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Functional Analysis - MT4515
Course Notes
Contents
Introductory Remarks about Functional Analysis ii
Chapter 1. Normed Spaces 1Useful Inequalities 1Normed Spaces 2Examples of Normed Spaces 3Convergence and Continuity 4Open, Closed and Dense Sets 5
Chapter 2. Completeness and the Contraction Mapping Theorem 8Basic Notions 8Important Complete Normed Spaces 9The Contraction Mapping Theorem 10
Chapter 3. Finite Dimensional Spaces 13
Chapter 4. The Space(C[K], ‖ · ‖∞
)15
Chapter 5. Operators on Normed Spaces 18The Space of Operators 19The Spectrum of an Operator 22
Chapter 6. Inner Products and Hilbert Space 25Basic Ideas 25Examples of Hilbert Spaces 26Orthonormal Systems 26
Chapter 7. The Spectral Theorem 30Sturm-Liouville Theory 33
i
Introductory Remarks about Functional Analysis
• In functional analysis, functions are considered as points or vectors in some big space. Westudy analysis (convergence, continuity, etc) in terms of these points rather like points or vectorsin R or Rn.
• Such function spaces are (generally infinite dimensional) normed spaces, in particular Banachspaces and Hilbert spaces, and we study the structure of such spaces in their own right.
• Transformations such as f 7→ dfdx and f 7→
∫ t0f(u) du are l inear mappings on such spaces, so
we develop linear algebra in this context. It is more involved than in the finite dimensional settingsince questions of convergence arise. Notions like eigenvalues and diagonalisation of matricesgeneralise (almost beyond recognition) to spectral theory.
• The flavour of the work is of analysis and calculus, but ideas from linear algebra are important.
• Functional analysis has many applications across mathematics, e.g. to differential equations,fluid dynamics, quantum theory, statistics. The language of functional analysis is standard inmany areas.
notation Note that when we write (xn) ∈ X, we mean a sequence of points xn in X. The := signis means that a quantity is being defined by the equation. The symbol indicates a contradiction.We use the usual ‘iff’ to stand for ‘if and only if’.
Some books
Rynne, B.P. & Youngson, M.A., Linear Functional Analysis, Springer, 2nd Ed 2007.
Young, N., An Introduction to Hilbert Space, Cambridge University Press, 1988.
Griffel, D.H., Applied Functional Analysis, Dover, 2nd Ed 2002.
Bollobas, B, Linear Analysis: An Introductory Course, Cambridge University Press, 1999.
Grateful thanks are due to Scott Thomson who typeset a large part of these notes and also to DonaldRobertson for his help with them.
ii
CHAPTER 1
Normed Spaces
Useful Inequalities
We establish some useful inequalities which we put in boxes for easy reference. The first inequalityis the triangle inequality on R:
∀x, y ∈ R |x+ y| 6 |x|+ |y|. (Triangle Inequality)
Cauchy’s Inequality or the Cauchy-Schwarz Inequality is frequently used:
N∑i=1
|aibi| 6
(N∑i=1
|ai|2)1/2( N∑
i=1
|bi|2)1/2
. (Cauchy’s Inequality)
Proof Note that 0 6∑i
(tai + bi)2 for real t. Multiplying out
t2∑i
a2i + 2t
∑i
aibi +∑i
b2i > 0.
This can be viewed as a quadratic in t. The discriminant condition requires that ‘B2 − 4AC 6 0’and by substituting the appropriate coefficients into this condition we get Cauchy’s Inequality.
As a corollary to this, we have the Distance inequality :(N∑i=1
(ai + bi)2
)1/2
6
(N∑i=1
a2i
)1/2
+
(N∑i=1
b2i
)1/2
. (Distance Inequality)
Proof ∑i
(ai + bi)2 =∑i
a2i + 2
∑i
aibi +∑i
b2i
(by Cauchy’s Inequality) 6
[(∑i
a2i
)1/2]2
+ 2(∑
i
a2i
)1/2(∑i
b2i
)1/2
+
[(∑i
b2i
)1/2]2
=
[(∑i
a2i
)1/2
+(∑
i
b2i
)1/2]2
.
Holder’s Inequality is a generalisation of Cauchy’s inequality. Suppose p, q > 1 are conjugate
indices, that is1p
+1q
= 1. Then for all ai, bi ∈ R,
N∑i=1
|aibi| 6( N∑i=1
|ai|p)1/p( N∑
i=1
|bi|q)1/q
. (Holder’s Inequality)
Cauchy’s inequality is the p = q = 2 case of this.
1
2 1. NORMED SPACES
Minkowski’s inequality is a generalisation of the distance inequality. Let p > 1. Then( N∑i=1
|ai + bi|p)1/p
6
( N∑i=1
|ai|p)1/p
+( N∑i=1
|bi|p)1/p
. (Minkowski’s Inequality)
Analogous inequalities, with similar derivations, hold for infinite sums and integrals. Forexample, Cauchy’s inequality can appear in the forms
∞∑i=1
|aibi| 6( ∞∑i=1
|ai|2)1/2( ∞∑
i=1
|bi|2)1/2
and∫|fg| 6
(∫|f |2
)1/2(∫|g|2)1/2
.
Moreover, such inequalities are also valid for complex numbers, with ‖ ‖ taken to mean the modulusof a complex number.
The following are useful when working with specific functions:
| cosx− cos y| 6 |x− y|, (1.1)
| sinx− sin y| 6 |x− y|, (1.2)
which follow easily from the mean value theorem.Also useful when working with integrals are:∣∣∣∣∫ f(x) dx
∣∣∣∣ 6 ∫ ∣∣f(x)∣∣ dx
and ∣∣∣∣∫ (f(x) + g(x)) dx∣∣∣∣ 6 ∫ ∣∣f(x)
∣∣dx+∫ ∣∣g(x)
∣∣dx.Normed Spaces
Throughout it is assumed that all vector spaces, or linear spaces as they are sometimes called, areover R or C. Most results will hold for either R or C unless otherwise stated.
Definition 1.1 (Norm, Normed Space). Let X be a vector space. A function ‖ · ‖ : X → R iscalled a norm if the following hold:
(1) ‖x‖ > 0 for every x ∈ X, and ‖x‖ = 0 when, and only when, x = 0 (Positivity);(2) ‖λx‖ = |λ| ‖x‖ for every x ∈ X and every λ in the field of scalars (Scalar property);(3) ‖x+ y‖ 6 ‖x‖ + ‖y‖ for evey x, y ∈ X (Triangle inequality).
We call the pair (X, ‖ · ‖) a normed space.
We immediately make the following observations:• By repeatedly applying the triangle inequality we get that∥∥∥∥∥
N∑i=1
xi
∥∥∥∥∥ 6N∑i=1
‖xi‖ (xi ∈ X).
• We have the reverse triangle inequality :∣∣‖x‖ − ‖y‖∣∣ 6 ‖x− y‖ ∀x, y ∈ X.Proof of the reverse triangle inequality Let x, y ∈ X. Then
‖x‖ = ‖(x− y) + y‖ 6 ‖x− y‖ + ‖y‖,so
‖x‖ − ‖y‖ 6 ‖x− y‖.Interchanging the roles of x and y completes the proof.
It is extremely useful to think of ‖x‖ as the length of the vector x and ‖x− y‖ as the distancebetween the two points x and y. In this way a normed space can be though of as a metric spacein a natural way.
EXAMPLES OF NORMED SPACES 3
Lemma 1.2. Define d : X ×X → R by
d(x, y) = ‖x− y‖,then d is a metric on X i.e., for every x, y, z ∈ X it satisfies
(1) d(x, y) > 0 ∀x, y ∈ X, and d(x, y) = 0 iff x = y,(2) d(x, y) = d(y, x) ∀x, y ∈ X,(3) and d(x, y) 6 d(x, z) + d(z, y) ∀x, y, z ∈ X.
Proof This is easy to verify from the defining properties of the norm, with (3) following from thenorm triangle inequality.
Examples of Normed Spaces
Standard examples of normed spaces include the following:(1) The space (R, | · |).(2) The space RN , where RN = x = (x1, x2, . . . , xN ) : xi ∈ R can be normed. In this case
we have a number of possible norms:(a) ‖x‖∞ = max
16i6N|xi| (maximum norm);
(b) ‖x‖1 =∑Ni=1 |xi| (1-norm);
(c) ‖x‖2 =(∑N
i=1 |xi|2)1/2
(2-norm or Euclidean norm);
(d) ‖x‖p =(∑N
i=1 |xi|p)1/p
(for p > 1 only) (p-norm).[By way of an example we check that (2c) is actually a norm: it is clearly non-negativeand zero if and only if x = 0, the scalar property holds from the definition and the triangleinequality is just the distance inequality.]
(3) The set C[0, 1] of real-valued continuous functions on [0, 1] forms a linear space under theobvious operations
(f + g)(t) = f(t) + g(t) and (λf)(t) = λf(t).
Possible norms on C[0, 1] include:(a) ‖f‖∞ := sup
t∈[0,1]
|f(t)| is known as the uniform norm or supremum norm.
To verify the triangle inequality:
‖f + g‖∞ = supt∈[0,1]
|f(t) + g(t)| 6 supt∈[0,1]
(|f(t)|+ |g(t)|)
6 supt∈[0,1]
|f(t)|+ supu∈[0,1]
|g(u)| = ‖f‖∞ + ‖g‖∞.
(b) ‖f‖1 :=∫ 1
0
|f |dt is known as the 1-norm. Note that a function could have a large
‖ · ‖∞-norm but a small ‖ · ‖1-norm.
(c) ‖f‖p :=(∫ 1
0
|f |p dt)1/p
is the p-norm.
(4) In order to establish the Lp spaces we first say that two functions f, g on [0, 1] areequivalent if they are equal ‘almost everywhere’, that is∫
X
|f − g|dx = 0.
Then the space Lp[0, 1] consists of all equivalence classes of functions on [0, 1] such that∫ 1
0|f |p dx is finite. We define the p-norm (for p > 1) on this space of functions by
‖f‖p :=(∫ 1
0
|f |p dx)1/p
.
4 1. NORMED SPACES
(It is necessary to regard such functions as equivalent to ensure that ‖f‖p = 0 implies f=0.The interested reader can find more detail on the Lebesgue Lp spaces in the literature onmeasure theory.)
(5) The space C1[0, 1] of continuously differentiable real-valued functions on [0, 1] becomes anormed space taking
‖f‖ = supt∈[0,1]
|f(t)|+ supt∈[0,1]
|f ′(t)|.
(6) The set of infinite sequences by x ≡ (x1, x2, . . .) is a vector space under the natu-ral componentwise operations (x1, x2, . . .) + (y1, y2, . . .) = (x1 + y1, x2 + y2 . . .) andλ(x1, x2, . . .) = (λx1, λx2, . . .). Certain subspaces are natural normed spaces,(a) l∞ is the space of bounded sequences with norm given by
‖x‖∞ := sup16i<∞
|xi|.
(b) lp (p > 1) is the space of sequences such that∞∑i=1
|xi|p <∞ which is a normed space
under the p-norm
‖x‖p :=( ∞∑i=1
|xi|p)1/p
.
Convergence and Continuity
Norms allows us to define notions such as convergence by requiring that the distance betweensequence elements and the limit tends to 0 (The reader unfamiliar with the formal definitions ofconvergence and continuity may consult the extensive literature on real and abstract analysis.)
Definition 1.3 (Convergence). Let (X, ‖ · ‖) be a normed space, let (xn) be a sequence in Xand let x ∈ X. We say that (xn) converges to a limit x iff
limn→∞
‖xn − x‖ = 0
and we write xn → x.
In other words, xn → x iff for every ε > 0, there exists n0 ∈ N such that ‖xn − x‖ < εwhenever n > n0.
Lemma 1.4. Let xn → x and yn → y in (X, ‖ · ‖). Let λ ∈ R. Then
(1) (xn + yn)→ (x+ y);(2) λxn → λx.
Proof For (1), we have
‖(xn + yn)− (x+ y)‖ 6 ‖xn − x‖ + ‖yn − y‖ → 0.
For (2), ‖λxn − λx‖ = |λ| ‖xn − x‖ → 0.
Example 1.5 (Convergence in different norms). Sometimes different norms will give dif-ferent convergent sequences. For, if
fn(t) :=
1− nt t ∈ [0, 1
n ]0 t ∈ [ 1
n , 1],
then ‖fn − 0‖1 = 1/2n→ 0, but ‖fn − 0‖∞ = 1 9 0.
For finite-dimensional spaces, all norms give the same convergent sequences, as will be seen inChapter 3.
