (where D(T ) ⊂ X and X and Y are normed spaces),is bounded if there is a real number c such that for all x ∈ D(T ),

‖Tx‖ ≤ c‖x‖. (2.0.4)

Further, a subset K in a normed space X is bounded if and only if there is a positivenumber a such that

‖x‖ ≤ a (2.0.5)

for every x ∈ K.Let M be a bounded set in X.

102

1. Show that if T is bounded ⇒ T (M) is a bounded set in Y :

Let x ∈M and Tx ∈ T (M), then from (2.0.4) and (2.0.5) we have

‖Tx‖ ≤ c‖x‖ ≤ ca (2.0.6)

where ca is a positive number. Since x ∈ M is arbitrary this proves T (M) is abounded set in Y (according to the definition (2.0.5) of a bounded subset).

2. Show that if T (M) is a bounded set in Y ⇒ T is bounded:

Let x ∈M and Tx ∈ T (M), then using the definition for the operator norm we have

‖T‖‖x‖ = supx∈M,x6=0

‖Tx‖‖x‖ ‖x‖ ≥

‖Tx‖‖x‖ ‖x‖ = ‖Tx‖, x 6= 0. (2.0.7)

Since ‖T‖ is a real number, call it c, we have

‖Tx‖ ≤ c‖x‖, x 6= 0. (2.0.8)

For the case when x = 0, since for linear operators T0 = 0, we have

‖Tx‖ = ‖T0‖ = 0 ≤ c‖x‖, x = 0. (2.0.9)

Since inequality (2.0.4) holds for all x ∈M , T is bounded.

Ex. 2.7.7

Given: T : X → Y is bounded and linear.Furthermore,

∃b > 0 3 ∀x ∈ X, ‖Tx‖ ≥ b‖x‖ ⇒ ‖x‖ ≤ ‖Tx‖b

(2.0.10)

We want to show that T−1 : Y → X exists and is bounded. According to [Theorem2.6-10(a)], T−1 exists if and only if Tx = 0 implies x = 0. In our case,

Tx = 0⇒ 0 ≥ b‖x‖Since b > 0 the only way for the latter inequality to be valid is that, x = 0. Hence, T−1

exists. Also, based on the definition of mapping we have,

∀x ∈ X ∃y ∈ Y 3 y = Tx ∧ x = T−1y (2.0.11)

⇒ ‖T−1y‖ = ‖x‖ ≤(1) ‖Tx‖b

=‖y‖b

(2.0.12)

⇒ ‖T−1y‖ ≤ 1

b‖y‖ (2.0.13)

therefore, according to [Definition 2.7-1], T−1 is bounded.

103

Ex. 2.7.8

If we consider T : l∞ → l∞, x 7→ y defined by

x = αj , y = βj , s.t. βj =αjj

the bounded set y = 1 ⊂ R (T ) is mapped by T−1 into x = jj∈N, that is not bounded.

Ex. 2.7.8

Let X be a normed space with points x = ξk being sequences of real or complex numbersthat has only finitely many non-zero terms. Further let ‖x‖ = supk |ξk|. Now define theoperator T : X → X as

Tx = (ξ1, ξ2/2, . . . ) = ξk/k.If y = ηk and α and β are numbers then

T (αx+ βy) = (αξk + βηk)/k = αξk+ βηk = αT (x) + βT (y)

so T is a linear operator. Also note that

‖Tx‖ = supk|ξk|/k ≤ ‖x‖ sup

k1/k ≤ ‖x‖

so T is bounded. Now, note that T−1x = kξk. Consider the sequence x = δk,n ∈ X.Then ‖x‖ = 1 and ‖T−1x‖ = n. Since n is an arbitrary integer and x and infinite sequence,we may let n→∞, which shows that T−1 is not bounded.

Ex. 2.8.6

The space C1[a, b] is the normed space of all continuously differentiable functions on J =[a, b] with norm defined by

‖x‖ = maxt∈J|x(t)|+ max

t∈J|x′(t)|. (2.0.14)

(N1) Since |x(t)| ≥ 0 and |x′(t)| ≥ 0, ‖x‖ ≥ 0.

(N2) ”⇒”: ‖x‖ = 0⇔ maxt∈J|x(t)| = −max

t∈J|x′(t)|⇔ x(t) = 0 and x′(t) = 0.

”⇐”: x = 0⇒ ‖x‖ = maxt∈J|0|+ max

t∈J|0| = 0

(N3) ‖αx‖ = maxt∈J|αx(t)|+ max

t∈J|αx′(t)| = |α|max

t∈J|x(t)|+ |α|max

t∈J|x′(t)| = |α‖|x‖

(N4) ‖x+ y‖ = differentiation is linear = maxt∈J|x(t) + y(t)|+ max

t∈J|x′(t) + y′(t)|

≤ maxt∈J|x(t)|+ max

t∈J|y(t)|+ max

t∈J|x′(t)|+ max

t∈J|y′(t)| = ‖x‖+ ‖y‖.

