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Lecture Notes on Functional Analysis
Review of Notation and Solutions to Homework Problems
Alberto Bressan
Review of main notation
R the field of real numbers.
C the field of complex numbers.
K a field of numbers, either R or C.
Re z, Imz the real and imaginary part of a complex number z.
z = a− ib the complex conjugate of the number z = a+ ib ∈ C.
[a, b] a closed interval, ]a, b[ an open interval, ]a, b], [a, b[ half-open intervals.
Rn the n-dimensional Euclidean space.
〈·, ·〉 scalar product on the Euclidean space Rn.
|v| .=√
(v, v) the Euclidean length of a vector v ∈ Rn.
A \B .= x ∈ A , x /∈ B a set-theoretic difference.
A the closure of a set A.
∂A the boundary of a set A.
χA
the indicator function of a set A. χA
(x) =
1 if x ∈ A ,0 if x /∈ A .
f : A 7→ B a mapping from a set A into a set B,
a 7→ b = f(a) the function f maps the element a ∈ A to the element b ∈ B.
.= equal, by definition.
⇐⇒ if and only if
C(E) = C(E,R) vector space of all continuous, real valued functions on the metric space E.
C(E,C) vector space of all continuous, complex valued functions on the metric space E.
BC(E) space of all bounded, continuous, real valued functions f : E 7→ R, with norm‖f‖ = supx∈E |f(x)|.
1
`1, `p, `∞ spaces of sequences of real (or complex) numbers.
L1(Ω), Lp(Ω), L∞(Ω) Lebesgue spaces.
W k,p(Ω) Sobolev space of functions whose weak partial derivatives up to order k lie in Lp(Ω),for some open set Ω ⊆ Rn.
Hk(Ω) = W k,2(Ω) Hilbert-Sobolev space.
Ck,γ(Ω) Holder space of functions u : Ω 7→ R whose derivatives up to order k are Holdercontinuous with exponent γ ∈ ]0, 1].
‖ · ‖ = ‖ · ‖X norm on a vector space X.
(·, ·) = (·, ·)H inner product on a Hilbert space H.
X∗ is the dual space of X, i.e. the space of all continuous linear functionals x∗ : X 7→ K.
〈x∗, x〉 = x∗(x) duality product of x∗ ∈ X∗ and x ∈ X.
xn → x strong convergence in norm; this means ‖xn − x‖ → 0.
xn x weak convergence.
ϕn∗ϕ weak-star convergence.
f ∗ g convolution of two functions f, g : Rn 7→ R.
∇u = (ux1 , ux2 , . . . , uxn) gradient of a function u : Rn 7→ R.
Dα =(
∂∂x1
)α1(
∂∂x2
)α2
· · ·(
∂∂xn
)αn= ∂α1
x1 ∂α2x2 · · · ∂
αnxn a partial differential operator of
order |α| .= α1 + α2 + · · ·+ αn.
meas(Ω) the Lebesgue measure of a set Ω ⊂ Rn.
−∫
Ωf dx
.=
1
meas(Ω)
∫Ωf dx the average value of f over the set Ω.
2
Solutions to Homework Problems
Chapter 2
1. (i) It is not a normed space. (N2) fails when λ < 0.
(ii) It is a normed space but not complete. The sequence xn =(
1, 12 ,
13 , . . . ,
1n , 0, 0, . . .
)is
Cauchy but does not have a limit in X.
(iii) It is a normed space but not complete. The sequence pn(x).=∑n
k=0xn
n! is Cauchy, butdoes not converge to any element of X. Indeed, its uniform limit on the interval [0, 1] is thefunction f(x) = ex, which is not a polynomial.
(iv) It is a normed space of dimension 3. It is also a Banach space. Every finite dimensionalnormed space is complete.
(v) It is a normed space, but not complete. Indeed, consider the functions
f(x).=
0 if x ∈
[0, 1
2
],
1 if x ∈]
12 , 1
],
fn(x).=
0 if x ∈[0, 1
2
],
n(x− 1
2
)if x ∈
[12 ,
12 + 1
n
],
1 if x ∈[
12 + 1
n , 1].
Then the sequence of continuous functions (fn)n≥1 is a Cauchy sequence without any limit inthe space X. Namely, ‖f − fn‖L1([0,1]) = 1
2n → 0 as n→∞, but f /∈ X.
(vi) For p < 1 this is not a norm, because (N3) fails. For example, in R2 take x = (1, 0),y = (0, 1). Then ‖x‖ = ‖y‖ = 1, but ‖x+ y‖ = 21/p > 2.
2. (N1) ‖(x, y)‖ = max‖x‖, ‖y‖ ≥ 0.
‖(x, y)‖ = 0 if and only if max‖x‖, ‖y‖ = 0 if and only if ‖x‖ = ‖y‖ = 0 if and only ifx = y = 0.
(N2) ‖(λx, λy)‖ = max‖λx‖, ‖λy‖ = max|λ| ‖x‖, |λ| ‖y‖ = |λ|max‖λx‖, ‖λy‖ =|λ| ‖(x, y)‖.
(N3) ‖(x+x, y+y)‖ = max‖x+x‖, ‖y+y‖ ≤ max‖x‖+‖x‖ , ‖y‖+‖y‖ ≤ max‖x‖, ‖y‖+max‖x‖, ‖y‖ = ‖(x, y)‖+ ‖(x, y)‖.