OPEN, CLOSED AND DENSE SETS 5
Definition 1.6 (Continuity). Let (X, ‖ ‖X) and (Y, ‖ ‖Y ) be normed spaces and let f : X → Y .We say that f is continuous at x ∈ X if for all (xn) with xn → x in ‖ ‖X we have f(xn) → f(x)in ‖ ‖Y . The function f is said to be continuous on X if f is continuous at every x ∈ X.
There are equivalent ‘ε-δ’ definitions of continuity but here this sequence characterisation isadequate for our purposes.
Open, Closed and Dense Sets
We need a few basic ideas from topology that generalise the notions of open and closed intervalson the real line to subsets of a normed space (X, ‖ · ‖).
Definition 1.7 (Open Set). A set S ⊆ X is called open if for every x ∈ X, there exists δ > 0such that
y ∈ X : ‖y − x‖ < δ ⊆ S.
Intuitively, an open set is one that does not contain its boundary points.
Example 1.8 (Open Ball). The unit open ball
BO := x ∈ X : ‖x‖ < 1is an open set.
Proof Fix x ∈ BO, so that ‖x‖ < 1. Let δ = 1− ‖x‖. If y is such that ‖y − x‖ < δ, then
‖y‖ = ‖(y − x) + x‖ < δ + (1− δ) = 1.
So y ∈ BO.
Definition 1.9 (Limit Point, Closed Set). Let S ⊆ X. Then x ∈ X is called a limit point ofS if there exists a sequence (xn) ∈ S such that xn → x. A set is closed if it contains all its limitpoints.
Example 1.10 (Closed Ball). The set
B := x ∈ X : ‖x‖ 6 1 is closed.
Proof Suppose x is a limit point of B; that is, there exists (xn) ∈ B with xn → x. By the reversetriangle inequality, we get
0 6∣∣‖xn‖ − ‖x‖∣∣ 6 ‖xn − x‖ → 0.
Hence ‖xn‖ → ‖x‖ in R, so ‖x‖ 6 1 and x ∈ B.
Note that ‘open’ and ‘closed’ are not mutually exclusive properties. A set could be neither, orboth. In particular the sets ∅ and X are both open and closed.
Proposition 1.11. Let S be a subset of the normed space (X, ‖ · ‖). Then S is closed iff thecomplement Sc := X \ S is open.
Proof (Only if): Let S be closed. Suppose for a contradiction that X \ S is not open. From thedefinition of openness, there exists x ∈ X \ S such that for every δ > 0
y : ‖y − x‖ < δ ∩ S 6= ∅.Choose a sequence (xn) ∈ S such that
‖xn − x‖ <1n
(taking δ = 1/n).
Then xn → x but x /∈ S. Hence S is not closed.(If): Assume that X \S is open. Let (xn) ∈ S be such that xn → x. If x /∈ S, i.e., x ∈ X \S,
there exists δ > 0 such thaty : ‖x− y‖ < δ ⊆ X \ S
6 1. NORMED SPACES
since X \ S is open. In particular, since (xn) ∈ S, ‖x− xn‖ > δ, a contradiction. We concludethat x ∈ S, so that S is closed.
Example 1.12. We showed that the unit closed ball B is closed. Hence
X \B = x : ‖x‖ > 1 is open.
Definition 1.13 (Closure). If S ⊆ X, then the closure S of the set S is the set of all limitpoints of S.
Proposition 1.14 (Properties of the Closure). The following hold:
(1) The closure S of a set S is closed.(2) S is the smallest closed set containing S. (That is, if T is closed and T ⊇ S, then T ⊇ S.)
Proof (1): To show S is closed, let x be a limit point of S, so that there exists xn ∈ S withxn → x. Since xn ∈ S, we may find yn ∈ S such that ‖xn − yn‖ < 1/n. Then
0 6 ‖yn − x‖ = ‖(yn − xn) + (xn − x)‖ 6 1n
+ ‖xn − x‖ → 0.
So x is a limit point of S, i.e., x ∈ S.(2): Let x ∈ S. Then we can find a sequence (xn) ∈ S with xn → x. But xn ∈ T so the limit
x is in T since T is closed. Thus S ⊆ T .
Definition 1.15 (Dense Set). Let S ⊆ T ⊆ (X, ‖ · ‖). We say that S is dense in T iff T ⊆ S.Equivalently if either of the following statements hold:
• Every x ∈ T is a limit point of S: there exists (xn) ∈ S such that xn → x.• For every x ∈ T and every ε > 0, there exists y ∈ S such that ‖x− y‖ < ε.
We think of a dense set S as one for which we can approximate points in T ⊇ S by pointslying in S.
Example 1.16. Q is dense in R with the | · |-norm.
We will derive the following more substantial density result in Chapter 4.
Theorem 1.17 (Weierstrass Approximation Theorem). The polynomials over [0, 1] aredense in
(C[0, 1], ‖ · ‖∞
). That is, given any continuous function f : [0, 1] → R and any ε > 0,
there exists a polynomial p(t) such that
‖f − p‖∞ = supt∈[0,1]
|f(t)− p(t)| < ε.
Proof This result is a corollary of Theorem 4.3.
The following result illustrates how the Weierstrass Approximation Theorem may be used totransfer a property that is easy to show for polynomials to all continuous functions.
Lemma 1.18 (Riemann-Lebesgue Lemma). Let f ∈ C[0, 1]. Then the Fourier cosine coefficients∫ 1
0
f(t) cos(2nπt) dt→ 0 as n→∞.
Proof Let ε > 0. By the Weierstrass Approximation Theorem, there exists a polynomial p(x)such that
‖f − p‖∞ <ε
2.
OPEN, CLOSED AND DENSE SETS 7
Using obvious inequalities and integration by parts,
I :=∣∣∣∣∫ 1
0
f(t) cos(2nπt) dt∣∣∣∣ =
∣∣∣∣∣∫ 1
0
(f(t)− p(t)
)cos(2nπt) dt+
∫ 1
0
p(t) cos(2nπt) dt
∣∣∣∣∣6
∣∣∣∣∫ 1
0
(f(t)− p(t)
)cos(2nπt) dt
∣∣∣∣+∣∣∣∣∫ 1
0
p(t) cos(2nπt) dt∣∣∣∣
6∫ 1
0
∣∣f(t)− p(t)∣∣ ∣∣cos(2nπt)
∣∣dt+
∣∣∣∣∣[p(t)
sin(2nπt)2nπ
]1
0︸ ︷︷ ︸=0
− 12nπ
∫ 1
0
sin(2nπt)p′(t) dt
∣∣∣∣∣6 ‖f − p‖∞ +
12nπ
∣∣∣∣∫ 1
0
p′(t) sin(2nπt) dt∣∣∣∣
6 ‖f − p‖∞ +1
2nπ
∫ 1
0
|p′(t)|dt.
If we choose n large enough, we can make the second term less than ε/2 and thus obtain I < ε.
CHAPTER 2
Completeness and the Contraction Mapping Theorem
Basic Notions
Completeness is a central concept in functional analysis. The following definition, which is familiarin real analysis, extends easily to normed spaces.
Definition 2.1 (Cauchy Sequence). A sequence (xn) in a normed space(X, ‖ · ‖
)is called a
Cauchy sequence (or just Cauchy) if for all ε > 0, there exists n0 ∈ N such that
‖xn − xm‖ < ε for every n,m > no.
Example 2.2. Let fn(t) = tn in(C[0, 1], ‖ · ‖1
).Then
‖fn − fm‖1 =∫ 1
0
|tn − tm|dt 6∫ 1
0
|tn|+ |tm|dt =1
n+ 1+
1m+ 1
.
Then if ε > 0, simply set n,m > 2/ε to obtain ‖fn − fm‖1 < ε. Thus, (fn) is Cauchy.
Proposition 2.3. In a normed space, every convergent sequence is Cauchy.
Proof Suppose xn → x and let ε > 0. There must exist n0 ∈ N such that
‖xn − x‖ <ε
2for n > n0. Then
‖xn − xm‖ 6 ‖xn − x‖ + ‖x− xm‖ < ε for every n,m > n0.
The converse of this proposition is in general not true. The Cauchy sequences may be thought ofsequences that are ’trying to converge’ and a space is complete if there is somewhere for them toconverge to.
Definition 2.4 (Complete Normed Space, Banach Space). Let(X, ‖ · ‖
)be a normed space.
If every Cauchy sequence in X converges to a limit in X, we call(X, ‖ · ‖
)a complete normed
space or a Banach space. Furthermore, if S is a subset of some normed space X and every Cauchysequence in S converges to a limit in S, we call S a complete subset of X.
Many important normed spaces are complete and we will give some of these below. It is usefulto have the completeness property in a normed space since we can assert that limits of sequencesexist without explicitly having to find the limit.
One way of demonstrating the completeness of sets such as R is via the following result:
Lemma 2.5. Every bounded sequence of real numbers has a convergent subsequence. (That is, ifthere exists M ∈ R such that |xn| 6 M for every n, then there exists x ∈ R and a subsequence(xni
)i∈N such that xni→ x.)
Proof Define
x := infα : (−∞, α) contains infinitely many terms of (xn)
. (2.1)
Clearly, −M 6 x 6M . Given ε > 0, there are infinitely many terms of the sequence in (x−ε, x+ε),since there are only finitely many in (−∞, x− ε) but infinitely many in (−∞, x+ ε). Now simply
8
IMPORTANT COMPLETE NORMED SPACES 9
let (xni)i∈N be such that
xni∈ (x− 1
i , x+ 1i ) and such that n1 < n2 < · · · .
Hence xni → x.
Note that in order to prove this result we have had to assume in (2.1) an axiomatic property ofreal numbers, namely that every bounded set of real numbers has a least upper bound or infimum.
Important Complete Normed Spaces
We give some important examples of complete normed spaces.
Proposition 2.6 (General Principle of Convergence). The space (R, | · |) is complete.
Proof Let (xn) be Cauchy. Then there is an n0 such that |xn − xm| < 1 for every n,m > n0. Ifn > n0 then, by the triangle inequality,
|xn| 6 |xn − xn0 |+ |xn0 | 6 1 + |xn0 |.So |xn| 6 max |x1|, |x2|, . . . , |xn|, 1 + |xn0 | for all n, so (xn) is bounded. By Lemma 2.5 we canfind a convergent subsequence (xni
)i∈N with limit x. Given ε > 0 we may find n0 such that forn,m > n0, we have |xn−xm| < ε
2 . We may then find i such that ni > n0 and |xni −x| < ε2 , giving
|xn − x| 6 |xn − xni|+ |xni
− x| < ε.
Hence, a sequence of real numbers converges if and only if it is Cauchy, by this proposition andProposition 2.3. Note that we proved completeness of R from Lemma 2.5, which in turn reliedon a bounded nonempty set having an infimum. It turns out that this property of subsets of R isequivalent to the completeness of R. A serious textbook on real analysis will have more to say onthis rather deep matter.
Completeness of other spaces often follows from completeness of R.
Example 2.7. The spaces(RN , ‖ · ‖∞
),(RN , ‖ · ‖1
)and
(RN , ‖ · ‖p
)(with p > 1) are all complete.
Sample proof for(RN , ‖ · ‖1
). Let (xn) = ((xn,1, . . . , xn,N )) be Cauchy in
(RN , ‖ · ‖1
). For each i,
|xn,i − xm,i| 6N∑i=1
|xn,i − xm,i| = ‖xn − xm‖1.
Thus (xn,i)i is a Cauchy sequence for each i and xn,i → xi for some xi ∈ R by completeness of(R, | · |). Then
‖xn − x‖1 =N∑i=1
|xn,i − xi| → 0,
so xn → x where x = (x1, . . . , xN ).
The next result is important for later work since many of the later results rely on this spacebeing complete.
Proposition 2.8. The space (C[0, 1], ‖ · ‖∞) is a Banach space.
Proof Suppose (fn)n is a Cauchy sequence in (C[0, 1], ‖ · ‖∞). Then for each t ∈ [0, 1] we have,
|fn(t)− fm(t)| 6 supu∈[0,1]
|fn(u)− fm(u)| = ‖fn − fm‖∞.
Hence, given ε > 0, there is n0 ∈ N such that, whenever n,m > n0, we have
|fn(t)− fm(t)| < ε for every t ∈ [0, 1]. (2.2)
So for each t, the sequence(fn(t)
)n
is Cauchy and by completeness of R converges to a limit f(t),say. We need to show that f(t) ∈ C[0, 1]. Letting m → ∞ in (2.2) it follows that if n > n0
then |fn(t) − f(t)| 6 ε for every t ∈ [0, 1]. Thus fn → f uniformly on [0, 1], that is in ‖·‖∞. A
10 2. COMPLETENESS AND THE CONTRACTION MAPPING THEOREM
standard result of real analysis states that the uniform limit of a sequence of continuous functionsis continuous, so since the fn are continuous f is continuous.