104

Define f(x) = x′(c), c = (a+ b)/2. This functional on C1[a, b] is bounded since

‖x‖ = maxt∈J|x(t)|+ max

t∈J|x′(t)| ≥

∣∣∣x(a+ b

2

)∣∣∣+∣∣∣x′(a+ b

2

)∣∣∣ ≥∣∣∣x′(a+ b

2

)∣∣∣⇔

|f(x)| ≤ ‖x‖. (2.0.15)

Further, f is linear since

f(αx+ βy) = differentiation is linear = αx′(a+ b

2

)+ βx′

(a+ b

2

)

= αf(x) + βf(y). (2.0.16)

If f is now considered as a functional on the subspace of C[a, b] which consists of allcontinuously differentiable functions, we want to show that f is not bounded. We considerthe sequence of continuous and differentiable functions (xε) = arctan( t

ε), where ε is a real

number. If we let ε −→ 0, then for c = 0,

limε−→0

x′ε(c) =∞,

see Figure 2.1.So, given a constant C, there exists an ε such that

|f(xε)| > C‖xε‖,

which proves that f is not bounded (since ‖xε‖ = maxt∈J| arctan( t

ε)| ≤ π

2independent of

ε).

Ex. 2.8.6

The space C1[a, b] is the normed space of all continuously differentiable functions on J =[a, b] with norm defined by

‖x‖ = maxt∈J|x(t)|+ max

t∈J|x′(t)|. (2.0.17)

Show that the axioms of a norm are satisfied. Show that f(x) = x′(c), c = (a + b)/2,defines a bounded linear functional on C1[a, b]. Show that f is not bounded, consideredas a functional on the subspace of C[a, b] which consists of all continuously differentiablefunctions.

Solution: First, we show that the four axioms of a norm are satisfied in this problem.(N1). By the definition of maximum and absolute function, ‖x‖ ≥ 0.(N2). If ‖x‖ = 0, as both max

t∈J|x(t)| and max

t∈J|x′(t)| are nonnegative, |x(t)| = |x′(t)| = 0.

If x(t) = 0 is a constant function, then |x′(t)| = 0. Therefore, ‖x‖ = 0.

105

−100 −80 −60 −40 −20 0 20 40 60 80 100−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

arct

an(x

/eps

)

Figure 2.1: The function arctan(xε ). Smaller and smaller values of ε leads to a steeper andsteeper function.

(N3).

‖αx‖ = maxt∈J|αx(t)|+ max

t∈J|αx′(t)|

= |α|(maxt∈J|x(t)|+ max

t∈J|x′(t)|)

= |α|‖x‖. (2.0.18)

106

(N4). The triangle inequality is satisfied

‖x+ y‖ = maxt∈J|x(t) + y(t)|+ max

t∈J|x′(t) + y′(t)|.

≤ maxt∈J|x(t)|+ max

t∈J|y(t)|+ max

t∈J|x′(t)|+ max

t∈J|y′(t)|.

= ‖x‖+ ‖y‖. (2.0.19)

Then, we show that functional f(x) is linear and bounded on C1[a, b]. As the spaceC1[a, b] is a normed space, it defines the algebraic operations. For x, y ∈ C1[a, b] and anyscalar α, the operation f satisfies

f(x+ y) = (x(c) + y(c))′ = f(x) + f(y), (2.0.20)

andf(αx) = (x(c) + y(c))′ = f(x) + f(y). (2.0.21)

This shows that f is linear. Consider the norm in C1[a, b],

|f(x)| = |x′(c)| ≤ maxt∈J|x′(t)| ≤ max

t∈J|x(t)|+ max

t∈J|x′(t)| = ‖x‖, (2.0.22)

followeding the definition in 2.8-2, f is bounded on C1[a, b].However, on the space C[a, b], where the norm is defined by

‖x‖ = maxt∈J|x(t)|, (2.0.23)

let xn(t) = sin(ntπ/c), where n ∈ N. Then ‖xn‖ ≤ 1 and

f(xn) = x′n(c) =nπ

ccos(nπ). (2.0.24)

So ‖f(x)‖ = nπ/c. As n ∈ N is arbitrary, this shows that there is no fixed number α suchthat ‖f(x)‖ ≤ α‖x‖, so f is unbounded.

Ex. 2.10.4

Let X and Y be normed spaces and Tn : X → Y , n = 1, 2, 3, .., be bounded linear operators.Show that convergence Tn → T implies that for every ε > 0 there is an N such that for alln > N and all x in any given closed ball we have ‖Tnx− Tx‖ < ε.