To prove that X×Y is a Banach space, let (xn, yn)n≥1 be a Cauchy sequence in X×Y . Then(xn)n≥1 is a Cauchy sequence in X and (yn)n≥1 is a Cauchy sequence in Y . Since X,Y arecomplete, there exist the limits xn → x, yn → Y . This implies (xn, yn)→ (x, y), showing thatX × Y is complete as well.
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3. (i) Let (x, y) ∈ X ×X and ε > 0 be given. Choose δ = ε/2. If ‖(x, y)− (x, y)‖ ≤ δ then
‖(x+ y)− (x+ y)‖ ≤ ‖x− x‖+ ‖y − y‖ ≤ δ + δ = ε.
This proves that the map (x, y) 7→ x+ y is continuous at the point (x, y).
(ii) Let α ∈ K, x ∈ X, and ε > 0 and be given. Choose
δ.= min
1,
ε/2
1 + ‖x‖,
ε/2
1 + |α|
.
Assume |β − α| ≤ δ, ‖y − x‖ ≤ δ. Then ‖y‖ ≤ 1 + ‖x‖ and
‖βy − αx‖ ≤ ‖βy − αy‖+ ‖αy − αx‖ ≤ |β − α| (1 + ‖x‖) + |α| ‖y − x‖ ≤ ε
2+ε
2= ε .
4. If V is a subspace of X it is clear that the properties (N1)–(N3) of the norm remain valid.Now assume that X is a Banach space and that V is a closed subset of X. If (vn)n≥1 is aCauchy sequence in V , then it is also a Cauchy sequence in X. Since X is complete, one hasthe convergence xn → x for some point x ∈ X. Since V is closed, x ∈ V , proving that V iscomplete as well.
5. Let X be a Banach space and assume that∑
n≥1 ‖xn‖ < ∞. Consider the sequence ofpartial sums yn
.=∑n
k=1 xk. This is a Cauchy sequence. Indeed, for m < n we have
‖ym − yn‖ =
∥∥∥∥∥∥∑
m<k≤nxk
∥∥∥∥∥∥ ≤∑
m<k≤n‖xk‖ → 0 as m,n→∞.
Since X complete, this sequence has a limit: yn → y =∑∞
k=1 xk.
Viceversa, assume that every absolutely convergent series of elements of X has a sum. Let(yn)n≥1 be a Cauchy sequence of elements in X. For each integer k ≥ 1, choose an index Nk
such that‖ym − yn‖ ≤ 2−k whenever m,n ≥ Nk .
It is not restrictive to assume N1 < N2 < N3 < · · · . Consider the series∑
k xk, where
x1 = yN1 , x2 = yN2 − yN1 , · · · xk = yNk −Nk−1, · · ·
This is absolutely convergent, because ‖xk‖ ≤ 21−k for all k ≥ 2. By assumption, there existthe sum y =
∑k xk. We claim that the sequence (yn)n≥1 converges to y. Indeed, let ε > 0 be
given and choose k large enough so that 22−k < ε. If n ≥ Nk then
‖yn − y‖ ≤ ‖y − yNk‖+ ‖yNk − y‖ ≤ 2−k +∑ν≥Nk
‖xν‖ ≤ 2−k + 21−k < ε.
6. (i) Let x′, x′′ ∈ coA, and assume λ ∈ [0, 1]. We need to prove that x = λx′+ (1−λ)x′′ ∈ A.By assumption we have
x′ =N ′∑k=1
θ′ka′k , x′′ =
N ′′∑k=1
θ′′ka′′k.
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Choosing
N = N ′ +N ′′ , θk =
λθ′k if k = 1, . . . , N ′,
(1− λ)θ′′k−N ′ if k = N ′ + 1, . . . , N ′ +N ′′,
we have the representation x =∑N
k=1 θkak ∈ coA.
(ii) Let x =∑N
k=1 θkak ∈ coA and let K be any convex set containing A. Then K containsall elements a1, . . . , aN . Being convex, K must also contain all convex combinations of theseelements. In particular, x ∈ K.
The converse inclusion is trivial: since coA is a convex set containing A, the intersection of allconvex sets that contain A cannot be larger than coA.
7. (i) Let A be open and consider any point x =∑N
k=1 θkak ∈ coA. We can assume θ1 > 0.Since a1 ∈ A and A is open, there exists a radius r > 0 such that B(a1, r) ⊂ A. Then
B(x, θ1r) =θ1y +
N∑k=2
θkak ; y ∈ B(a1, r)⊂ coA
This shows that x lies in the interior of coA, hence coA is open.
(ii) Let A be bounded, say A ⊆ B(0, R) for some radius R > 0. Then coA ⊆ B(0, R) becauseB(0, R) is a bounded convex set containing A. Hence coA is bounded.
(iii) If A ⊆ B(0, r1) and B ⊆ B(0, r2), then A + B ⊆ B(0, r1 + r2), showing that A + B isbounded.
(iv) Let (xn)n≥1 be a sequence of points in A+B, with xn → x as n→∞. We need to provethat x ∈ A + B. By assumption xn = an + bn, for some an ∈ A and bn ∈ B. Since B iscompact, by possibly choosing a subsequence and relabeling we can assume bn → b for someb ∈ B. We now have
lim supm,n→∞
‖am−an‖ = lim supm,n→∞
‖xm−bm−(xn−bn)‖ ≤ lim supm,n→∞
(‖xm−xn‖+‖bm−bn‖
)= 0.
Indeed, the sequences (xn)n≥1 and (bn)n≥1 are both convergent, hence they are Cauchy se-quences. This shows that the sequence (an)n≥1 is Cauchy as well. Hence there exists the limitan → a for some a ∈ A, because X is a Banach space and A ⊆ X is closed. We conclude thatx = a+ b ∈ A+B, proving that the sum A+B is closed.