Note that the spaces(C[0, 1], ‖ · ‖1
)and
(C[0, 1], ‖ · ‖p
)are not Banach spaces.
We list a few more complete spaces for reference:
Example 2.9. The spaces(L1[0, 1], ‖ · ‖1
),(L2[0, 1], ‖ · ‖2
)and
(Lp[0, 1], ‖ · ‖p
)are all Banach
spaces.
Example 2.10. The sequence spaces(l1[0, 1], ‖ · ‖1
),(l2[0, 1], ‖ · ‖2
)and
(lp[0, 1], ‖ · ‖p
)are Ba-
nach spaces.
Completeness of subsets is inherited from the space provided that the subset is closed:
Proposition 2.11. Let(X, ‖ · ‖
)be a complete normed space, and Y be a closed subset of X.
Then Y is complete.
Proof Let (xn) be a Cauchy sequence in Y . Then (xn) is a Cauchy sequence in X, and thusxn → x ∈ X. Since Y is closed, x ∈ Y ; hence Y is complete.
The Contraction Mapping Theorem
Notation 2.12. If T : X → X we write Tn : X → X to denote the nth iterate of T , given by
Tn(x) = T (T (T (· · · (T︸ ︷︷ ︸n times
x) · · · )))
(i.e., Tn+1x = T (Tnx) where T 1(x) = T (x)).
Definition 2.13 (Contraction Mapping). Let(X, ‖ · ‖
)be a normed space. Let T : X → X.
We call T a contraction mapping if there exists k ∈ (0, 1) such that
‖Tx− Ty‖ 6 k‖x− y‖ for every x, y ∈ X.
Thus contractions reduce distances by a factor k or less. The following theorem (which in factholds for any complete metric space) is sometimes called Banach’s fixed point theorem.
Theorem 2.14 (Contraction Mapping Theorem). Let T : X → X be a contraction mappingon a Banach space
(X, ‖ · ‖
). Then T has a unique fixed point, that is there exists a unique x ∈ X
such that
Tx = x.
Moreover, for every y ∈ X, the sequence Tny → x as n → ∞. This theoem remains true if X istaken to be a complete subset of a normed space.
Proof For every n ∈ N and every y ∈ X, we have ‖Tny − Tn−1y‖ = ‖T (Tn−1)y − T (Tn−2)y‖ 6k‖Tn−1y − Tn−2y‖, so by induction we get ‖Tny − Tn−1y‖ 6 kn−1‖Ty − y‖. Then if n > m
‖Tny − Tmy‖ = ‖(Tny − Tn−1y) + (Tn−1y − Tn−2y) + · · ·+ (Tm+1y − Tmy)‖6 ‖Tny − Tn−1y‖ + ‖Tn−1y − Tn−2y‖ + · · ·+ ‖Tm+1y − Tmy‖6 [kn−1 + kn−2 + · · ·+ km]‖Ty − y‖
6km
1− k‖Ty − y‖ (2.3)
where the last step comes from summing the geometric series. If ε > 0, there exists n0 ∈ N suchthat
km
1− k‖Ty − y‖ < ε if m > n0
THE CONTRACTION MAPPING THEOREM 11
so by (2.3) the sequence (Tny)n∈N is Cauchy. The completeness of X implies that Tny → x forsome x ∈ X. To show that x is a fixed point of T , note that
‖Tx− x‖ = ‖(Tx− Tny) + (Tny − x)‖ 6 k‖x− Tn−1y‖ + ‖Tny − x‖ → 0.
So ‖Tx− x‖ = 0, that is Tx = x.To show uniqueness, suppose that x and x′ are both fixed points. Then
0 6 ‖x− x′‖ = ‖Tx− Tx′‖ 6 k‖x− x′‖;
since 0 6 k < 1 this implies ‖x− x′‖ = 0 so x = x′.Finally, for any z ∈ X, the sequence Tnz converges to some fixed point of T , as in the first
part of the proof; since this point is unique Tnz → x for all z ∈ X.
Note that for all z ∈ X, we have ‖Tnz − x‖ 6 kn‖z − x‖. If k is small then we will have rapidconvergence.
The contraction mapping theorem is a very powerful tool with applications in a wide range ofsettings.
Example 2.15 (Roots of Equations). Investigate the real roots of the equation x3+3x−1 = 0.
Rearranging gives
x =1
x2 + 3.
The mapping T : R→ R such that x 7→ 1/(x2 + 3) is a contraction on the complete space (R, | · |),since
|Tx− Ty| =∣∣∣∣ 1x2 + 3
− 1y2 + 3
∣∣∣∣ =∣∣∣∣ (y − x)(y + x)(x2 + 3)(y2 + 3)
∣∣∣∣6 |x− y|
(|x|
x2 + 31
y2 + 3+|y|
y2 + 31
x2 + 3
)6 2
3 |x− y|.
(We could have also used the mean value theorem thus
|f(x)− f(y)| 6 |x− y|maxc|f ′(c)|
where c ranges over the interval in question.) By the CMT the equation has a unique real solution.Moreover, for any y ∈ R, the sequence (Tny)n converges to the solution, so taking y = 0.5 we getthe sequence of approximations
0.5, 0.308, 0.323, 0.3221, 0.32219 . . . .
We next turn to the question of whether a given differential equation has a solution and if sowhether it is unique. This is not always clear, for example, the equation
dydx
= 2|y|1/2 with y(0) = 0
has y = 0 and y = x2 both as solutions, in fact, it has infinitely many solutions. However, theCMT may used to show that that a wide variety of differential equations have unique solutions.
Example 2.16 (Differential Equations). The differential equation
dfdt
= 14 cos
(f(t)
)+ t3 with f(0) = 0.
has a unique solution for 0 6 t 6 1.
This differential equation is equivalent to the integral equation
f(t) = 14
∫ t
0
cos(f(u)
)du+ 1
4 t4,
12 2. COMPLETENESS AND THE CONTRACTION MAPPING THEOREM
which includes the initial conditions. Define the mapping T on (C[0, 1], ‖ · ‖∞) by
(Tf)(t) = 14
∫ t
0
cos(f(u)
)du+ 1
4 t4.
Then, ∣∣(Tf)(t)− (Tg)(t)∣∣ 6 1
4
∫ t
0
∣∣cos(f(u))− cos(g(u))∣∣du
6 14
∫ t
0
∣∣f(u)− g(u)∣∣ du
6 14 t‖f − g‖∞ 6
14‖f − g‖∞ since t ∈ [0, 1].
In particular, ‖Tf − Tg‖∞ 6 14‖f − g‖∞. Hence T is a contraction on the complete normed space
(C[0, 1], ‖ · ‖∞). This means that there is a unique ‘fixed point’ f of T that is the solution tothe integral equation and thus to the differential equation. We could even use iterative numericalmethods to find a sequence of approximations Tn(g) to this f for any initial g ∈ C([0, 1]).
CHAPTER 3
Finite Dimensional Spaces
In this short chapter we will show that all norms on a given finite dimensional space are equiv-alent in that they define the same convergent sequences, etc. We first prove a lemma which is ageneralisation of Lemma 2.5 to RN .
Lemma 3.1. Every bounded sequence in(RN , ‖ · ‖∞
)has a convergent subsequence.
Proof. Recall (Lemma 2.5) that every bounded sequence in R has a convergent subsequence.So let (xn)n be a bounded sequence in RN . Then there is a real M such that ‖xn‖∞ 6 M forevery n ∈ N. We write members of the sequence in coordinate form xn = (xn,1, xn,2, . . . , xn,N ).Apply the established result for R to each coordinate:
Since (xn,1) is bounded it has a convergent subsequence (xnk(1),1)k(1) → x1; since (xnk(1),2) isbounded it has a convergent subsequence (xnk(2),2)k(2) → x2, and so on. At each stage we are con-sidering only the points left in the sequence (xn)n after deleting to obtain a convergent subsequencein the previous coordinate. We are left with all coordinates converging, that is (xnk(N),i)k(N) → xifor all i = 1, . . . , N .
Definition 3.2 (Equivalent Norms). Two norms ‖ · ‖A and ‖ · ‖B on the same space X arecalled equivalent if there are a, b > 0 such that
a‖x‖A 6 ‖x‖B 6 b‖x‖A for every x ∈ X.
We note that if a sequence converges in one of the norms, it converges in the other, sincea‖xn − x‖A 6 ‖xn − x‖B 6 b‖xn − x‖A, so xn → x in ‖·‖A iff xn → x in ‖·‖B . Similarly,if a sequence is Cauchy in one norm it is Cauchy in the other, etc. It can be shown that theequivalence of norms is an equivalence relation.
In the case of finite-dimensional spaces all norms on a space are equivalent:
Theorem 3.3. Let X be a vector space with finite dimension N . Then all norms on X areequivalent.
Proof. Let e1, e2, . . . , eN be a basis for X, and define∥∥∥∥∥N∑i=1
xiei
∥∥∥∥∥∞
= max16i6N
|xi|.
We will show that any other given norm ‖ · ‖ is equivalent to this norm.First, ∥∥∥∥∥∑
i
xiei
∥∥∥∥∥ 6∑i
|xi| ‖ei‖ 6 maxi|xi|
∑i
‖ei‖ =
∥∥∥∥∥∑i
xiei
∥∥∥∥∥∞
∑i
‖ei‖︸ ︷︷ ︸b
.
Thus ‖∑i xiei‖ 6 b‖
∑i xiei‖∞.
Secondly, to show that ‖x‖∞ 6 c‖x‖ we argue by contradiction. Suppose that there is no cwith this property: then for every n ∈ N, there is an xn ∈ X such that
‖xn‖∞ > n‖xn‖.
Put yn = xn/‖xn‖∞, so that ‖yn‖∞ = 1 and ‖yn‖∞ > n‖yn‖, so ‖yn‖ 6 1/n.
13
14 3. FINITE DIMENSIONAL SPACES
Since (yn)n is bounded, by Lemma 3.1 it has a convergent subsequence (yni)i with yni → y inthe ‖ · ‖∞-norm. Then ∣∣‖yni
‖∞ − ‖y‖∞∣∣ 6 ‖yni
− y‖∞ → 0which implies ‖y‖∞ = 1. However, using the triangle inequality,
‖y‖ 6 ‖y − yni‖ + ‖yni‖ 6 b‖y − yni‖∞ +1ni→ 0.
Thus ‖y‖ = 0, so y = 0 which contradicts that ‖y‖∞ = 1.
Corollary 3.4. In a finite dimensional space X, Cauchy sequences, convergent sequences, limitpoints, closed sets etc. are independent of the norm chosen. In particular, every finite dimensionalnormed space (X, ‖ · ‖) is complete since, as we showed in Chapter 2, (X, ‖ · ‖1) is complete.
Example 3.5. Norms on PN =a0 + a1t+ a2t
2 + . . .+ aN tN : ai ∈ R
, the space of polynomials
of degree at most N on [0, 1].
The space PN has dimension N + 1. We define ‖ · ‖ on PN by
‖a0 + a1t+ a2t2 + . . .+ aN t
N‖ = maxi|ai|.
which can be checked is a norm. Another possible norm is
‖a0 + a1t+ a2t2 + . . .+ aN t
N‖∞ = supt∈[0,1]
|p(t)|.
These norms are equivalent by Theorem 3.3. Hence a sequence (pn)n of polynomials convergesin one norm iff it converges in the other. In other words, a sequence of polynomials (pn) in PNconverges to p uniformly, that is in ‖ · ‖∞, if and only each sequence of individual coefficientsconverges.
CHAPTER 4
The Space(C[K], ‖ · ‖∞
) Notation 4.1. We write (C[K], ‖ · ‖∞) to denote the vector space of continuous functions on K,
where K is a compact topological or metric space, equipped with the norm
‖f‖∞ = supt∈K|f(t)|. (4.1)
For our purposes K will normally be the closed interval [0, 1], or [a, b], or the interval [0, 2π] with 0and 2π identified (thus forming a circle), or the square [0, 1]× [0, 1]. We will develop the theory for(C[0, 1], ‖ · ‖∞) but identical arguments hold in the other cases. We will state the Stone-Weierstrasstheorem for real-valued continuous functions on [0, 1]. We work with the supremum norm (4.1)where the space K is [0, 1].
Definition 4.2 (Algebra). A vector subspace A of C[0, 1] is called an algebra if fg ∈ A wheneverf ∈ A and g ∈ A. (By fg is meant the function defined by (fg)(t) = f(t)g(t) for every t ∈ C[0, 1].)
For example C[0, 1] is itself an algebra and the polynomials form a subalgebra. As usual the closureA of A consists of the limit points of A; if A is a subalgebra of C[0, 1] then so is A.
The Stone-Weierstrass Theorem gives simple conditions for an algebra to be dense in C[0, 1],in other words tells us when every f ∈ C[0, 1] can be approximated arbitraily closely by membersof A.