Solution: Denote the given closed ball by B(x0; r) ⊂ X. Then, for all x ∈ B(x0; r),there exists a positive number C such that ‖x‖ ≤ C. From the convergence of Tn, for agiven ε > 0, there exists N ∈ N such that for all n > N ,

‖Tn − T‖ ≤ε

C. (2.0.25)

Therefore, for all x ∈ B(x0; r)

‖Tnx− Tx‖ ≤ ‖Tn − T‖‖x‖ ≤ ε. (2.0.26)

107

Ex. 2.10.8

Since ek∞k=1 with ek = (δkj)∞j=1 is a Schauder basis for c0, for every x ∈ c0 we can write

x =∞∑

k=1

λkek,

where x = (λk) and limn→∞ λk = 0 and we can approximate x by

sn =n∑

k=1

λkek,

then limn→∞ ‖ sn − x ‖∞= 0. Let f ∈ c′0, which is the dual space of c0. This means thatf : c0 → R is a bounded linear functional, so f(sn)→ f(x), as n→∞. So

f(x) =∞∑

k=1

λkf(ek).

Now we show that f(ek) is in l1. Define the sequence γn = f(en)|f(en)| for f(en) 6= 0 and γn = 0

for f(en) = 0. Consider the sequence xn = (γ1, γ2, · · · , γn, 0, 0, · · · ) = γ1e1 + · · ·+ γnen. Itis obvious that ‖ xn ‖∞≤ 1, so

|f(xn)| =n∑

k=1

|f(ek)| ≤‖ xn ‖∞‖ f ‖≤‖ f ‖ .

So∑n

k=1 |f(ek)| ≤‖ f ‖ for all n. Since this inequality holds for all n, we have

∞∑

k=1

|f(ek)| ≤‖ f ‖ . (2.0.27)

This means that (f(ek)) ∈ l1. Now we want to show that for every y = (λk) ∈ l1, we canobtain a corresponding bounded linear operator g : c0 → R. Let

g(x) =∞∑

k=1

λkµk, x = (µk).

Clearly g is linear and

|g(x)| = |∞∑

k=1

λkµk| ≤ supj|µj|

∞∑

k=1

|λk| ≤‖ x ‖∞∞∑

k=1

|λk|.

So g is bounded. Taking the supremum over all x with norm 1, we get

|g(x)| ≤∞∑

k=1

|λk|.

108

This inequality and the inequality (2.0.27) gives

∞∑

k=1

|f(ek)| =‖ f ‖ .

So the mapping f : c′0 → l1 with f → (f(ek)) is bijective and preserves norms and henceisomorphism.

Ex. 2.10.8

We consider f a bounded linear functional on c0. Since a Schauder basis for c0 is ekk∈N,with ek = δkj, then every x = αk ∈ c0 has a unique representation

x =∞∑

k=1

αkek

and, from the linearity of f , we know also that

f (x) =∞∑

k=1

αkf (ek) .

We can write

|f (x)| =

∣∣∣∣∣∞∑

k=1

αkf (ek)

∣∣∣∣∣ ≤∞∑

k=1

|αk| |f (ek)| ≤ ‖x‖∞∞∑

k=1

|f (ek)| = c ‖x‖∞ ,

where c = |f (ek)| < ∞ because f (x) is bounded. This implies that f (ek) ⊂ l1 andtherefore c

′0 ⊆ l1.

On the other hand, for every b = βkk∈N ∈ l1, we can obtain a bounded linear functionalg in c0. In fact, we may define g on c0 by

f (x) =∞∑

k=1

αkβk.

Then g is

• Linear

g (λx+ µy) =∞∑

k=1

(λαk + µγk) βk =

= λ

∞∑

k=1

αkβk + µ

∞∑

k=1

γkβk = λg (x) + µg (y) , λ, µ ∈ R

where y = γkk∈N ∈ c0.

109

• Bounded

|g (x)| =∣∣∣∣∣∞∑

k=1

αkβk

∣∣∣∣∣ ≤∞∑

k=1

|αkβk| ≤ ‖x‖∞∞∑

k=1

βk = ‖x‖∞ ‖b‖1 .

Then g ∈ c′0 and therefore l1 ⊆ c′0.

We finally prove that ‖g‖ = ‖b‖1. Indeed, since∑∞

k=1 |βk| converges, then

∀ε > 0, ∃N = N (ε) s.t.

∞∑

k=n+1

|βk| < ε, ∀n > N.

Let us consider x s.t.

αk =

|βk|βk, k < N,

0, k ≥ N.

x ∈ c0, and

|g (x)− ‖b‖1| =∣∣∣∣∣∞∑

k=1

αkβk −∞∑

k=1

|βk|∣∣∣∣∣ =

∣∣∣∣∣∞∑

k=N+1

|βk|∣∣∣∣∣ < ε.

This shows that the bijective linear mapping on l1 onto c′0 defined by g is an isomorphism.

Ex. 2.10.8

Show that the dual space of the space c0 is l1.c0 is the space of all sequences of scalars converging to zero. The norm on c0 is ‖ · ‖∞.