(v) In R2, consider the two closed subsets
A.= (x, y) ; x > 0, y > 0, xy ≥ 1 , B
.= (x, y) ; x < 0, y > 0, xy ≤ −1 .
Then the setA+B =
(x, y) ; y > 0
is open, not closed.
(vi) The inclusion A + A ⊆ 2A is always trivially true. On the other hand, if A is convex,then any element a+ a′ ∈ A+A can be written as a+a′
2 + a+a′
2 ∈ 2A. Hence A+A ⊆ 2A.
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(vii) Assume that A is closed and A + A = 2A. Let a, b ∈ A. By assumption there existsc ∈ A such that a+ b = 2c. This implies a+b
2 ∈ A. Repeating the argument, we obtain that
1
2· a+ b
2+
1
2· b =
1
4a+
3
4b ∈ A,
1
2· a+
1
2· a+ b
2=
3
4a+
1
4b ∈ A .
By induction, we obtain
k
2na+
(1− k
2n
)b ∈ A for all k = 0, 1, . . . 2n. (1)
Since the set of all convex combinations of the form (1) are dense in the set of all convexcombinations, and we are assuming that A is closed, we conclude that A is convex.
Notice that the result would fail if A were not closed. For example, let A be the set of allrational numbers contained inside [0, 1].
8. (i) Let S be convex. Let x, y ∈ S and let θ ∈ [0, 1]. We need to show that θx+(1−θ)y ∈ S.By assumption, there exist sequences xn → x, yn → y, with xn, yn ∈ S for every n ≥ 1. Byconvexity, θxn+ (1− θ)yn ∈ S. We now have the convergence θxn+ (1− θ)yn → θx+ (1− θ)y,showing that S is convex.
(ii) Let S be symmetric and assume x ∈ S. Then there exists a sequence xn → x , with xn ∈ Sfor every n ≥ 1. Therefore −xn ∈ S and −xn → −x. This shows that −x ∈ S, hence S issymmetric.
9. (i) Let S be convex. Assume y1, y2 ∈ Λ(S) and θ ∈ [0, 1]. Then there exist x1, x2 ∈ S suchthat y1 = Λ(x1), y2 = Λx2. Since Λ is linear, this implies
θy1 + (1− θ)y2 = θΛ(x1) + (1− θ)Λ(x2) = Λ(θx1 + (1− θ)x2) ∈ Λ(S),
because S is convex.
(ii) Let S be symmetric. Assume y ∈ Λ(S). Then y = Λ(x) for some x ∈ S. This implies−y = Λ(−x) ∈ Λ(S) because S is symmetric.
10. If Λ is not continuous, then it is not bounded. Hence one can find a sequence of pointsxn ∈ X with ‖yn‖ ≤ 1 but ‖Λ(yn)‖ ≥ n for every n ≥ 1. Defining the sequence xn
.= n−1/2yn
we have xn → 0 but ‖Λ(xn)‖ ≥ n1/2 →∞, against the assumption.
11. (i) If S is a finite set, then span(S) is a finite dimensional space. By Theorem 2.20, it iscomplete. Being complete, it must be a closed subset of X.
(ii) Since X is a Banach space, the closure Y.= span(S) is complete, hence it is a Banach space
as well. For each n ≥ 1, consider the finite dimensional subspace Vn = spanv1, . . . , vn ⊂ Y .Being finite dimensional, this subspace is complete, hence it must be closed. It is easy tosee that Vn has empty interior. Indeed, since by assumptions the vectors v1, . . . , vN+1 arelinearly independent, for every y ∈ Vn the sequence y+ 1
mvn+1 does not lie in Vn and convergesto y as m→∞.
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According to Baire’s category theorem, the complement
Y \⋃n≥1
Vn =⋂n≥1
(Y \ Vn)
is everywhere dense in Y . Observing that span(S) = ∪n≥1Vn, this shows that span(S) 6= Y =span(S).
12. If x, y ∈ X and α ∈ R, then
Re(Φ(x+ y)) = Re(Φ(x)) +Re(Φ(y)) , Re(Φ(αx)) = αRe(λ(x)) .
Moreover, Im(Φ(x)) = Re(−iΦ(x)) = −Re(Φ(ix)). Hence
Φ(x) = Re(Φ(x)) + Im(Φ(x)) = φ(x)− iφ(ix) .
13. If X is finite dimensional, let e1, . . . , en be a basis. Then every linear functional iscontinuous and is a linear combination of the n linear functionals e∗1, . . . , e
∗n where e∗i (ej) = δij .
On the other hand, if X is infinite dimensional, let e1, e2, . . . be a countable family oflinearly independent vectors. For each n, consider the subspace Vn = spane1, . . . , en. Letϕn : Vn 7→ R be the linear functional such that
ϕn(en) = 1 , ϕn(ek) = 0 for all k < n .
Since Vn is finite dimensional, ϕn is continuous. By Hahn-Banach, it can be extended to acontinuous linear functional ϕn : X 7→ R.
It remains to check that the countable set of functionals ϕn ; n ≥ 1 is linearly independent.If
N∑k=1
akϕk(x) = 0 for all x ∈ X ,
let j be the smallest index such that aj 6= 0. Then
0 =
N∑k=1
akϕk(ej) = aj ,
reaching a contradiction.