Theorem 4.3 (Stone-Weierstrass Theorem). Let A be a subalgebra of (C[0, 1], ‖ · ‖∞) suchthat
(1) A contains the constant functions,(2) A ‘separates the points of [0, 1]’, i.e., for all t 6= u ∈ [0, 1] we may find f ∈ A such that
f(t) 6= f(u).
Then A is dense in (C[0, 1], ‖ · ‖∞), that is A = C[0, 1].
The proof uses several lemmas.
Lemma 4.4. Given ε > 0 there exists a polynomial p on [−1, 1] such that∣∣p(t) − |t|∣∣ < ε for all
−1 6 t 6 1. (Thus |t| is approximable by polynomials in (C[−1, 1], ‖ · ‖∞).)
Proof We use an iterative trick. Define a sequence of polynomials by p1(t) = 0, and
pn+1(t) = 12 t
2 + pn(t)− 12pn(t)2 (n = 1, 2, . . .). (4.2)
Clearly pn(t) is a polynomial for all n. We claim that pn(t)→ |t| in (C[−1, 1], ‖ · ‖∞).Note that if 0 6 a 6 1 then 0 6 a − 1
2a2 6 1
2 by a simple calculus check. Hence from (4.2),if 0 6 pn(t) 6 1 for all t ∈ [−1, 1], then 0 6 pn+1(t) 6 1 for all t ∈ [−1, 1]. Thus by induction,0 6 pn(t) 6 1 for all t ∈ [−1, 1] and all n. Moreover
|pn+1(t)− |t|| =∣∣ 1
2 t2 + pn(t)− 1
2pn(t)2 − |t|∣∣
=∣∣pn(t)− |t|
∣∣ ∣∣1− 12 (pn(t) + |t|)
∣∣6∣∣pn(t)− |t|
∣∣(1− 12 |t|)
for t ∈ [−1, 1]. Iterating this,∣∣pn(t)− |t|∣∣ 6 ∣∣p1(t)− |t|
∣∣ (1− 12 |t|)n−1 = |t|
(1− 1
2 |t|)n−1
. (4.3)
15
16 4. THE SPACE`C[K], ‖ · ‖∞
´So given 0 < ε < 1, choose N large enough to get
(1− 1
2 |ε|)n−1
< ε for all n > N . It follows thatif |t| < ε, then ∣∣pn(t)− |t|
∣∣ 6 |t| < ε, by (4.3),and if ε 6 |t| 6 1, then∣∣pn(t)− |t|
∣∣ 6 (1− 12 |t|)n−1
<(1− 1
2 |ε|)n−1
< ε; again by (4.3).
Hence∣∣pn(t)− |t|
∣∣ < ε for all t ∈ [−1, 1] and n > N , i.e., ‖pn − |t| ‖∞ < ε.
Lemma 4.5. Let A be a closed subalgebra of C[0, 1]. If f ∈ A then |f | ∈ A.
Proof We may assume that ‖f‖∞ 6 1 on multiplying f by a scalar. Given ε > 0 choose a poly-nomial p as in Lemma 4.4 with
∣∣p(t)− |t|∣∣ < ε for −1 6 t 6 1. Then∣∣p(f(u))− |f(u)|
∣∣ < ε for allu ∈ [0, 1], i.e.,
∥∥p(f) − |f |∥∥∞ < ε. But p(f) is just the sum of powers of f so is in the algebra A
since f ∈ A. As ε is arbitrary, |f | ∈ A = A since A is closed.
Lemma 4.6. Let A be a closed subalgebra of C[0, 1]. If f, g ∈ A then maxf, g ∈ A andminf, g ∈ A.
Proof We have
maxf, g = 12
((f + g) + |f − g|
)∈ A,
minf, g = 12
((f + g)− |f − g|
)∈ A;
by using Lemma 4.5 and that A is an algebra.
Proof of the Stone-Weierstrass Theorem: To prove the Theorem we assume that A is a closedsubalgebra and show that A = C[0, 1]. (If A is an algebra satisfying (1) and (2) then so is A.)
Note that (1) and (2) in the statement of the Theorem imply that given t 6= u ∈ [0, 1] therealways exists h ∈ A such that h(t) and h(u) take any prescribed values. (Take h(s) = af(s) + bfor suitable constants a, b, with f as in (2).)
Let f ∈ C[0, 1] and ε > 0; we will find g ∈ A such that ‖f − g‖∞ < ε. For t 6= u ∈ [0, 1] takeft,u ∈ A such that ft,u(s) = f(s) when s = t and s = u. Then
Ot,u = s ∈ [0, 1] : ft,u(s) < f(s) + εis an open subset of [0, 1] containing t and u. Thus for each t we have
⋃u∈[0,1]Ot,u = [0, 1]. A
standard result (the Heine-Borel theorem or the compactness of [0, 1]) states that any cover of[0, 1] by a collection of open sets has a finite subcover. Thus there is a f inite collection of pointsu1, . . . , uk such that
k⋃i=1
Ot,ui= [0, 1].
Define ft = mini ft,ui, which is in A, by Lemma 4.6. Then
ft(t) = f(t) and ft(s) < f(s) + ε for all s ∈ [0, 1], (4.4)
since if s ∈ [0, 1] then s ∈ Ot,ui for some i.In the same way, Ot = s ∈ [0, 1] : ft(s) > f(s) − ε is an open subset of [0, 1] containing t.
Then⋃t∈[0,1]Ot = [0, 1] so by the standard result there is a f inite collection of points t1, . . . , tm
such that⋃mi=1Oti = [0, 1]. Define g = maxi fti which is in A, by Lemma 4.6. Then if s ∈ [0, 1]
we haveg(s) 6 max
ifti(s) < f(s) + ε by (4.4),
but also s ∈ Oti for some i, sog(s) > fti(s) > f(s)− ε.
Thus ‖g − f‖∞ < ε. It follows that A is dense in C[0, 1], so since A is closed, A = C[0, 1].
4. THE SPACE`C[K], ‖ · ‖∞
´17
Corollary 4.7 (to Stone-Weierstrass Theorem). The following follow immediately fromthe Stone-Weierstrass Theorem:
(1) Weierstrass’s Theorem: The polynomials on [0, 1] are dense in (C[0, 1], ‖ · ‖∞).(2) The infinitely differentiable functions on [0, 1] are dense in (C[0, 1], ‖ · ‖∞).(3) The two variable polynomials p(x, y) on [0, 1]× [0, 1] are dense in (C([0, 1]× [0, 1]), ‖ · ‖∞).(4) The trigonometric polynomials on [0, 2π], i.e., functions of the form
n∑j=1
aj sin(jt) +m∑j=0
bj cos(jt),
are dense in (C[0, 2π], ‖ · ‖∞) (with 0 and 2π identified).
Proof One only needs to show that these sets satisfy the assumptions of the theorem which isstraightforward. To show the last set is an algebra, note that
sin(rt) cos(st) = 12
(sin((r + s)t) + sin((r − s)t)
).
These approximation theorems often allow us to obtain extend results which are easy to provefor polynomials, say, to general continuous functions.
Example 4.8. Functions determined by moments.
As above, the polynomials are dense in (C[0, 1], ‖ · ‖∞). Suppose f ∈ C[0, 1] is such that its‘moments’ are all zero, i.e.∫ 1
0
f(t)tk dt = 0 for every integer k > 0. (4.5)
We claim that f(t) = 0 for all t ∈ [0, 1]. To see this, first let ε > 0, and choose a polynomial p suchthat ‖f − p‖∞ 6 ε. Note that
f2 = f(f − p) + fp.
We then have ∫ 1
0
f2(t) dt =∫ 1
0
f(t)(f(t)− p(t)
)dt+
∫ 1
0
f(t)p(t) dt︸ ︷︷ ︸=0 by (4.5)
so∣∣∣∣∫ 1
0
f2(t) dt∣∣∣∣ 6 ∫ 1
0
∣∣f(t)∣∣∣∣f(t)− p(t)
∣∣dt6 ε
∫ 1
0
|f(t)|dt.
Thus∫ 1
0f2(t) dt = 0 so f = 0.
We may conclude that a function f ∈ C[0, 1] is uniquely determined by its moments, that isby the values given for every integer k > 0 by
Mk(f) :=∫ 1
0
f(t)tk dt.
If for every k we have Mk(g) = Mk(h), then∫ 1
0
(g(t)− h(t)
)tk dt = 0,
so from the above, g = h, that is we have uniqueness.
CHAPTER 5
Operators on Normed Spaces
In this chapter we generalise the notions of linear mappings, eigenvalues and eigenvectors to moregeneral spaces. This eventually leads us to Spectral Theory. We begin with some basic terminology:
Definition 5.1 (Linear, Continuous and Bounded Operators). Let(X, ‖ · ‖X
)and
(Y, ‖ · ‖Y
)be normed spaces where X,Y are vector spaces over either R or C. We call a mapping T : X → Yan operator. We say that T is linear if
T (λx+ µx′) = λT (x) + µT (x′)
for every x, x′ ∈ X and all scalars λ, µ. We call T continuous if given a sequence (xn) in X wehave
xn → x⇒ T (xn)→ T (x).T is called bounded iff there exists c > 0 such that
‖Tx‖Y 6 c‖x‖X for every x ∈ X.
Note that we often omit the subscript on the norm when it is clear what the domain of the normis. We also write Tx and T (x) interchangeably when the meaning is clear. In cases where T islinear and has a matrix representation this allows us to write Tx where T is the matrix; but thismeans more or less the same as T (x), for our purposes at least.
Example 5.2. Let A : RN → RN be given by x 7→ Ax where A is an N ×N matrix. A is clearly
linear. If the matrix A has entries (aij) then (Ax)i =∑Nk=1 aikxk (where the subscripts refer to
the ith and kth coordinates). We have that if (xn,1, . . . , xn,N )→ (x1, . . . , xN ) in the ‖ · ‖∞-norm,then ((Axn)1, . . . , (Axn)N )→ ((Ax)1, . . . , (AxN )) in ‖ · ‖∞, so A is continuous. We also have that
‖Ax‖∞ 6(
maxi
N∑j=1
|aij |︸ ︷︷ ︸c
)‖x‖∞,
and the constant c is only dependent on A, so A is bounded.
Example 5.3. The following are linear, continuous and bounded on (C[0, 1], ‖ · ‖∞)→ (C[0, 1], ‖ · ‖∞):(1) T : (Tf)(t) = (t2 + 2)f(t);
(2) T : (Tf)(t) =∫ t
0
u2f(u) du;
(3) T : (Tf)(t) =∫ 1
0
g(t, u)f(u) du where g(t, u) is continuous.
Proof of (2) Linearity of T follows from linearity of the integral. To see that T is continuous, letfn → f in ‖·‖∞. Then for every nwe have∣∣(Tfn)(t)− (Tf)(t)
∣∣ 6 ∫ t
0
u2|fn(u)− f(u)|du 6∫ t
0
u2‖fn − f‖∞ du =13‖fn − f‖∞.
Then ‖Tfn − Tf‖∞ = supt∈[0,1]
∣∣(Tfn)(t)− (Tf)(t)∣∣→ 0, establishing continuity.
For boundedness, we have ∣∣(Tf)(t)∣∣ 6 ∫ t
0
u2‖f‖∞ du 6 13‖f‖∞
18
THE SPACE OF OPERATORS 19
for all t ∈ [0, 1], so in particular ‖Tf‖∞ 6 13‖f‖∞.
Example 5.4. Let D : C1[0, 1] → C[0, 1] with D : f 7→ dfdt
, where C1[0, 1] denotes the space of
functions on [0, 1] with continuous derivative and the norm in both spaces is the ‖ · ‖∞-norm. Bythe usual linearity of derivatives, D is linear. It is not continuous or bounded, for if fn(t) :=n−1/2 sin(nt), then fn → f := 0 in ‖ · ‖∞. But Dfn = n1/2 cos(nt) 6→ 0 = Df . Clearly D is notbounded.
The following property, that a linear operator is bounded if and only if it is continuous isfundamental to all that follows.
Proposition 5.5. Let T :(X, ‖ · ‖X
)→(Y, ‖ · ‖Y
)be linear. Then T is continuous if and only if
it is bounded.
Proof (⇒): Assume T is continuous but not bounded. Then choose 0 6= xn ∈ X such that
‖Txn‖Y > n2‖xn‖X .
By replacing xn by xn/‖xn‖X we may assume ‖xn‖X = 1. Considering the sequence xn/n, wehave ∥∥∥xn
n
∥∥∥X
=1n→ 0. (5.1)
We also have ∥∥∥T (xnn
)∥∥∥Y
=1n‖T (xn)‖Y > n‖xn‖X = n→∞. (5.2)
The equations (5.1) and (5.2) imply that T is not continuous. (⇐): Let T be bounded and let xn → x in ‖ · ‖X . Then, we have by linearity and boundedness,
‖Txn − Tx‖Y = ‖T (xn − x)‖Y 6 c‖xn − x‖X → 0.