Let c′0 be the dual space of c0. We want to show that c

′0 is isomorphic with l1. The norm

on l1 is ‖ · ‖1. We prove it step by step.

Step 1: Let x ∈ c0, then x has a unique representation

x =∞∑

k=1

ξkek,

where (ek) = (δkj) is a Schauder basis for c0, (δkj) has one in the kth place and zerosotherwise, and ξk → 0 as k → ∞. Since f is linear, we have f(x) =

∑∞k=1 ξkf(ek).

Since f is bounded on c0, we have

‖f‖ = supx∈c0

|f(x)|‖x‖∞

<∞.

110

We construct s sequence (xk) such that

xk =

|f(ek)|f(ek)

k ≤ n and f(ek) 6= 0

0 k > n or f(ek) = 0

Obviously, (xk) ∈ c0. Therefore, |f(x)|‖x‖∞ <∞. Since the elements of (xk) can only take

values 0 , 1, or -1, we have ‖x‖∞ = 1. We notice that

f(x) =∞∑

k=1

xkf(ek) =n∑

k=1

|f(ek)|,

f(x) ≤ ‖f‖‖x‖∞ = ‖f‖

Since n is arbitrarily large, we have

∞∑

k=1

|f(ek)| ≤ ‖f‖. (2.0.28)

Step 2: Take any x ∈ c0, we have

|f(x)| = |∞∑

k=1

ξkf(ek)|

≤ ‖x‖∞∞∑

k=1

|f(ek)|.

Taking all x such that ‖x‖∞ = 1 gives

‖f‖ ≤∞∑

k=1

|f(ek)|. (2.0.29)

By (2.0.28) and (2.0.29), we get that ‖f‖ =∑∞

k=1 |f(ek)|. Therefore, f is in l1. This alsoshows that given a sequence (ηk) in l1, there exists a linear bounded operator g on c0 suchthat g(x) =

∑∞k=1 ξkηk, where x = ξk ∈ c0. g is linear by construction. Boundedness can

be proved similar to (2.0.29).We conclude that c

′0 is isomorphic with l1.

111

Chapter 3

Inner product spaces

112

Ex. 3.3.4

A X is a Hilbert space (complete) and M ⊂ X is closed. A closed subset of a complete oneis also complete, so M is complete. M is also a vector space, which is convex by definition.Hence, M has all the properties needed for Theorem 3.3-1. B Appolonius identity is

||x− yn||2 + ||x− ym||2 =1

2||yn − ym||2 + 2||x− 1

2(yn + ym)||2. (3.0.1)

Since M is convex, we have

2||x− 1

2(yn + ym)||2 ≥ 4δ2

where δ = ||x− y|| the minimum distance. Putting vn = yn − x and using this in (3.0.1),one obtains

||vn||2 + ||vm||2 ≥1

2||vn − vm||2 + 2δ2. (3.0.2)

Using the parallelogram equality on the left-hand side of (3.0.2), one can conclude that

||vn − vm||2 ≥ 4δ2

which is the first part of the proof. The rest of the proof can then proceed as usual.

Ex. 3.3.4

(a) Show that the conclusion of Theorem 3.3-1 also holds if X is a Hilbert space andM ⊂ X is a closed subspace. (b) How could we use Appolonius’ identity (Sec. 3.1, Prob.5) in the proof of Theorem 3.3-1?

Solution The problem states that: Let X be a Hilbert space and a closed subspaceM ⊂ X. For every given x ∈ X there exists a unique y ∈M , such that δ = inf

y∈M‖x− y‖ =

‖x− y‖.1) Existence. From theorem 3.2-4, we know that M is complete.From theorem 3.3-4, the Hilbert space X can be represented as X = M ⊕M⊥. Then

for any given x ∈ X there exists y ∈ M and z ∈ M⊥ such that x = y + z. Then we provesuch a y satisfies ‖x− y‖ ≤ ‖x− y‖ for any y ∈M . Since y − y ∈M , the norm

‖x− y‖2 = ‖x− y + y − y‖2

= ‖z + y − y‖2

= ‖z‖2+ < z, y − y > + < y − y, z > +‖y − y‖2

= ‖z‖2 + ‖y − y‖2

≥ ‖z‖2.

113

2) Uniqueness.Suppose there are two different elements in M satisfy the minimizing condition, denote

as y1 and y2 such that δ = infy∈M‖x− y‖ = ‖x− y1‖ = ‖x− y2‖. By using the Appolonius’

identity,

‖y1 − y2‖2 = 2‖x− y1‖2 + 2‖x− y2‖2 − 4‖x− y1 + y2

2‖2

= 2δ2 + 2δ2 − 4‖x− y1 + y2

2‖2. (3.0.3)

As y1 6= y2, we have ‖x − y1+y22‖2 < δ2. Then, y = y1+y2

2is a better approximation for

the norm ‖x − y‖ in M ., contradiction. Therefore, there is a unique y ∈ M satisfies theminimizing condition.