14. If x = (x1, x2, . . .) ∈ `p, then
‖x‖`∞ = supn|xn| =
(supn|xn|p
)1/p≤( ∞∑n=1
|xn|p)1/p
= ‖x‖`p .
Hence x ∈ `∞ and the embedding `p ⊆ `∞ is bounded, with operator norm 1. Moreover, ifp < q <∞, then
‖x‖`q =
( ∞∑n=1
|xn|q)1/q
=
( ∞∑n=1
|xn|p|xn|q−p)1/q
≤
( ∞∑n=1
|xn|p · ‖x‖q−p`∞
)1/q
= ‖x‖p/q`p · ‖x‖(q−p)/q`∞ ≤ ‖x‖p/q`p · ‖x‖
(q−p)/q`p = ‖x‖`p .
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15. Consider first the case 1 ≤ p < ∞. Let x = (x1, x2, . . .) ∈ `p. Define the sequence oflinear combinations
yn.= (x1, x2, . . . , xn, 0, 0, . . .) =
n∑k=1
xkek ∈ V.= spanek ; k ≥ 1 . (2)
Then‖x− yn‖p`p =
∑k>n
|xk|p → 0 as n→∞ .
Next, assume x = (x1, x2, . . .) ∈ `∞. Define the sequence of approximations yn as in (2). Ifx ∈ c0, i.e. if limk→∞ xk = 0, then
‖x− yn‖`∞ = supk>n|xk| → 0 as n→∞ .
Hence V.= spanek ; k ≥ 1 is dense on c0. To prove that c0 coincides with the closure of V , it
remains to prove that c0 is closed. Consider a sequence of elements xm = (xm,1, xm,2, xm,3, . . .),with xm ∈ c0 for every m ≥. Assume that limm→∞ ‖xm − x‖`∞ = 0. We claim that x ∈ c0
as well. Indeed, let ε > 0 be given. By assumption, there exists an integer m large enough sothat ‖xm − x‖`∞ < ε/2. Since xm ∈ c0, there exists N large enough so that |xm,n| < ε/2 forall n > N . Hence
|xn| ≤ |xn − xm,n|+ |xm,n| ≤ ‖x− xm‖`∞ + |xm,n| <ε
2+ε
2= ε .
Since ε > 0 was arbitrary, this proves limk→∞ xk = 0.
16. To fix the ideas, we assume 1 ≤ p <∞. The case p =∞ is similar.
(i) If M.= supk |λk|, then
‖Λx‖`p =
(∑k
|λkxk|p)1/p
≤ M ·
(∑k
|xk|p)1/p
= M‖x‖`p .
This shows that ‖Λ‖ ≤ M . On the other hand, given any ε > 0, choose k such that |λk| >M − ε. Then
‖Λek‖`p = ‖λkek‖`p = |λk|‖ek‖`p ≥ (M − ε) ‖ek‖`p .
This shows that ‖Λ‖ ≥M − ε. Since ε > 0 was arbitrary, we conclude that ‖Λ‖ = M .
To prove (ii), assume that the sequence (λk)k≥1 is unbounded. Then
‖Λ‖ ≥ supk≥1‖Λek‖`p ≥ sup
k≥1|λk| = ∞ .
In this case, we can extract a subsequence with |λnk | ≥ 3k. Consider the vector v =∑∞
k=1 2−kenk .Observe that this sum is well defined, because the series is absolutely convergent:∑
k≥1
‖2−kenk‖`p =∑k≥1
|2−k| = 1 .
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We now haveΛv =
∑k
λnk2−kenk /∈ `p
Indeed,
limm→∞
m∑k=1
|λnk2−k| ≥ limm→∞
m∑k=1
(3/2)k = ∞ .
17. To fix the ideas, assume 1 ≤ p <∞.
(i) Let g ∈ L∞(Ω), say with ‖g‖L∞ = K. If f ∈ Lp(Ω), then
‖fg‖Lp =
(∫Ω|fg|p dx
)1/p
≤(∫
ΩKp|f |p dx
)1/p
≤ K‖f‖Lp .
Hence the norm of the multiplication operator satisfies ‖Mg‖ ≤ K. On the other hand, forany ε > 0 there exists a bounded measurable subset A ⊆ Ω with strictly positive measuresuch that |g(x)| ≥ K − ε for all x ∈ A. Let f = χ
Abe the characteristic function of the set
A. Namely, f(x) = 1 if x ∈ A and f(x) = 0 if x /∈ A. Then
‖Mgf‖Lp = ‖fg‖Lp =
(∫A|g|p dx
)1/p
≥(∫
A(K − ε)p dx
)1/p
= (K − ε) ·(meas(A)
)1/p= (K − ε) · ‖f‖Lp .
This proves that ‖Mg‖ ≥ K − ε.
(ii) Next, assume g /∈ L∞. Then for every constant K ≥ 1 we can find a bounded measurableset A with strictly positive measure such that |g(x)| ≥ K− ε for all x ∈ A. Arguing as before,taking f = χ
Awe conclude that ‖Mg‖ ≥ K. Since K is now arbitrary, we conclude that the
operator Mg is unbounded.
To obtain a function f ∈ Lp such that fg /∈ Lp, we first construct a sequence of disjoint,bounded, measurable subsets An ⊂ Ω with the following properties.
1) Each An has strictly positive measure.
2) |g(x)| ≥ 3n for every x ∈ An.
3) For every n ≥ 1 one has g /∈ L∞(Ω \ ∪nj=1Aj).