This implies T is continuous.
Example 5.6. Let T : (C[0, 1], ‖ · ‖∞)→ (C[0, 1], ‖ · ‖∞) be given by
(Tf)(t) =∫ 1
0
(t2 + u2)f(u) du.
Then T is linear and bounded, for it is easy to show that ‖Tf‖∞ 6 43‖f‖∞. Continuity of T
follows from the last lemma.
The Space of Operators
Definition 5.7 (Space of Operators, Induced Norm). We write B(X,Y ) to denote thespace of bounded linear operators from X to Y . If Y = X we simply write B(X) for the space ofbounded operator T : X → X on X. If T ∈ B(X,Y ), define ‖T‖ to be the smallest (i.e., infimum)number c such that ‖Tx‖ 6 c‖x‖ for all x ∈ X. Thus
‖T‖ := supx 6=0
‖Tx‖Y‖x‖X
= supx∈X‖x‖=1
‖Tx‖Y‖x‖X
= supx∈X‖x‖=1
‖Tx‖.
In particular we have for every x ∈ X that ‖Tx‖ 6 ‖T‖ ‖x‖. We call this norm of T the inducednorm on T .
The definition does not make it clear that B(X,Y ) is actually a normed linear space. We willgo on to show that, in fact, it is.
20 5. OPERATORS ON NORMED SPACES
Example 5.8. We find the induced norm for T where T : (C[0, 1], ‖ · ‖∞) → (C[0, 1], ‖ · ‖∞) isdefined by
(Tf)(t) =∫ t
0
uf(u) du.
First,∣∣(Tf)(t)
∣∣ 6 ∫ t
0
|u| ‖f‖∞ du 6 12‖f‖∞. So in particular, for all f ∈ C[0, 1]),
‖(Tf)(t)‖∞ 6 12‖f‖∞
i.e., ‖T‖ 6 12 .Furthermore if f(t) = 1 for every t ∈ [0, 1], then we have ‖Tf‖∞ = 1
2 . This meansthat 1
2 is the smallest number such that ‖Tf‖∞ 6 12‖f‖∞ for arbitrary f so we conclude that
‖T‖ = 12 .
Definition 5.9. Let T,U ∈ B(X,Y ) and let λ be a scalar. Then we define T + U and λT by
(T + U)(x) = Tx+ Ux and (λT )(x) = λT (x). (5.3)
Also, if V ∈ B(Y,Z) then we write V T for the composition V T : X → Z defined by
(V T )(x) = V (T (x)) for every x ∈ X.
We will show that our definitions are ‘good’ ones in the results that follow.
Theorem 5.10. The set B(X,Y ) forms a linear space under the operations (5.3) and this isa normed space with the induced norm ‖ · ‖. Furthermore if Y is a Banach space, then so is(B(X,Y ), ‖ · ‖
).
Proof To verify B(X,Y ) is a linear space, we must do a few routine checks. So suppose T,U ∈B(X,Y ) and that λ is a scalar. It is a standard result that T +U and λT are linear mappings. Wemust check that these are both bounded. So,
‖(T + U)(x)‖ 6 ‖Tx‖ + ‖Ux‖ 6 ‖T‖ ‖x‖ + ‖U‖ ‖x‖ =(‖T‖ + ‖U‖
)‖x‖, (5.4)
and‖(λT )(x)‖ = ‖λ(Tx)‖ = |λ| ‖Tx‖ 6 |λ| ‖T‖ ‖x‖. (5.5)
Thus, T + U ∈ B(X,Y ) and λT ∈ B(X,Y ). We must also check that B(X,Y ) is normed. Clearly‖T‖ > 0 and ‖0‖ = 0. If ‖T‖ = 0, then ‖Tx‖ 6 0‖x‖ for every x ∈ X, so T = 0. The triangleinequality follows from (5.4). The scalar property follows from (5.5) noting that
‖λT‖ = sup‖x‖=1
‖λTx‖ = sup‖x‖=1
|λ|‖T (x)‖ = |λ|‖T‖.
To check the rest of the linear space axioms is routine.Now suppose Y is complete and let Tn be a Cauchy sequence in B(X,Y ). So if ε > 0, there
exists n0 ∈ N such that ‖Tn − Tm‖ < ε whenever n,m > n0. If x ∈ X, we have
‖Tnx− Tmx‖ = ‖(Tn − Tm)x‖ 6 ‖Tn − Tm‖ ‖x‖ < ε‖x‖ whenever n,m > n0. (5.6)
Then, since Y is complete, the sequence (Tnx)n is Cauchy in Y and converges to a limit which wecall Tx ∈ Y . It is easy to check that T is a linear operator. Also, if we let m→∞ in (5.6) we getthat if n > n0, then ‖Tnx− Tx‖ < ε‖x‖. Thus
‖Tx‖ 6 ‖Tx− Tnx‖ + ‖Tnx‖ 6(ε+ ‖Tn‖
)‖x‖.
Hence, T is bounded. Moreover, if n > n0 then ‖(T − Tn)x‖ 6 ε‖x‖; that is
‖T − Tn‖ 6 ε whenever n > n0,
so Tn → T in ‖ · ‖.
Lemma 5.11. Let T ∈ B(X,Y ) and let U ∈ B(Y, Z). Then
UT ∈ B(X,Z) and ‖TU‖ 6 ‖T‖ ‖U‖.
THE SPACE OF OPERATORS 21
Proof It is easy to check that UT is linear. If x ∈ X, then
‖(UT )(x)‖ = ‖U(Tx)‖ 6 ‖U‖ ‖Tx‖ 6 ‖U‖ ‖T‖ ‖x‖.
Definition 5.12 (Identity Operator, Invertible Operator). We define the identity oper-ator in B(X) by
I(x) = x for every x ∈ X.We say that T ∈ B(X) is invertible with inverse T−1 iff there exists T−1 ∈ B(X) such thatTT−1 = T−1T = I.
Note that the condition T−1 ∈ B(X) requires T−1 to be bounded as well as being the set-theoreticinverse of T . Since ‖Ix‖ = ‖x‖ for all x it is clear that ‖I‖ = 1.
We give an example of an operator and its inverse:
Example 5.13. Let T : (C[0, 1], ‖ · ‖∞)→ (C[0, 1], ‖ · ‖∞) be such that (Tf)(t) = (t2 + 2)f(t). Ifwe define the operator T−1 by
(T−1g)(t) =g(t)t2 + 2
,
then we see that T−1 : (C[0, 1], ‖ · ‖∞)→ (C[0, 1], ‖ · ‖∞) and that T−1T = I. That is, T−1 is theinverse of T . But we must also check that it is bounded. It is, for |(T−1g)(t)| 6 1
2 |g(t)| and thus
‖T−1g‖∞ 6 12‖g‖∞.
Example 5.14. Let T : (C[0, 1], ‖ · ‖∞) → (C[0, 1], ‖ · ‖∞) be given by (Tf)(t) = (3t2 − 1)f(t).Then T is not invertible (in our sense), since it has no bounded inverse.
Proposition 5.15. Let T,U ∈ B(X). Then TU is invertible iff T and U are both invertible. Theinverse of TU , if it exists, is U−1T−1.
Proof (⇐): If both T and U are invertible, then the product U−1T−1 must be bounded. It is theinverse, since U−1T−1TU = I.
(⇒): If TU is invertible, and V is its inverse, then (TU)V = I = T (UV ). Thus UV = T−1.(Note that right inverses are always equal to left inverses.) Similarly for U−1.
Lemma 5.16. Let T ∈ B(X). Then T is invertible iff both the following hold:
(1) T is a surjection.(2) There exists c > 0 such that ‖Tx‖ > c‖x‖ for every x ∈ X.
Proof (⇒): If T is invertible, then T must be surjective. As T−1 is bounded we must have forsome c′ > 0 that ‖T−1y‖ 6 c′‖y‖ for every y ∈ X. So, taking y = Tx we get ‖x‖ 6 c′‖Tx‖; i.e.,‖Tx‖ > c‖x‖ where c = 1/c′.
(⇐): If ‖Tx‖ > c‖x‖ for every x ∈ X, then
‖Tx− Tx′‖ > c‖x− x′‖.This means that if x 6= x′, then Tx 6= Tx′, so T is injective. T is then bijective since (by hypothesis)it is a surjection. We must now show that the set-theoretic inverse T−1 is linear and bounded. Forlinearity, choose y, y′ ∈ X and let x, x′ be the unique elements of X such that
Tx = y and Tx′ = y′.
Then, since T is linear,
y + y′ = T (x+ x′) giving T−1(y + y′) = T−1y + T−1y′. (5.7)
For a scalar λ,λy = T (λx) so T−1(λy) = λx = λT−1(y). (5.8)
22 5. OPERATORS ON NORMED SPACES
The equations (5.7) and (5.8) imply that T−1 is linear. Finally, T−1 is bounded, for
‖T−1y‖ = ‖x‖ 6 1c‖Tx‖ = 1
c‖y‖.
Proposition 5.17. Let(X, ‖ · ‖
)be a Banach space, and let T ∈ B(X) with ‖T‖ < 1. Then
I − T is invertible and
(I − T )−1 = I + T + T 2 + T 3 + · · · .
This series converges in B(X).
Proof We first prove that the series does actually converge in B(X). First, we have, if n > m:∥∥∥∥∥n∑k=0
T k −m∑k=0
T k
∥∥∥∥∥ =
∥∥∥∥∥n∑
k=m+1
T k
∥∥∥∥∥ 6n∑
k=m+1
‖T k‖
6n∑
k=m+1
‖T‖k 6∞∑
k=m+1
‖T‖k
=‖T‖m+1
1− ‖T‖,
summing a geometric series. Then, given ε > 0, we can choose n0 such that
‖T‖n0+1
1− ‖T‖< ε.
This means( n∑k=0
T k)n
is Cauchy in(B(X), ‖ · ‖
). Since B(X) is complete by Theorem 5.10, this
sequence of partial sums converges. Now we set
U = limn→∞
n∑k=0
T k
and check that this is the inverse of I − T . So,
U(I − T ) =(U − (I + T + T 2 + · · ·+ Tn) + (I + T + T 2 + · · ·+ Tn)
)(I − T
)=(U − (I + T + T 2 + · · ·+ Tn)
)(I − T
)+ I − Tn+1
and hence ‖U(I − T )− I‖ 6∥∥(U − (I + T + T 2 + · · ·+ Tn)
)(I − T
)∥∥+ ‖Tn+1‖6∥∥(U − (I + T + T 2 + · · ·+ Tn)
)∥∥∥∥(I − T )∥∥+ ‖T‖n+1 → 0.
This means ‖U(I − T )− I‖ = 0; that is, U(I − T ) = I. The proof that U is also the right inverseof I − T is similar.
The Spectrum of an Operator
From this point onwards we will take(X, ‖ · ‖
)to be a complex Banach space and consider operators
in B(X).
Definition 5.18 (Eigenvalue, Eigenfunction, Spectrum). A number λ ∈ C is called aneigenvalue of T ∈ B(X) with corresponding eigenvector (or eigenfunction) x ∈ X iff
(T − λI)x = 0 and x 6= 0.
For a given λ, the set of all eigenvectors corresponding to it is called the eigenspace of λ (and iseasily checked to be a subspace of X). The set of all eigenvalues of T is called the point spectrumof T . The spectrum σ(T ) of T is the set
σ(T ) = λ ∈ C : T − λI is not invertible .
THE SPECTRUM OF AN OPERATOR 23
We note that if λ is an eigenvalue of T , then T − λI annihilates its eigenspace (of nonzerovectors), so cannot be injective. Thus T − λI is not invertible and λ is in the spectrum of T .It is not always the case that the spectrum is contained in the set of eigenvalues, though in thefinite-dimensional case it is. We have the following three possibilities for λ ∈ σ(T ):
(1) T − λI is not injective (λ is an eigenvalue);(2) T − λI is injective but not surjective;(3) T − λI is a bijection but (T − λI)−1 is not bounded.
Note that if dim(X) <∞, then (1) and (2) are equivalent and correspond exactly to λ ∈ σ(T ). Ina Banach space, (3) is in fact impossible, though this is quite hard to prove. For finite dimensionalspaces, the theory of eigenvalues and the spectrum does not depend on the norms and is coveredin the basic linear algebra course. Things are rather more subtle for infinite dimensional spaceswhere the role of the norm is crucial.