Ex. 3.3.4

(a) Show that the conclusion of Theorem 3.3-1 (minimizing vector) holds also if X is aHilbert space and M ⊂ X is a closed subspace. (b) How could we use Appolonius’ identityin the proof of Theorem 3.3-1?

Solution: (a) The assumptions in Theorem 3.3-1 are that X is an inner product spaceand M 6= ∅ is a convex subset which is complete. We now assume that X is a Hilbert spaceand M is a closed subspace. Since M is closed and X is a Hilbert space, M is complete.Since M is a subspace, M is also convex. Thus, all the assumptions in Theorem 3.3-1 arefulfilled and the conclusions of the Theorem follow.

(b) Appolonius’ identity reads

‖c− a‖2 + ‖c− b‖2 =1

2‖a− b‖2 + 2‖c− 1

2(a+ b)‖2

or, equivalently,

‖a− b‖2 = 2 ‖c− a‖2 + 2 ‖c− b‖2 − 4‖c− 1

2(a+ b)‖2. (3.0.4)

To prove uniqueness of the minimizing vector in Theorem 3.3-1, assume that both y ∈Mand y0 ∈ M satisfy ‖y − x‖ = δ and ‖y0 − x‖ = δ. By Appolonius’ identity (3.0.4) witha = y, b = y0 and c = x, we obtain

‖y − y0‖2 = 2 ‖x− y‖2 + 2 ‖x− y0‖2 − 4‖x− 1

2(y + y0)‖2

= 2δ2 + 2δ2 − 4‖x− 1

2(y + y0)‖2.

Because M is convex, 12(y + y0) ∈M , so that ‖x− 1

2(y + y0)‖ ≥ δ, which implies

‖y − y0‖2 ≤ 2δ2 + 2δ2 − 4δ2 = 0.

114

But since ‖y − y0‖2 ≥ 0, we must have ‖y − y0‖ = 0 and thus y = y0, which proves thatthe minimizing vector is unique.

Ex. 3.3.10

We want to show that if Y ⊂ H is a closed subspace such that M ⊂ Y implies that

M⊥⊥ ⊂ Y.

As a first step let y⊥ ∈ Y ⊥. This implies 〈y⊥,m〉 = 0 ∀m ∈M since M ⊂ Y , which showsus that y⊥ ∈M⊥. This means that

Y ⊥ ⊂M⊥ (3.0.5)

since y⊥ was arbitrary.Let now x ∈M⊥⊥. Since Y is a closed subspace we have that X = Y

⊕Y ⊥. So x can

be decomposed as x = y + y⊥ with y ∈ Y and y⊥ ∈ Y ⊥. But from (3.0.5) we have that〈x, y⊥〉 = 0 so we get

0 =⟨x, y⊥

⟩= 〈y + y⊥, y⊥〉 = ‖y⊥‖2 ,

which means that y⊥ = 0. Thus x = y ∈ Y , and since x was arbitrary we get

M⊥⊥ ⊂ Y.

Ex. 3.3.10

If M ⊂ H, show that M⊥⊥ is contained in any closed subspace Y ⊂ H such that M ⊂ Y .There are two cases: Case I) if M is closed; then, M = M⊥⊥ by Theorem (3.3-6). Since

M ⊂ H, then M⊥⊥ is the smallest subset of H that contains M .Case II) if M is not closed; we know that M⊥⊥ contains M ; i.e.,

∀x ∈M, x⊥M⊥ ⇒ x ∈M⊥⊥ ⇒M ⊂M⊥⊥

If there exists a closed subspace Y ⊂ H, then,

M ⊂ Y ⇒ Y ⊥ ⊂M⊥ ⇒M⊥⊥ ⊂ Y ⊥⊥ = Y

Y was arbitrary, so M⊥⊥ is the smallest closed subset of H which contains M .

115

Ex. 3.4.6

(Minimum property of Fourier coefficients) Let e1, . . . , en be an orthonormal set in aninner product space X, where n is fixed. Let x ∈ X be any fixed element and y =β1e1 + . . . + βnen. Then ‖x− y‖ depends on β1, . . . , βn. Show by direct calculation that‖x− y‖ is minimum if and only if βj = 〈x, ej〉, where j = 1, . . . , n.