This sequence of subsets can be constructed by induction on n. Taking
f(x).=
∞∑n=1
2−nχAn
‖χAn‖Lp
we have that f ∈ Lp(Ω) but fg /∈ Lp(Ω).
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18. For any given a < b, the map
f 7→ ϕ(f).=
∫ b
af(x) dx
is a bounded linear functional on Lp(R). By definition, the weak convergence fn f impliesϕ(fn)→ ϕ(f).
19. (i) A straightforward computation yields
‖fn‖L∞ =1
n, ‖fn‖Lp = n
1−pp .
As n→∞, for every p > 1 the Lp norm of fn approaches zero.
(ii) On the other hand, ‖fn‖L1 = 1 for every n.
Consider any subsequence, say fnk , with n1 < n2 < n3 < · · · By choosing a further subse-quence and relabeling, it is not restrictive to assume nk+1 > 3nk.
Define the function g ∈ L∞(R) by setting
g(x) = 0 if n2k−1 < x ≤ n2k
g(x) = 1 if n2k < x ≤ n2k+1 .
Then the map f 7→ ϕ(f).=∫fg dx is a bounded linear functional linear on L1(R). The above
definition implies
ϕ(fn2k) ≥ 2
3, ϕ(fn2k+1
) ≤ 1
3.
Therefore the subsequence ϕ(fnk) has no limit. As a consequence, the original subsequencedoes not converge weakly.
20. For every y ∈ Y , let Λy = (Λ1y,Λ2y, . . . ,Λny) ∈ Rn. By the Hahn-Banach extensiontheorem, each linear functional Λj : Y 7→ R can be extended to a bounded linear functional
Λj : X 7→ R with norm ‖Λj‖ = ‖Λj‖ ≤ ‖Λ‖. For x ∈ X, define the extension Λx =
(Λ1x, Λ2x, . . . , Λnx) ∈ Rn. This is a bounded linear operator, with norm ‖Λ‖ ≤ n1/2‖Λ‖.Indeed, if ‖x‖ ≤ 1, then
|Λx| =
n∑j=1
|Λjx|21/2
≤(n‖Λ‖2
)1/2= n1/2‖Λ‖ .
21. Let φ(x1, x2) = ax1 + bx2. To prove (i) we write
‖φ‖∞.= sup
|x1|+|x2|≤1|ax1 + bx2| ≤ sup
|x1|+|x2|≤1(|ax1|+ |bx2|) ≤ max|a|, |b|.
Choosing
(x1, x2) =
( a|a|, 0)
if |a| ≥ |b| ,
(0,
b
|b|
)if |a| ≤ |b| ,
10
one obtains the converse inequality
‖φ‖∞ ≥ |ax1 + bx2| = max|a| , |b| .
To prove (ii) we write
‖φ‖∞.= sup
|x1|≤1, |x2|≤1|ax1 + bx2| ≤ sup
|x1|≤1, |x2|≤1(|ax1|+ |bx2|) ≤ |a|+ |b| .
Choosing
(x1, x2) =( a|a|,b
|b|
),
one obtains the converse inequality
‖φ‖∞ ≥ |ax1 + bx2| = |a|+ |b| .
Using the discrete Holder inequality, (iii) is proved by
‖φ‖p.= sup
(|x1|q+|x2|q)1q≤1
|ax1 + bx2| ≤ sup
(|x1|q+|x2|q)1q≤1
(|ax1|+ |bx2|) ≤ (|a|p + |b|p)1p .
Choosing
(x1, x2) =(|a|q + |b|q
) 1−qq ·
(|a|q−1sign a , |b|q−1sign b
),
one obtains the converse inequality
‖φ‖p ≥ |ax1 + bx2| = (|a|p + |b|p)1p .
22. By assumption, B ⊂ X is a convex set with the following property. For every non-zerovector x ∈ X, there exists ηx > 0 such that
B ∩ t x ; t ∈ K = t x ; |t| ≤ ηx . (3)
This clearly implies B = −B and 0 ∈ B. Moreover, the functional
‖x‖ .= min r ≥ 0 ; x ∈ rB =
0 if x = 0 ,
1
ηxif x 6= 0 ,
(4)
is well defined and satisfies
‖x‖ ≤ 1 if and only if x ∈ B .
It remains to prove the properties (N1)–(N3) of a norm. From (3)-(4) one immediately obtains
‖x‖ = 0 if and only if x = 0 ,
λαx ∈ B ⇐⇒ |λ| |α| ≤ ηx ⇐⇒ |α| ≤ θλx = θ|λ|x .
11
Henceθ|λ|x =
ηx|λ|
, ‖λx‖ = |λ| ‖x‖ .
To prove the sub-additivity of the norm, consider any x, y 6= 0. Observe that
x ∈ ‖x‖B , y ∈ ‖y‖B .
Since B is a convex set containing the origin, this implies
x+ y ∈ (‖x‖+ ‖y‖)B ,
‖x+ y‖ .= minr ; x+ y ∈ rB ≤ ‖x‖+ ‖y‖ .
23. (i) =⇒ (iii) Let (fj)j≥1 be a Cauchy sequence for the distance d. Let K ⊂ Ω be anycompact subset and let ε > 0 be given. Since the open sets Aj cover K, there exists N large
enough such that K ⊂⋃Nk=1Ak. This implies
lim supj,`→∞
(supx∈K|fj(x)− f`(x)|
)≤ 2N lim
j,`→∞d(fj , f`) = 0.
Therefore, the sequence (fj)j≥1 converges uniformly on the compact set K.An identical argument shows that (ii) =⇒ (iii).