Example 5.19. Let T : (C[0, 1], ‖ · ‖∞) → (C[0, 1], ‖ · ‖∞) be given by (Tf)(t) = (t2 − 3)f(t).Then we have that (T − λI)f(t) = (t2 − 3 − λ)f(t). If (t20 − 3 − λ) = 0 for some t0 ∈ [0, 1], thenT − λI is not surjective since (T − λI)f(t0) = 0 for every f ∈ C[0, 1]. Now if λ ∈ [−3,−2], thensuch a t0 does exist, so [−3,−2] ⊂ σ(T ). In fact, [−3,−2] = σ(T ), since for λ /∈ [−3,−2],
(T − λI)−1g(t) =g(t)
t2 − 3− λ,
and this is bounded.
Example 5.20. Let T :(l∞, ‖ · ‖∞
)→(l∞, ‖ · ‖∞
)be given by
T : (x1, x2, . . . , ) 7→(x1,
x2
2,x3
22, . . . ,
xi2i−1
, . . .
).
Then the set Λ =
1, 12 ,
122 , . . .
is a set of eigenvalues since
T (0, . . . , 0, 1,↑
nth place
0, . . .) =1
2n−1(0, . . . , 0, 1, 0, . . .).
If λ /∈ Λ and λ 6= 0, then (T − λI)(x1, x2, . . .) =((1− λ)x1, ( 1
2 − λ)x2, . . .), so
(T − λI)−1(y1, y2, . . .) =(
y1
1− λ,
y212 − λ
, . . .
).
This operator will be bounded if supn=0,1,2,...
∣∣∣∣ 1(1/2n)− λ
∣∣∣∣ <∞, which is the case if λ /∈ Λ and λ 6= 0.
Thus the spectrum is σ(T ) = Λ ∪ 0.
Theorem 5.21. Let T ∈ B(X) where(X, ‖ · ‖
)is a Banach space. Then
(1) σ(T ) is bounded; in fact if λ ∈ σ(T ) then |λ| 6 ‖T‖;(2) σ(T ) is a closed subset of C;(3) σ(T ) 6= ∅.
Proof (1): Let |λ| > ‖T‖. We show that this would imply λ /∈ σ(T ). Note that (T − λI) =−λ(I − 1
λT ). Now, ∥∥∥∥Tλ∥∥∥∥ =
1|λ|‖T‖ < 1 since |λ| > ‖T‖.
Hence, by Proposition 5.17 we have that
(T − λI)−1 = − 1λ
(I − T
λ
)−1
(since this inverse exists).
So T − λI is invertible and thus λ /∈ σ(T ).(2): We show that the complement C \ σ(T ) is open. Let λ0 /∈ σ(T ). We show that if λ is
close enough to λ0, then λ /∈ σ(T ). Note that T − λ0I is invertible. Also, identically,
(T − λI) = (T − λ0I)(I − (T − λ0I)−1(λ− λ0)
). (5.9)
24 5. OPERATORS ON NORMED SPACES
Again by Proposition 5.17 we have that
if |λ− λ0| <1
‖(T − λ0I)−1‖, then (5.9) is invertible.
Then, by definition C \ σ(T ) is open, so σ(T ) is closed.(3): (Sketch Proof) Replacing λ by z in (5.9) and taking inverses, we get
(T − zI)−1 =(I − (T − z0I)−1(z − z0)
)−1
︸ ︷︷ ︸Ω
(T − z0I)−1.
If z0 /∈ σ(T ) then for any z close to z0 we have a convergent power series expression (by Proposition5.17) for the bracket Ω, thus
(T − zI)−1 =(I + (T − z0I)−1(z − z0) + (T − z0I)−2(z − z0)2 + · · ·
)(T − z0I)−1.
Hence if z0 /∈ σ(T ) then (T − zI)−1 has a convergent power series in z about z0: this means that(T − zI)−1 is analytic for z ∈ C near z0. If we assume, for a contradiction, that σ(T ) = ∅, then(T − zI)−1 is analytic in C. But by (1) we have that
(T − zI)−1 = −1z
(I − T
z
)−1
,
for large z, so the function ‖(T − zI)−1‖ is bounded. By an analogue of Liouville’s theorem incomplex analysis (which states that a function that is analytic and bounded on C is constant), weconclude that (T − zI)−1 is constant, which it clearly is not. Thus σ(T ) 6= ∅.
Theorem 5.22 (Spectral Mapping Theorem). Let p(t) be a polynomial and let T ∈ B(X).Then
σ(p(T )
)= p(σ(T )
).
Proof. Let p(z) = c(z − z1)(z − z2) · · · (z − zn) where c 6= 0. Then
0 /∈ σ(p(T ))⇔ (p(T ))−1 exists
⇔(c(T − z1I)(T − z2I) · · · (T − znI)
)−1 exists
⇔ (T − ziI)−1 exists for each i = 1, . . . , n
⇔ z1, z2, . . . , zn /∈ σ(T )
⇔ 0 /∈ p(σ(T )). (5.10)
To complete the proof,
λ /∈ σ(p(T ))⇔ (p(T )− λI)−1 exists i.e.,((p(T )− λI)− 0I
)−1 exists
⇔ 0 /∈ σ(p(T )− λI)
⇔ 0 /∈ p(σ(T ))− λ by (5.10)
⇔ λ /∈ p(σ(T )).
CHAPTER 6
Inner Products and Hilbert Space
Basic Ideas
Definition 6.1. An inner product space is a vector space X over R or C, together with real orcomplex inner product 〈x, y〉 defined for all y, x ∈ X, such that
(1) 〈λx+ µy,w〉 = λ〈x,w〉+ µ〈y, w〉 for every x, y, w ∈ X and all scalars µ, λ;
(2) 〈x, y〉 = 〈y, x〉 (where the bar denotes complex conjugation);(3) 〈x, x〉 is real, 〈x, x〉 > 0, and 〈x, x〉 = 0 iff x = 0.
We write(X, 〈 · , · 〉
)to denote the space and its inner product.
Note that (1) and (2) above imply that
〈w, λx+ µy〉 = λ〈w, x〉+ µ〈w, y〉,
and
〈x+ y, w + z〉 = 〈x, z〉+ 〈x,w〉+ 〈y, z〉+ 〈y, w〉.
We can easily turn X into a normed space by setting
‖x‖ =√〈x, x〉. (6.1)
To prove that this is a norm, we need the following well-known and important result:
Lemma 6.2 (Schwarz’ Inequality). Let x, y ∈ X where X is an inner product space withnormgiven by (6.1). Then
|〈x, y〉| 6 ‖x‖ ‖y‖. (Schwarz’ Inequality)
Proof Assume y 6= 0. Then for all scalars λ,
0 6 〈x− λy, x− λy〉 = 〈x, x〉 − λ〈y, x〉 − λ〈x, y〉+ λλ〈y, y〉.
In particular we can choose λ = 〈x, y〉/〈y, y〉, which gives
0 6 〈x, x〉 − 〈x, y〉〈x, y〉〈y, y〉
− 〈x, y〉〈x, y〉〈y, y〉
+|〈x, y〉|2
〈y, y〉,
so
0 6 ‖x‖2 − |〈x, y〉|2
‖y‖2,
and rearranging gives the Schwarz inequality.
Definition 6.3 (Hilbert Space). let(X, 〈 · , · 〉
)be an inner product space with norm on X is
given by (6.1). Then we call(X, 〈 · , · 〉
)a Hilbert space if X is complete when equipped with this
norm.
25
26 6. INNER PRODUCTS AND HILBERT SPACE
Examples of Hilbert Spaces
Example 6.4. The simplest example of a Hilbert space is RN with the usual scalar product
〈(x1, x2, . . . , xN ), (y1, y2, . . . , yN )〉 =N∑i=1
xiyi.
The norm is the ‖ · ‖2-norm.
Example 6.5. The space l2 over either R (or alternatively C)
l2 =
(x1, x2, . . .) : xi ∈ R (orC) and
∞∑i=1
|xi|2 <∞
is a complete inner product space with the inner product
〈(x1, x2, . . .), (y1, y2, . . .)〉 =∞∑i=1
xiyi.
Example 6.6. The space C[0, 1] of continuous functions becomes an inner product space with
〈f, g〉 =∫ 1
0
f(t)g(t) dt (6.2)
and the norm is the ‖ · ‖2-norm. This space is not complete.
However the completion of this space is illustrated in the next example:
Example 6.7. The space L2[0, 1] of square-integrable functions is a Hilbert space with innerproduct
〈f, g〉 =∫ 1
0
f(t)g(t) dt (6.3)
which gives the ‖ · ‖2-norm.
We can extend this last inner product to include ‘weighted’ inner products.
Example 6.8. Let w be a real positive continuous ‘weight function’ . Let
H =f ∈ L2[0, 1] :
∫ 1
0
w(u)|f(u)|2 du <∞
and set
〈f, g〉 =∫ 1
0
w(u)f(u)g(u) du.
Then H is a Hilbert space under this inner product.
Orthonormal Systems
Lemma 6.9 (Parallelogram Law). For every x, y in an inner product space,
‖x+ y‖2 + ‖x− y‖2 = 2‖x‖2 + 2‖y‖2.
Proof
‖x+ y‖2 + ‖x− y‖2 = 〈x+ y, x+ y〉+ 〈x− y, x− y〉= 〈x, x〉+ 〈y, y〉+ 〈x, y〉+ 〈y, x〉
+ 〈x, x〉+ 〈y, y〉 − 〈x, y〉 − 〈y, x〉= 2‖x‖2 + 2‖y‖2.
ORTHONORMAL SYSTEMS 27
Definition 6.10 (Orthogonal, Orthonormal). Let(H, 〈 · , · 〉
)be a Hilbert space. We say
that x, y ∈ H are orthogonal or perpendicular if 〈x, y〉 = 0. A set of elements eii in H is calledorthonormal if 〈ei, ei〉 = ‖ei‖2 = 1 for all i and 〈ei, ej〉 = 0 if i 6= j.
Definition 6.11 (Span, Complete Orthonormal System). The span of a set xi of vectorsinH, written span xi, is the set of all finite linear combinations of the xi, that is
∑ni aiei : ai ∈ R.
If ei is an orthonormal set and span ei is dense in H, then we call ei a complete orthonormalsystem.
(Note that ‘complete’ in this context is not the same as the completeness property of a normedspace.)
Example 6.12 (Two Complete Orthonormal Systems). The following orthonormal systemsunderly the theory of Fourier series:
(1) The set 1∪√
2 cos(2πjt)∞j=1∪√
2 sin(2πjt)∞j=1
is a complete orthonormal system
in L2[0, 1].(2) The set
e2πint
∞n=−∞ is a complex complete orthonormal system.
Here is a useful check for a vector to be zero depending on orthogonality.
Lemma 6.13. Let S be a dense subset of a Hilbert space H. Let x ∈ H and suppose that 〈x, s〉 = 0for all s ∈ S. Then x = 0. In particular, if ei is an complete orthonormal set and 〈x, ei〉 = 0 forall i then x = 0.
Proof If S is dense in H, then we can find (xn)n ∈ S such that xn → x. Then,
0 6 |〈x, x〉| = |〈x− xn, x〉+ 〈xn, x〉| 6 |〈x− xn, x〉+ 0| (since xn ∈ S)
(by Schwarz) 6 ‖x− xn‖ ‖x‖ → 0.
So 〈x, x〉 = 0 and hence x = 0.
Theorem 6.14 (Parseval’s Equality). Let ei be a (countable) complete orthonormal systemin the Hilbert space H. Then if x ∈ H, we have
x =∞∑i=1
〈x, ei〉ei,
with this series convergent in H. Also,
‖x‖2 =∞∑i=1
∣∣〈x, ei〉∣∣2. (Parseval’s Equality)
Proof Let x ∈ H. Write
xn =n∑i=1
〈x, ei〉ei.
Then〈x− xn, ei〉 = 〈x, ei〉 − 〈xn, ei〉 = 0 where i = 1, 2, . . . , n.
This gives 〈x− xn, xn〉 = 0 since xn is a linear combination of e1, . . . , en. Then
‖x‖2 = 〈x, x〉= 〈(x− xn) + xn, (x− xn) + xn〉= ‖x− xn‖2 + ‖xn‖2 + 〈x− xn, xn〉︸ ︷︷ ︸
=0
+ 〈xn, x− xn〉︸ ︷︷ ︸=0
;
28 6. INNER PRODUCTS AND HILBERT SPACE
hence ‖x‖2 > ‖xn‖2. Thus for each n we haven∑i=1
∣∣〈x, ei〉∣∣2 = ‖xn‖2 6 ‖x‖2,
which means that∑ni=1 |〈x, ei〉|2 is convergent.
Moreover, suppose ε > 0, then if n and m are large enough,
‖xn − xm‖2 =
∥∥∥∥∥n∑
i=m+1
〈x, ei〉ei
∥∥∥∥∥2
=n∑
i=m+1
∣∣〈x, ei〉∣∣2 < ε,
since∑∞i=1 |〈x, ei〉|2 converges. Since H is complete, there must be y ∈ H such that xn → y.