Solution:

‖x− y‖2 = 〈x− y, x− y〉 = 〈x, x〉 − 〈x, y〉 − 〈y, x〉+ 〈y, y〉= ‖x‖2 + ‖y‖2 − 〈x, y〉 − 〈x, y〉 = ‖x‖2 + ‖y‖2 − 2Re 〈x, y〉≥ ‖x‖2 + ‖y‖2 − 2|〈x, y〉|

= ‖x‖2 +n∑

k=1

|βk|2 − 2|〈x,n∑

k=1

βkek〉|

≥ ‖x‖2 +n∑

k=1

|βk|2 − 2n∑

k=1

|βk||〈x, ek〉|

= ‖x‖2 +n∑

k=1

(|βk| − |〈x, ek〉|)2 − |〈x, ek〉|2 ≥ ‖x‖2 −n∑

k=1

|〈x, ek〉|2

By choosing βk = 〈x, ek〉, k = 1, . . . , n, we obtain

‖x− y‖2 = ‖x‖2 + ‖y‖2 − 2Re 〈x, y〉

= ‖x‖2 +n∑

k=1

|βk|2 − 2Re

〈x,

n∑

k=1

βkek〉

= ‖x‖2 +n∑

k=1

|βk|2 − 2Re

n∑

k=1

βk〈x, ek〉

= ‖x‖2 +n∑

k=1

|〈x, ek〉|2 − 2Re

n∑

k=1

|〈x, ek〉|2

= ‖x‖2 −n∑

k=1

|〈x, ek〉|2

which shows that the choice βk = 〈x, ek〉, k = 1, . . . , n minimizes ‖x− y‖

116

Ex. 3.4.6

Since X is an inner product space and x, y ∈ X, x − y is also in X, and we can use theBessel inequality for x− y:

‖x− y‖2 ≥n∑

k=1

|〈x− y, ek〉|2

=n∑

k=1

|〈x, ek〉 − 〈y, ek〉|2

=n∑

k=1

|〈x, ek〉 −n∑

j=1

βj〈ej, ek〉|2

=n∑

k=1

|〈x, ek〉 − βk|2.

Thus, the minimun of ‖x− y‖ is obtained if and only if βk = 〈x, ek〉.

Ex. 3.4.6

Let e1, ..., en be an orthonormal set in an inner product space X, where n is fixed. Letx ∈ X be any fixed element and y = β1e1 + ... + βnen. We want to show that ‖x − y‖ isminimum if and only if βi = 〈x, ei〉.

The set e1, ..., en is a basis of a closed finite vectorial subspace Y ⊂ X, i.e.

Y = span(e1, ..., en), dim(Y ) = n.

which is a convex and complete (see Theorem 2.4-2) subspace of X (see Exercise 3.3.4).Then we can say that y = β1e1 + ...+ βnen ∈ Y for any list of n scalars (β1, ..., βn). FromTheorem 3.3-1, we know that for every given x ∈ X there exists a unique y ∈ Y such that

‖x− y‖ = infy∈Y‖x− y‖, (3.0.6)

which in our context we can rephrase as follows: there exists a unique list of n scalars(β1, ..., βn) such that y = β1e1 + ... + βnen minizes ‖x − y‖. From Lemma 3.3-2 we alsoknow that a such y must be the orthogonal projection of x on Y , i.e. z = x−y ∈ Y ⊥. Thenwe need to show that if we choose the coeffincients of y to be βi = 〈x, ei〉, then z = x− yis orthogonal to any other element of Y . We take v ∈ Y ,

∑nk=1 αkek and we multiply it by

117

z with respect to the inner product of X:

〈z, v〉 = 〈x− y, v〉

= 〈x, v〉 − 〈y, v〉

= 〈x,∑nk=1 αkek〉 − 〈

∑nk=1 βkek,

∑nk=1 αkek〉

=∑n

k=1 αk 〈x, ek〉 −∑n

k=1 βk 〈ek,∑n

k=1 αkek〉

=∑n

k=1 αk 〈x, ek〉 −∑n

k=1 βk∑n

j=1 αj 〈ek, ej〉

=∑n

k=1 αk 〈x, ek〉 −∑n

k=1 βkαk

=∑n

k=1 αk [〈x, ek〉 − βk]

which prove that 〈z, v〉 = 0 iff βi = 〈x, ei〉 . Note that in the second last equality we haveused that

∑nj=1 αj 〈ek, ej〉 = αk since 〈ek, ej〉 = 1 only when k = j and zero otherwise.

3.4.8

From Bessel’s inequality, we have

∞∑

j=1

|〈x, ei〉|2||x||2 ≤ 1. (3.0.7)

Assume we have nm inner products 〈x, ei〉 satisfying |〈x, ei〉| > 1m

. We have

nm∑

k=1

|〈x, ek〉|2||x||2 >

nm∑

k=1

1

m2||x||2 =nm

m2||x||2 .

We must then (at least) havenm < m2||x||2,

since (3.0.7) cannot hold otherwise. This proves the statement.

Ex. 3.5.4

If (xj) is a sequence in an inner product space X such that the series

‖x1‖+ ‖x2‖+ ‖x3‖+ . . . (3.0.8)

converges, show that Sn is a Cauchy sequence, where Sn = x1 + . . .+ xn

118

Solution: We letSn = x1 + x2 + x3 + . . .+ xn (3.0.9)

Then

‖Sn − Sm‖ = ‖xm + xm+1 + · · ·+ xn‖ ≤ ‖xm‖+‖xm+1‖+. . .+‖xn‖ → 0 as n,m→∞(3.0.10)

Since the series (3.0.8) converges.