(iii)=⇒ (ii) Assume that the sequence (fj)j≥1 converges uniformly on compact subsets of Ω.Let ε > 0 be given. Choose N so that 2−N < ε. Then the set K
.= ∪1≤k≤NAk is a compact
subset of Ω. We thus have the estimate
lim supj,`→∞
d(fj , f`) = lim supj,`→∞
N∑k=1
2−kpk(fj − f`)
1 + pk(fj − f`)+ lim sup
j,`→∞
∞∑k=N+1
2−kpk(fj − f`)
1 + pk(fj − f`)
≤ 0 + 2−N < ε .
Hence the sequence (fj)j≥1 is Cauchy w.r.t. the distance d(·, ·).An identical argument shows that (iii) =⇒ (ii).
24. (i) There is no guarantee that the series∑∞
k=1 2−kpk(f) converges. For example, takeΩ = ]0, 1[ and define
pk(f).= max
|f(x)| ; x ∈ Ak =
[1
k,k − 1
k
].
If f(x) = e1/x then∞∑k=1
2−kpk(f) =∞∑k=1
2−kek = ∞ .
(ii) Choose N large enough so that 2−N < ε. Then, if p1(f) = · · · = pN (f) = 0, then for anyλ > 0 we have
∞∑n=1
2−npn(λ−1f)
1 + pn(λ−1f)≤
∞∑n=N+1
2−npn(λ−1f)
1 + pn(λ−1f)< 2−N < ε .
12
Hence λ−1f ∈ Bε for every λ > 0. By definition, this implies ‖f‖ε = 0, showing that theproperty (N1) of the norm can fail.
25. Assume x ∈ span(S), so that
x =n∑k=1
αkxk , αk ∈ K, xk ∈ S .
Choose an integer m > n∑
k ‖αxk‖. Then
x =(α1
mx1 +
α2
mx2 + · · · +
αmmxn
)+ · · · +
(α1
mx1 +
α2
mx2 + · · · +
αmmxn
),
where the right hand side contains m equal groups of summands. By the choice of m, eachpartial sum has norm ≤ 2‖x‖. We thus achieve the desired representation (2.42), with N =mn.
26. The implication (ii) =⇒ (i) is clear.
To prove the converse implication (i) =⇒ (ii) we proceed as follows. Assume x ∈ span(S),so that x = limn→∞ yn with yn ∈ span(S) for every n. It is not restrictive to assume‖yn+1 − yn‖ < 2−n for every n ≥ 1. We can write x as an absolutely convergent sum:
x = y1 +∞∑n=1
(yn+1 − yn) , yn+1 − yn ∈ span(S).
Using problem 25, we can write
yn+1 − yn =
N(n)∑i=1
cnixni , cni ∈ K , xni ∈ S ,
with ∥∥∥∥∥k∑i=1
cnixni
∥∥∥∥∥ ≤ 2‖yn+1 − yn‖ ≤ 21−n for every k ∈ 1, . . . , N(n). (5)
We now arrange all terms in the double series1
∞∑n=1
N(n)∑i=1
cnixni
into a single sequence, namely
c11x11 + · · ·+ c1,N(1)x1,N(1) + c21x21 + · · ·+ c2,N(2)x2,N(2) + c31x31 + · · ·
Thanks to (5) we obtain a series converging to x.
1Note that, in general, this double series may not be absolutely convergent. Hence different rearrangementsof its terms may yield different limits. It is the property (5) that guarantees that the sequence converges to x.
13
27. (i) Let V ⊂ `∞ be the subspace of all sequences x = (x1, x2, . . .) such that limn→∞ xnexists. For x ∈ V define the linear operator F (x)
.= limn→∞ xn. Observing that |F (x)| =
| limn→∞ xn| ≤ supn |xn| = ‖x‖∞` . it is clear that f : V 7→ R is a continuous linearfunctional with norm ‖F‖ = 1. Using the Hahn-Banach extension theorem, we can extend Fto the entire space `∞, still with the same norm ‖F‖ = 1.
(ii) Assume m = lim inf xn < lim supxn = M . If F (x) = M + δ for some δ > 0, acontradiction is achieved as follows. Choose N large enough so that m− δ
2 < xn < M + δ2 for
all n > N . Consider the sequence y = (x1, x2, . . . , xN ,m,m,m, . . .). Then
F (x)− F (y) = (M + δ)−m > M +δ
2−m ≥ ‖x− y‖`∞ .
This is a contradiction, because ‖F‖ = 1, hence we should have F (x) − F (y)| ≤ ‖x − y‖`∞ .The case F (x) < m is handled in a similar way.
(iii) Using linearity and then (ii), we obtain
F (y)− F (x) = F (y − x) ≥ lim infn→∞
yn − xn ≥ 0 .
(iv) If a = (a1, a2, . . .) ∈ `1, then there exists N large enough so that∑
n>N |an| < 1. Considerthe sequence x = (0, 0, . . . , 0, 1, 1, 1, . . .), where the first N entries are zero and all other entriesare equal to 1. Then
∞∑n=1
anxn =∑n>N
an < 1 = limn→∞
xn = F (x).
(v) The functional F satisfies (2.43)-(2.44), but it is not linear. For example, take x =(1, 0, 0, 1, 0, 0, 1, 0, 0, . . .), y = (0, 0, 1, 0, 0, 1, 0, 0, 1, . . .). Then x+ y = (1, 0, 1, 1, 0, 1, 1, 0, 1, . . .)and
F (x+ y) =1
26= F (x) + F (y) =
1
2+
1
2.