Hence, for each i,〈x− y, ei〉 = 〈x− xn, ei〉︸ ︷︷ ︸
=0 if n>i
+ 〈xn − y, ei〉︸ ︷︷ ︸→0 by Schwarz
→ 0.
Thus 〈x− y, ei〉 = 0 for every i. Since ei is complete, the previous lemma gives x = y, and
‖x‖2 = limn→∞
‖xn‖2 =∞∑i=1
∣∣〈x, ei〉∣∣2.
If we drop the requirement in Theorem 6.14 that the set ei is complete (i.e., that its span isdense in H), we get Bessel’s Inequality :
∞∑i=1
∣∣〈x, ei〉∣∣2 6 ‖x‖2. (Bessel’s Inequality)
Example 6.15 (Fourier Series). Recall that
spane2πint
∞n=−∞
is a complete o.n. system in L2[0, 1].
Thus if f ∈ L2[0, 1] we get
f(t) =∞∑
n=−∞
(∫ 1
0
f(t)e−2πint dt)e2πint,
and
∫ 1
0
|f(t)|2 dt =∞∑
n=−∞
∣∣∣∣∫ 1
0
f(t)e−2πint dt∣∣∣∣2.
The following theorem is of theoretical significance and its proof provides a computationalmethod for finding orthonormal bases.
Theorem 6.16 (Gram-Schmidt Orthonormalisation). Let x1, x2, . . . be a linearly inde-
pendent set in H (i.e. if∑N
1 λixi = 0 then λi = 0 for each i). Then we may find an orthonormalsequence e1, e2, . . . ⊂ H such that for every n = 1, 2, . . ., we have
span e1, e2, . . . = span x1, x2, . . . .
Proof Lete1 =
x1
‖x1‖.
Then define the remaining terms in the sequence by the formula
en+1 =xn+1 −
∑ni=1〈xn+1, ei〉ei
‖xn+1 −∑ni=1〈xn+1, ei〉ei‖
.
Clearly ‖ei‖ = 1 for each i, and 〈en, ei〉 = 0 for i 6= n.
ORTHONORMAL SYSTEMS 29
Definition 6.17 (Basis). An orthonormal set ei is called a basis for H if for every x ∈ H wemay find scalars a1, a2, . . . such that
x =∞∑i=1
aiei,
and the sum converges (i.e.,∑i |ai|2 <∞).
Corollary 6.18 (to Gram-Schmidt Theorem). If H is a Hilbert space and there exists asequence (xn)n in H such that span x1, x2, . . . is dense in H, then H has an orthonormal basis.
Proof The Gram-Schmidt process gives an orthonormal set ei with span ei = span xi whichmust be dense in H. By Theorem 6.14 the set ei is a basis.
CHAPTER 7
The Spectral Theorem
The aim of this chapter is to prove the spectral theorem for compact self-adjoint operators on aninfinite dimensional complex Hilbert space H. This should be thought if as being analogous todiagonalising symmetric matrices by finding a suitable basis.
Definition 7.1 (Self-adjoint Operator). An operator T ∈ B(H) is self-adjoint or Hermitianiff
〈Tx, y〉 = 〈x, Ty〉 for all x, y ∈ H.
(Self-adjoint operators correspond to symmetric or Hermitian matrices.) Note that if T isself-adjoint then 〈Tx, x〉 is real, since 〈Tx, x〉 = 〈x, Tx〉 = 〈Tx, x〉.
Example 7.2. The following operators are easily seen to be self-adjoint.
(1) T : l2 → l2 given by T (x1, x2, x3, . . .) = (x1,12x2,
13x3, . . .).
(2) T : L2[0, 1]→ L2[0, 1] given by (Tf)(t) = f(t)(t3 + 3).(3) T : L2[0, 1]→ L2[0, 1] given by (Tf)(t) =
∫ 1
0f(u)(t2 + u2) du.
Lemma 7.3. If T is self-adjoint then
(1) the eigenvalues of T are all real,(2) eigenvectors corresponding to different eigenvalues are orthogonal.
Proof (1): Suppose that Tx = λx. Then
λ〈x, x〉 = 〈λx, x〉 = 〈Tx, x〉 = 〈x, Tx〉 = 〈x, λx〉 = λ〈x, x〉.If x 6= 0 then λ = λ, and hence λ is real.
(2): Suppose that Tx = λx and Ty = µy. Then
λ〈x, y〉 = 〈λx, y〉 = 〈Tx, y〉 = 〈x, Ty〉 = 〈x, µy〉 = µ〈x, y〉 = µ〈x, y〉,since µ is real. Thus if λ 6= µ then 〈x, y〉 = 0.
It is useful to have an alternative form for the norm of a self-adjoint operator:
Lemma 7.4. Let T be self-adjoint. Then ‖T‖ = sup‖x‖=1
|〈Tx, x〉|.
Proof Write m = sup‖x‖=1
|〈Tx, x〉|. First note that if ‖x‖ = 1 then
|〈Tx, x〉| 6 ‖Tx‖ ‖x‖ 6 ‖T‖ ‖x‖ ‖x‖ = ‖T‖using Schwarz’ inequality; so m 6 ‖T‖.
For the opposite inequality, let x be such that ‖x‖ = 1 and Tx 6= 0, and write y = Tx/‖Tx‖so that ‖y‖ = 1. Then 〈Tx, y〉 = 〈Tx, Tx〉/‖Tx‖ = ‖Tx‖. By expanding the inner product andnoting that m = sup
‖z‖6=0
|〈Tz, z〉|/‖z‖2, we have
‖Tx‖ = 〈Tx, y〉 = 12
(〈Tx, y〉+ 〈y, Tx〉
)= 1
2
(〈Tx, y〉+ 〈Ty, x〉
)= 1
4
(〈T (x+ y), (x+ y)〉 − 〈T (x− y), (x− y)〉
)6 1
4m(‖x+ y‖2 + ‖x− y‖2
)= 1
4m× 2(‖x‖2 + ‖y‖2) = m,
30
7. THE SPECTRAL THEOREM 31
using the parallelogram law. Thus ‖T‖ 6 m as required.
We now turn our attention to compact operators, which may be thought of as operators withimages that are ‘nearly finite-dimensional’.
Definition 7.5 (Compact Operator). An operator T ∈ B(H) is compact if for every boundedsequence (xn) the sequence (Txn) has a convergent subsequence.
In a finite dimensional space every bounded sequence has a convergent subsequence so anyoperator on a finite dimensional space is necessarily compact. Checking that operators on infinitedimensional spaces are compact can be more awkward. For the Hilbert space L2 the followingproposition is helpful.
Proposition 7.6. Let T : L2[0, 1]→ L2[0, 1] be such that the following ‘equicontinuity condition’holds: for all ε > 0 there exists δ > 0 such that |Tf(x) − Tf(y)| < ε whenever |x − y| < δ for allf ∈ L2[0, 1] such that ‖f‖2 6 1. Then T is compact.
Proof This follows from the Arzela-Ascoli theorem which gives a condition for a family of continu-ous functions to have a convergent subsequence. The proof of this is technical but not particularlydifficult, see, for example, the book by Griffel.
Example 7.7. The following operators are compact:
(1) T : l2 → l2 given by T (x1, x2, x3, . . .) = (x1,12x2,
13x3, . . .).
(2) T : L2[0, 1]→ L2[0, 1] given by (Tf)(t) =∫ 1
0f(u)(t2 + u2) du.
(3) T : L2[0, 1]→ L2[0, 1] given by (Tf)(t) =∫ t
0f(u)(u2 + 1) du.
Proposition 7.6 may be sued to verify that (2) and (3) are compact.
Lemma 7.8. If T is compact and self-adjoint and r > 0, then there are at most finitely manyeigenvalues (counted according to multiplicity of independent eigenvectors) such that |λ| > r.
Proof Suppose to the contrary. Then we may find an infinite sequence of eigenvectors (ei) andcorresponding eigenvalues λi all with |λi| > r. For each distinct eigenvalue we may select a basisof the corresponding eigenspace consisting of orthonormal eignevectors (using the Gram-Schmidtprocess), so together with Lemma 7.3 (2) this means we can assume that the infinite sequence (ei)is orthonormal. Then for i 6= j we have
〈Tei, T ej〉 = 〈λiei, λjej〉 = λiλj〈ei, ej〉 = 0,
so expanding,
‖Tei − Tej‖2 = 〈Tei − Tej , T ei − Tej〉 = ‖Tei‖2 − 〈Tei, T ej〉 − 〈Tej , T ei〉+ ‖Tej‖2
= |λi|2 − 0− 0 + |λj |2 > 2r2.
Thus (Tei)i cannot have a convergent subsequence (since no subsequence can be a Cauchy se-quence), which contradicts the compactness of T .
The key property that we need in order to prove the spectral theorem is that the operator hasat least one non-zero eignevalue. We show this (and more) in the following lemma.
Lemma 7.9. Let T be a compact self-adjoint operator on H. Then either ‖T‖ or −‖T‖ is aneigenvalue of T .
Proof Assume that ‖T‖ > 0 (otherwise the result is trivial). Writing m = sup‖x‖=1
|〈Tx, x〉| = ‖T‖
by Lemma 7.4, we show that T has an eigenvalue λ with |λ| = m . Noting that 〈Tx, x〉 is real,suppose that sup
‖x‖=1
〈Tx, x〉 = m. (A symmetrical argument applies if inf‖x‖=1
〈Tx, x〉 = −m.) Choose
32 7. THE SPECTRAL THEOREM
a sequence (xn) with ‖xn‖ = 1 such that 〈Txn, xn〉 → m. Then expanding
‖Txn −mxn‖2 = 〈Txn −mxn, Txn −mxn〉= ‖Txn‖2 − 2m〈Txn, xn〉+m2‖xn‖2
6 ‖T‖2‖xn‖2 − 2m〈Txn, xn〉+m2‖xn‖2
6 2m2 − 2m〈Txn, xn〉 → 0 as n→∞.
Since T is compact, we may assume, by taking a subsequence if necessary, that Txn → y forsome y 6= 0. Then
mxn = Txn + (mxn − Txn)→ y + 0.Thus xn → y/m and by continuity Txn → (Ty)/m so (Ty)/m = y giving that m is an eigenvaluewith eigenvector y.
We put together the above properties to get the spectral theorem.
Theorem 7.10 (Spectral theorem for compact self-adjoint operators). Let T be acompact self-adjoint operator on H. Then there is an orthonormal sequence (en) of eigenvectorsin H, with Ten = λnen, say, with λn real, λn 6= 0 and λn → 0, such that for all x ∈ H we maywrite
x =∞∑n=1
anen + y,
where an = 〈x, en〉, with∑∞n=1 |an|2 < ∞, and y ∈ H satisfying 〈y, en〉 = 0 for all n and Ty = 0.
In particular
Tx =∞∑n=1
anλnen.
Proof For each eigenvalue λ 6= 0 let Hλ be the eigenspace Hλ = x : Tx = λx. By Lemma 7.8,each Hλ has finite dimension so has a finite orthonormal basis (by the Gram-Schmidt process).Let en be the aggregate of the basis vectors obtained in this way from all λ 6= 0; by Lemma 7.3(2) the en are all orthonormal (though they need not form a basis of H).
Suppose Ten = λnen for each n. As T is self-adjoint the λn are real (Lemma 7.3 (1)) and sinceT is compact λn → 0 (Lemma 7.8). Given x ∈ H, we may write
x =∞∑n=1
anen + y where an = 〈x, en〉
for some y ∈ H; since the en are orthonormal∑∞n=1 |an|2 <∞ and the series converges by Bessel’s
inequality. Taking inner products with en we see that 〈y, en〉 = 0 for all n.To show that Ty = 0, let H0 = y : 〈y, en〉 = 0 for all n. If H0 = 0 there is nothing left to
show. Otherwise, H0 is closed so may itself be regarded as a Hilbert space. Moreover, if y ∈ H0
then〈Ty, en〉 = 〈y, Ten〉 = 〈y, λnen〉 = λn〈y, en〉 = 0,
so Ty ∈ H0. Thus T maps H0 into itself, so we may regard T as a compact self-adjoint operator
on H0. By Lemma 7.9 there is an eigenvalue λ with |λ| = sup06=x∈H0
‖Tx‖‖x‖
and a (nonzero) eigen-
vector w ∈ H0 such that Tw = λw. But w ∈ H0 is orthogonal to all eigenvectors with non-zeroeigenvalues, so λ = 0, and hence Ty = 0 for all y ∈ H0.
We conclude that
Tx =∞∑n=1
anλnen.