Ex. 3.5.6

Given: x =∑∞

i=1 αiei and y =∑∞

j=1 βjej where, (ei) is an orthonormal sequence. Then,

〈x, y〉 = 〈∞∑

i=1

αiei,∞∑

j=1

βjej〉 =∞∑

i=1

∞∑

j=1

〈αiei, βjej〉 =∞∑

i=1

∞∑

j=1

αiβj〈ei, ej〉

=∞∑

i=1

∞∑

j=1

αiβjδij =∞∑

i=1

αiβi

It must be added that the series of x and y are absolutely convergent. For example, for x:

‖x‖2 = 〈x, x〉 = 〈∞∑

i=1

αiei,∞∑

i=1

αjej〉 =∞∑

i=1

αiαi =∞∑

i=1

|αi|2

Ex. 3.6.4

We study〈x+ y, x+ y〉 = 〈x, x〉+ 〈x, y〉+ 〈y, x〉+ 〈y, y〉

which can be re-written by using Parsevals identity

∑〈x+ y, ek〉〈x+ y, ek〉 =

∑〈x, ek〉〈x, ek〉+

∑〈y, ek〉〈y, ek〉+ 〈x, y〉+ 〈y, x〉. (3.0.11)

By using the linear properties of the inner product and rearranging, (3.0.11) can be written

〈x, y〉+ 〈y, x〉 =∑

k

〈y, ek〉〈x, ek〉+∑

k

〈x, ek〉〈y, ek〉.

Since 〈x, y〉+〈y, x〉 = 2Re(〈x, y〉) and∑

k〈y, ek〉〈x, ek〉+∑

k〈x, ek〉〈y, ek〉 = 2Re(∑

k〈y, ek〉〈x, ek〉)we can conclude that

Re(〈x, y〉) = Re(∑

k

〈x, ek〉〈y, ek〉).

119

As for the imaginary part, we have

〈x+ iy, x+ iy〉 = 〈x, x〉 − i〈x, y〉+ i〈y, x〉+ 〈y, y〉,

where i is now the usual imaginary number. Repeating the same procedure on this, onearrives at

i(−〈x, y〉+ 〈y, x〉) = i(∑

k

〈y, ek〉〈x, ek〉 −∑

k

〈x, ek〉〈y, ek〉).

By using i(−〈x, y〉 + 〈y, x〉) = 2Im(〈x, y〉) and i(∑

k〈y, ek〉〈x, ek〉 −∑

k〈x, ek〉〈y, ek〉) =

2Im(∑

k〈y, ek〉〈x, ek〉), one can conclude

Im(〈x, y〉) = Im(∑

k

〈x, ek〉〈y, ek〉).

Since both the real and imaginary parts are equal, we must have

〈x, y〉 =∑

k

〈x, ek〉〈y, ek〉,

which completes the proof.

Ex. 3.6.4

Let M be a total orthonormal set in a Hilbert space H. Then ∀ x, y ∈ H we know that

x =∑

m

〈x, em〉 em and y =∑

n

〈y, en〉 en (3.0.12)

where en and em are the elements of M which generate a countable set of Fourier coeffin-cients (Lemma 3.5-3) for x and y respectively. In the next step we consider (ek) = (el) =(em) ∪ (en) which is a countable set since it is the countable union of countable sets.

〈x, y〉 = 〈∑k 〈x, ek〉 ek,∑

l 〈y, el〉 el〉

=∑

k 〈x, ek〉 〈ek,∑

l 〈y, el〉 el〉

=∑

k

∑l 〈x, ek〉 〈y, el〉 〈ek, el〉

=∑

k 〈x, ek〉 〈y, ek〉

(3.0.13)

which is another version of the Parseval relation.

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Ex. 3.6.4

We have Parseval’s relation∑

k |〈x, ek〉|2 = ‖x‖2 and wish to show that 〈x, y〉 =∑

k〈x, ek〉〈y, ek〉.Consider ‖x− y‖. We have

‖x− y‖2 = 〈x− y, x− y〉 = ‖x‖2 + ‖y‖2 − 2<[〈x, y〉]but also from Parseval’s relation that

‖x− y‖2 =∑|〈x− y, ek〉|2 =

∑|〈x, ek〉 − 〈y, ek〉|2

=∑|〈x, ek〉|2 +

∑|〈y, ek〉|2 − 2<[

∑〈x, ek〉〈y, ek〉]

= ‖x‖2 + ‖y‖2 − 2<[∑〈x, ek〉〈y, ek〉],

where we again have used Parseval’s relation in the final equality. Comparing the twoexpressions tells us that

<[〈x, y〉] = <[∑

k

〈x, ek〉〈y, ek〉].