28. The map f 7→ f(0) is a bounded linear operator on V , with norm
supf continuous, ‖f‖L∞≤1
|f(0)| = 1 .
By the Hahn-Banach theorem it can be extended to a linear functional Λ defined on the entirespace L∞(R), with norm ‖Λ‖ = 1.
Next, assume that there exists g ∈ L1(R) such that Λf =∫fg dx for all f ∈ L∞(R). The
dominated convergence theorem yields
limε→0
∫ ε
−ε|g(x)| dx = 0.
Hence we can choose δ > 0 so that∫ δ−δ |g(x)| dx ≤ 1/2. Let φ : R 7→ [0, 1] be a smooth
function, with φ(0) = 1 and with support contained in the interval [−δ, δ]. We now reach acontradiction by writing
1 = φ(0) = Λ(φ) =
∫Rg(x)φ(x) dx =
∫ δ
−δg(x)φ(x) dx ≤ max
x|φ(x)| ·
∫ δ
−δ|g(x)| dx ≤ 1
2.
14
29. (i) Assuming that x ∈ λ1Ω, y ∈ λ2Ω for some λ1, λ2 > 0, we claim that x+y ∈ (λ1 +λ2)Ω.Indeed, by assumption there exist ω1, ω2 ∈ Ω such that x = λ1ω1, y = λ2ω2. Using theassumption that Ω we obtain
x+ y = λ1ω1 + λ2ω2 = (λ1 + λ2) ·(
λ1
λ1 + λ2ω1 +
λ2
λ1 + λ2ω2
)∈ (λ1 + λ2)Ω .
The identity p(tx) = tp(x) for t > 0 follows from the fact that x ∈ λΩ if and only if tx ∈ tλΩ.
(ii) We have
p(x) = infλ > 0 ; x ∈ λΩ ≤ infλ > 0 ; x ∈ λBr =‖x‖r.
(iii) If Ω = B1, then
p(x) = infλ > 0 ; x ∈ λB1 ≤ infλ > 0 ; ‖x‖ < λ = ‖x‖ .
30. (i) If (fn)n≥1 is a sequence of continuous functions such that fn → f uniformly on [0, 1]and fn(0) = 0 for every n ≥ 1, then f(0) = 0. Hence X is a closed subspace.
(ii) The map f 7→ Λf.=∫ 1
0 f(x) dx is a linear functional on X. for every f ∈ X with
‖f‖ = maxx∈[0,1] |f(x)| ≤ 1 we have |Λf | = |∫ 1
0 f(x) dx| < 1, because f(0) = 0, hence
|f(x)| < 1/2 for x in some interval [0, δ]. On the other hand, choosing fn = x1/n we havefn ∈ X, ‖fn‖ = 1 and Λfn = n
n+1 → 1 as n→∞. Hence
‖Λ‖ .= sup
f∈X , ‖f‖≤1
∣∣∣∣∫ 1
0f(x) dx
∣∣∣∣ = 1,
but the supremum is not attained.
31. Being the intersection of a family of closed, convex sets, the set Ω∗∗ is also closed andconvex. It remains to show that Ω∗∗ is contained in the closure of Ω. Assume, on the contrary,that there exists x0 ∈ Ω∗∗ and a radius r > 0 such that the open ball B(x0, r) does notintersect Ω. By the separation theorem, there exists a bounded linear functional F : X 7→ Rand c > 0 such that
F (x) ≤ c for all x ∈ Ω, F (x) > c for all x ∈ B(x0, r) .
Defining φ(x) = 1cF (x) we obtain
φ(x) ≤ 1 for all x ∈ Ω .
This yields a contradiction, because φ ∈ Ω∗ but φ(x0) > 1.
32. On the one dimensional space U = tξ ; t ∈ R define the functional F (tξ) = t. Thisis a bounded linear functional with norm ‖F‖ = ‖ξ‖. Using the Hahn-Banach Theorem, weextend F to a bounded linear functional F : X 7→ R with the same norm. Let V = ker(F ).We claim that all conclusions are satisfied.
15
Since F is continuous linear functional, V is a closed subspace. Given x ∈ X, consider theprojections
πUx.= F (x) ξ , πV x
.= x− F (x)ξ .
These are bounded linear operators, and satisfy
πUx ∈ U , πV x ∈ V , πUx+ πV x = x .
33. Consider the subspace Y = ax+ b ; a, b ∈ R of all polynomials of degree 1. Restrictedto Y , the linear functional ϕ is continuous and has norm
‖ϕ‖ = supf∈Y, f 6=0
|ϕ(f)|‖f‖L1
= sup(a,b)6=(0,0)
a∫ 1
−1|ax+ b| dx
= 1
Indeed, for a given a, the minimum of the denominator is achieved when b = 0.
Using the Hahn-Banach theorem, the functional ϕ can then be extended to the entire spaceL1([−1, 1]), with the same norm.
An explicit form of this functional is
ϕ(f) =
∫ 1
0f(x) dx−
∫ 0
−1f(x) dx .
34. It is clear that each pk(·) is a seminorm. If f ∈ Cm(Ω), f 6= 0, then f(x) 6= 0 for somepoint x ∈ Ω. If x ∈ Ak, then pk(f) > 0.