To check this formally, note that∥∥∥∥∥Tx−N∑n=1
anλnen
∥∥∥∥∥ =
∥∥∥∥∥T(x−
N∑n=1
anen
)∥∥∥∥∥ =
∥∥∥∥∥T(y +
∞∑n=N+1
anen
)∥∥∥∥∥
STURM-LIOUVILLE THEORY 33
=
∥∥∥∥∥T( ∞∑n=N+1
anen
)∥∥∥∥∥ 6 ‖T‖∥∥∥∥∥∞∑
n=N+1
anen
∥∥∥∥∥ = ‖T‖∞∑
n=N+1
|an|2 → 0.
Corollary 7.11. Let T be a compact self-adjoint operator on H. Then H has an orthonormalbasis of eigenvectors of T .
Proof Take en as in the spectral theorem together with an orthonormal basis of H0.
If T is compact self-adjoint and x =∞∑n=1
anen+y, then applying the spectral theorem repeatedly
gives that
T kx =∞∑n=1
λknanen.
In particular, if λ1 is the unique largest eigenvalue (in absolute value), then
T kx
λk1→ a1e1 and
T kx
‖T kx‖→ e1,
giving an iterative method of finding the eigenvector corresponding to the largest eigenvalue λ1.If T is compact self-adjoint and λ 6= 0 is not an eigenvalue, then we may solve (T − λI)x = w
given w =∑∞n=1 anen + y, by noting that
(T − λI)( ∞∑n=1
anenλn − λ
− y
λ
)=∞∑n=1
anen + y.
This solution is in H since ∥∥∥∥∥∞∑n=1
anenλn − λ
∥∥∥∥∥ =∞∑n=1
|an|2
|λn − λ|26 c
∞∑n=1
|an|2
where c = maxn 1/|λn − λ|2 <∞.
Sturm-Liouville Theory
Sturm-Liouville theory applies the spectral theorem to the eigenvalue structure of linear dif-ferential equations.
We illustrate this with the eigenvalue problem for the homogeneous differential equation onthe interval [0, 1] with boundary conditions:
y′′ + µy = 0 y(0) = y(1) = 0 (7.1)
where µ is a number, and its inhomogeneous counterpart:
y′′ + f = 0 y(0) = y(1) = 0 (7.2)
for a given function f . We proceed using several steps.
(A): Define a continuous function g by
g(s, t) =
s(1− t) 0 6 s 6 tt(1− s) t 6 s 6 1
where s, t ∈ [0, 1]. (7.3)
Then g is called the Green’s function for the problem (7.2). We check that integrating f(s) ∈ C[0, 1]against g(s, t) solves (7.2). Let
y(t) =∫ 1
0
g(s, t)f(s) ds
=∫ t
0
s(1− t)f(s) ds+∫ 1
t
t(1− s)f(s) ds.
34 7. THE SPECTRAL THEOREM
Clearly y(0) = y(1) = 0. Differentiating:
y′(t) = t(1− t)f(t)−∫ t
0
sf(s) ds− t(1− t)f(t) +∫ 1
t
(1− s)f(s) ds
= −∫ t
0
sf(s) ds+∫ 1
t
(1− s)f(s) ds,
y′′(t) = −tf(t)− (1− t)f(t)
= −f(t).
(B): With g as in (7.3) we define the Green’s function operator G by
(Gf)(t) =∫ 1
0
g(s, t)f(s) ds. (7.4)
We also have the differential operator L, with
(Ly)(t) = −y′′(t).
We need to consider the spaces on which these operators act. As well as using C[0, 1] and L2[0, 1],we introduce C2
B [0, 1] for the space of functions f on [0, 1] with continuous second derivative whichsatisfy the boundary conditions f(0) = f(1) = 0.
It follows from (A) that G : C[0, 1] → C2B [0, 1] and also that LG : C[0, 1] → C[0, 1] is the
identity on C[0, 1] (since LG(f) = −(Gf)′′ = f for f ∈ C[0, 1]).Moreover L : C2
B [0, 1] → C[0, 1] is injective (since if 0 = Ly = −y′′ then y(t) = at + b, so ify(0) = y(1) = 0 then y(t) = 0), and surjective (since LG, and thus L, maps onto the whole ofC[0, 1]).
Thus L : C2B [0, 1]→ C[0, 1] is invertible with inverse G, that is LG = IC[0,1] and GL = IC2
B [0,1],and we have the picture:
C[0, 1]G−→←−L
C2B [0, 1].
Note that G is a much easier operator to work with than L since G is bounded and L is not.
(C): We may also regard G as a bounded operator on the Hilbert space L2[0, 1]. We claim that
G : L2[0, 1] −→ C[0, 1] ⊂ L2[0, 1].
To see this note that if f ∈ L2[0, 1] then by Cauchy’s inequality∣∣Gf(t)∣∣ =
∣∣∣∣ ∫ 1
0
g(s, t)f(s) ds∣∣∣∣ 6 (∫ 1
0
g(s, t)2 ds)1/2(∫ 1
0
f(s)2 ds)1/2
(7.5)
and ∣∣Gf(t)−Gf(u)∣∣ =
∣∣∣∣ ∫ 1
0
(g(s, t)− g(s, u)
)f(s) ds
∣∣∣∣6
(∫ 1
0
∣∣g(s, t)− g(s, u)∣∣2 ds
)1/2(∫ 1
0
f(s)2 ds)1/2
,
(7.6)
so Gf(t) is bounded and continuous. From (7.5)
‖Gf‖2 6 ‖Gf‖∞ 6(∫ 1
0
g(s, t)2 ds)1/2
‖f‖2.
Theorem 7.12. The map G, regarded as a bounded operator on L2[0, 1], is compact and self-adjoint.
Proof (Compact): If ‖f‖2 6 1 and ε > 0, then by (7.6),∣∣Gf(t)−Gf(u)∣∣ 6 (∫ 1
0
∣∣g(s, t)− g(s, u)∣∣2 ds
)1/2
6 sups∈[0,1]
∣∣g(s, t)− g(s, u)∣∣ < ε
STURM-LIOUVILLE THEORY 35
provided |t−u| is small enough, since g is continuous. By the equicontinuity criterion for a compactoperator (Proposition 7.6 above), G is compact.
(Self-adjoint): Since g(s, t) is real, we have for f, h ∈ L2[0, 1] that
〈Gf, h〉 =∫ 1
0
Gf(t)h(t) dt =∫ 1
0
(∫ 1
0
g(s, t)f(s) ds)h(t) dt =
∫ 1
0
∫ 1
0
f(s)g(s, t)h(t) dsdt
=∫ 1
0
f(s)(∫ 1
0
g(s, t)h(t) dt)
ds = 〈f,Gh〉,
so G is self-adjoint.
(D): We now apply the spectral theorem to the compact self-adjoint operator G on the Hilbertspace L2[0, 1].
Theorem 7.13. We have the following:
(1) The number 0 is not an eigenvalue of G.(2) The space L2[0, 1] has an orthonormal basis fn of eigenfunctions ofG, withGfn = λnfn,
where λn are real and |λn| → 0 as n→∞. Moreover fn ∈ C2B [0, 1] for all n.
Proof (1): Suppose that Gf = 0 for some f ∈ L2[0, 1]. If h ∈ C[0, 1], then 0 = 〈Gf, h〉 = 〈f,Gh〉.But by (B), Gh : h ∈ C[0, 1] = C2
B [0, 1], which is a space that is ‖ · ‖2-dense in L2[0, 1]. (Thismay be established by showing that C2
B [0, 1] is dense in C[0, 1] which in turn is dense in L2[0, 1]in the ‖ · ‖2-norm.) Thus f = 0, so 0 is not an eigenvalue of G.
(2): Applying the spectral theorem to G on L2[0, 1] the first part of (2) follows immediately,since by (1), y : Ty = 0 = 0 (in the notation used in the spectral theorem), so that theeigenvectors with non-zero eigenvalues form a basis.
If fn ∈ L2[0, 1] is an eigenfunction of G then
λnfn = Gfn ∈ C[0, 1], since G : L2[0, 1]→ C[0, 1],
so fn ∈ C[0, 1] since λn 6= 0, and
λnfn = Gfn ∈ C2B [0, 1], since G : C[0, 1]→ C2
B [0, 1],
so fn ∈ C2B [0, 1].
(E): We now transfer these results to the differential operator L : C2B [0, 1]→ C[0, 1].
We call µ an eigenvalue of L with eigenfunction f if Lf = µf where 0 6= f ∈ C2B [0, 1]. Note
in particular that the eigenfunctions of L satisfy the boundary conditions.
Lemma 7.14. The number λ is an eigenvalue of G : L2[0, 1]→ L2[0, 1] with eigenfunction f if andonly if µ = 1/λ is an eigenvalue of L : C2
B [0, 1]→ C[0, 1] with eigenfunction f .
Proof (⇒): If Gf = λf with 0 6= f ∈ L2[0, 1], then f ∈ C2B [0, 1] and λ 6= 0 by (D). Thus
f = LGf = L(λf) = λLf , so λ−1 is an eigenvalue of L.(⇐): If Lf = µf with 0 6= f ∈ C2
B [0, 1], then µ 6= 0 (since L is injective; see (B)) andf = GLf = G(µf) = µGf , so µ−1 is an eigenvalue of G.
(F): The Sturm-Liouville Theorem collects together these conclusions in the context of differentialequations.
Theorem 7.15 (Sturm-Liouville Theorem). The eigenvalue problem (7.1), which may bewritten as
Ly = µy (y(0) = y(1) = 0),has countably many eigenvalues µn 6= 0, all real, with |µn| → ∞. For each eigenvalue there isa unique eigenfunction (to within a scalar multiple) fn ∈ C2
B [0, 1], which may be taken to be areal-valued function, with Lfn = µnfn. These eigenfunctions form an orthonormal basis of L2[0, 1].
36 7. THE SPECTRAL THEOREM
Proof Most of this is immediate from (D) translated from G to L using the lemma in (E).To see that we may assume that the eigenfunctions are real, note that if µf = Lf = −f ′′ then,
since µ is real and L maps real functions to real functions, we may take the real and imaginaryparts Re f and Im f as two separate real eigenfunctions, with µ (Re f) = L(Re f) = −(Re f)′′ andµ (Im f) = L(Im f) = −(Im f)′′.
Finally, to see that the eigenfunction for each eigenvalue µ is essentially unique, note that theinitial value problem µf = Lf = −f ′′ with f(0) = 0 and f ′(0) = a has a unique solution. (Oneway to see this is to observe that the solution must satisfy f(t) = µ
∫ t0f(s)(s− t) ds+ at and show
uniqueness by a contraction mapping method.) Thus any non-zero solution of the boundary valueproblem µf = Lf = −f ′′ with f(0) = f(1) = 0 is uniquely determined by the value of f ′(0) withall solutions scalar multiples of each other.
For our original problem (7.1), elementary calculus gives the eigenvalues as λ−1n = µn = n2π2
with orthonormal eigenfunctions fn(t) =√
2 sin(πnt), and these form a basis for L2[0, 1] by theSturm-Liouville Theorem. Thus, going back to the spectral theorem, we may express every f ∈L2[0, 1] as
f(t) =∞∑n=1
2(∫ 1
0
f(s) sin(πns) ds)
sin(πnt)
where this ‘Fourier series’ converges in the ‖ · ‖2-norm.
(G): The importance of Sturm-Liouville theory is that all the above applies to every secondorder linear differential equation boundary value problem. The reader may have noticed that afterparagraph (A) we worked entirely with the operators G and L (until we specialised again at theend). If (7.1) is replaced by
a(t)y′′(t) + b(t)y′(t) + c(t)y + µy = 0 with y(0) = y(1) = 0 (7.7)
(subject to natural requirements on a(t), b(t), c(t) such as continuity and differentiability), then wedefine the differential operator
(Ly)(t) = −(a(t)y′′(t) + b(t)y′(t) + c(t)y
).
We may solve the initial value problem (by turning it into an integral equation) to find real valuedfunctions u and v such that
Lu = 0 with u(0) = 0, u′(0) = 1 and Lv = 0 with v(1) = 0, v′(1) = 1.
We then define the Green’s function g(s, t) by
g(s, t) =
cu(s)v(t) 0 6 s 6 tcv(s)u(t) t 6 s 6 1
where s, t ∈ [0, 1],
and where c is a suitable normalising constant. The Green’s function operator G is then definedjust as in (7.4). Then L and G operate on exactly the same spaces as before and satisfy the samerelationships LG = IC[0,1] and GL = IC2
B [0,1]. The theory goes through in just the same way, andthe statement of the Sturm-Liouville theorem is exactly the same with (7.1) replaced by (7.7). Wealso note that this theory can work for different sets of boundary conditions, e.g., y′(0) = y′(1) = 0.