Repeating the exercise using ‖x+ y‖ yields the analogous result for the imaginary parts,

=[〈x, y〉] = =[∑

k

〈x, ek〉〈y, ek〉].

Consequently the desired result follows.

Ex. 3.6.4

Let x ∈ X, M = ek ⊂ X be a total orthonormal set and

x =∑

j

〈x, ej〉 ej.

Since〈x− x, ej〉 = 0 ∀i

we havex− x ⊥M.

This implies x = x by Theorem 3.6.2 a) since M is total in X. Thus we can write anyelement y ∈ X as

y =∑

k

〈y, ek〉 ek.

This now gives

〈x, y〉 =

⟨∑

j

〈x, ej〉 ej,∑

k

〈y, ek〉 ek⟩

=∑

j

∑

k

〈x, ej〉 〈y, ek〉 〈ek, ej〉 =

=∑

k

〈x, ek〉 〈y, ek〉 .

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Ex. 3.6.10

Let M be a subset of a Hilbert space H, and let u, v ∈ H. Suppose that 〈v, x〉 = 〈w, x〉for all x ∈M implies v = w. If this holds for all v, w ∈ H, show that M is total in H.

Solution: By the linearity of the inner product

〈v, x〉 = 〈w, x〉 ⇐⇒ 〈v − w, x〉 = 0.

Thus,〈v, x〉 = 〈w, x〉 for all x ∈M implies v = w, for all v, w ∈ H ⇐⇒〈v − w, x〉 = 0 for all x ∈M implies v = w, for all v, w ∈ H ⇐⇒〈u, x〉 = 0 for all x ∈M implies u = 0⇐⇒u⊥M implies u = 0⇐⇒M is total in H.

In the last step we used Theorem 3.6-2. Here it is important that H is complete.

Ex. 3.6.10

Let us consider M subset of H Hilbert space and v, w ∈ H. Moreover we admit that

〈v, x〉 = 〈w, x〉 , ∀x ∈M ⇒ v = w

holds ∀v, w ∈ H.We want to prove that M is total: in order to do this we consider z ⊥M , that satisfies forconstruction 〈z, x〉 = 0, ∀x ∈M .But since also 〈0, x〉 = 0, ∀x ∈M , then we have

〈z, x〉 = 〈0, x〉

and this implies that z = 0. Thus, by Theorem 3.6.2 the claim follows.

Ex. 3.8.10

Sesquilinearity follows by

h(x1 + x2, y) = 〈x1 + x2, y〉 = 〈x1, y〉+ 〈x2, y〉 = h(x1, y) + h(x2, y)

h(x, y1 + y2) = 〈x, y1 + y2〉 = 〈x, y1〉+ 〈x, y2〉 = h(x, y1) + h(x, y2)

h(αx, y) = 〈αx, y〉 = α 〈x, y〉 = αh(x, y)

h(x, βy) = 〈x, βy〉 = β 〈x, y〉 = βh(x, y).

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From the Cauchy-Schwartz inequality we get:

‖h‖ = supx 6=0

supy 6=0

| 〈x, y〉 |‖x‖ ‖y‖ ≤ sup

x 6=0supy 6=0

‖x‖ ‖y‖‖x‖ ‖y‖ = 1.

But note that for x = y we have

h(x, x)

‖x‖2 =‖x‖2

‖x‖2 = 1

so we get ‖h‖ = 1.

Ex. 3.8.14

Schwarz inequality. Let X be a vector space and h a Hermitian form on X×X. This formis said to be positive semidefinite if h(x, x) ≥ 0 for all x ∈ X. Show that then h satisfiesthe Schwarz inequality |h(x, y)|2 ≤ h(x, x)h(y, y).

Since h is positive semidefinite, we have

h(x− λy, x− λy) ≥ 0.

By using the fact that h is a Hermitian sesquilinear form, we obtain

h(x, x)− λh(x, y)− λh(y, x) + λλh(y, y) ≥ 0. (3.0.14)

Choose

λ =h(x, y)

h(y, y).

Thus,

λ =h(x, y)

h(y, y)=h(y, x)

h(y, y).

Substituting this into inequality (3.0.14), we obtain

0 ≤ h(x, x)− h(y, x)

h(y, y)h(x, y)− h(x, y)

h(y, y)h(y, x) +

h(x, y)h(y, x)

h(y, y)

= h(x, x)− h(y, x)

h(y, y)h(x, y)

= h(x, x)− |h(x, y)|2h(y, y)

.

Therefore, |h(x, y)|2 ≤ h(x, x)h(y, y).Remark: It may not be straightforward to get the correct λ value. The idea is that theequality in the Schwarz inequality is taken when x and y are linear dependent. Therefore,we use λ to project y so that λy and x are linear dependent.

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