To show that the space Cm(Ω) is complete for the corresponding distance
d(f, g).=
∞∑k=1
2−kpk(f − g)
1 + pk(f − g),
let (fj)j≥1 be a Cauchy sequence. Then for every multi-index α with |α| ≤ m the sequenceof continuous functions (Dαfj)j≥1 is uniformly convergent on each open set Ak. Therefore, ithas a continuous limit, say
f(x)→ g(x) , Dαfj(x)→ gα(x) for all x ∈ Ω, |α| ≤ m,
for some continuous functions g, gα. Since the limit is uniform on compact sets, it followsgα = Dαg for every |α| ≤ m. Hence d(fj , g) → 0, proving that this Cauchy sequence has alimit.
35. Consider the subspace Y.= y = y + tx0 ; y ∈ Y, t ∈ R be the space spanned by Y
together with x+ 0. Let α.= d(x0, Y ) > 0. Define the functional ϕ : Y 7→ R by setting
ϕ(y + tx0).= αt .
This clearly implies ϕ(x0) = α = d(x0, Y ) and ϕ(y) = 0 for y ∈ Y .
16
We claim that ϕ is a bounded operator with norm ‖ϕ‖ = 1. Indeed, assume that ‖y0 + tx0‖ <αt for some choice of y0 ∈ Y and t 6= 0. This leads to a contradiction, because
‖ − y0 − tx0‖ = ‖y0 + tx0‖ < αt ,∥∥∥− y0
t− x0
∥∥∥ < α = infy∈Y‖y − x0‖ .
Using the Hahn-Banch theorem, the functional ϕ can then be extended to the entire space X.
36. (i) If 1 ≤ p < ∞, every continuous linear functional on `p has the form x 7→ ϕ(x) =∑n anxn, for some sequence a = (an)n≥1 ∈ `q. In this case,
limk→∞
ϕ(ek) = limk→∞
ak = 0 .
proving the weak convergence ek 0 in the space `p.
(ii) Now consider the case p = ∞. Take any subsequence (enk)k≥1. Define the elementa = (a1, a2, . . .) ∈ `∞ as follows.
aj =
1 if j = nk for some odd integer k,0 otherwise.
Consider the bounded linear functional on `1 defined by ϕ(x).=∑
j ajxj . Then
lim infk→∞
ϕ(enk) = 0 < 1 = lim supk→∞
ϕ(enk) ,
showing that the subsequence (enk)k≥1 is not weakly convergent.
37. In the case 1 ≤ p <∞, assume φ′(x) ≥ c > 0 for all x ∈ R. For every f ∈ Lp(R), settingy = φ(x) one obtains(∫
|f(φ(x))|p dx)1/p
=
(∫|f(y)|p dy
φ′(x)
)1/p
≤ 1
c1/p· ‖f‖Lp .
Hence, in this case, Λ is a bounded linear operator on Lp with norm ‖Λ‖ ≤ c−1/p.
In the case p = ∞, it suffices to assume that φ′(x) > 0. Then Λ is a bounded operator with
norm ‖Λ‖ = 1. Indeed, if f ∈ L∞, let M.= ‖f‖L∞ . Consider the set A
.=x ; f(φ(x)) >
M
. If meas(A) > 0, we obtain a contradiction by writing
0 = meas(φ(A)
)=
∫Aφ′(x) dx > 0.
Hence ‖Λf‖L∞ ≤ M . This proves that ‖Λ‖ ≤ 1. On the other hand, taking the functiong(x) ≡ 1, one has ‖Λg‖L∞ = ‖g‖L∞ = 1.
38. If the Banach space X is infinite dimensional, the closed unit ball is not precompact. Inparticular, we can find a sequence of points xn ∈ X such that ‖xn‖ ≤ 1, ‖xm − xn‖ ≥ 1/2whenever m 6= n.
17
Choose x ∈ Ω0 and r > 0 such that the ball B(x, r) is contained in Ω0. By assumption, thisimplies 0 < µ(B(x, r)) ≤ µ(Ω0) <∞.
Consider the countably many open balls Bn.= B
(r2xn ,
r8
). These are mutually disjoint, and
all contained in the open ball B(0, r). Hence, by countable additivity we have
µ(B(0, r)) ≥ µ
⋃n≥1
Bn
=∑n≥1
µ(Bn) = ∞
because all balls Bn have the same radius and the same strictly positive measure. This yieldsa contradiction, because by translation invariance µ(B(0, r)) = µ(B(x, r)) <∞.
39. (i) This is an immediate consequence of Theorem 2.33 (ii), taking A = y and B = S.
(ii) Assume y /∈ S. Then there exists a bounded linear functional ϕ ∈ X∗ such that ϕ(y) <infx∈S ϕ(x).
The weak convergence xn y implies ϕ(xn)→ ϕ(y). But this yields a contradiction because
limn→∞
ϕ(xn) ≥ infx∈S
ϕ(x) > ϕ(y).
40. (i) Let V ⊂ L∞(R) be the subspace of all functions f such that ess-limx→0 f(x) exists. Weclaim that the functional Φ(f)
.= ess-limx→0 f(x) is a bounded linear functional on V . The
linearity is obvious. If Φ(f) = λ, then for any ε > 0 there exists δ > 0 such that |f(x)−λ| < εfor a.e. x ∈ [−δ, δ]. In particular, this implies ‖f‖L∞ ≥ |λ| − ε. Since ε > 0 is arbitrary, weconclude
|Φ(f)| ≤ ‖f‖L∞ (6)
for all f ∈ V . By the Hahn-Banach theorem, we can extend the linear functional Φ to theentire space L∞(R), still satisfying the bound (6) for all f ∈ L∞(R).
(ii) In the space L1(R) the conclusion fails because the functional Φ : V 7→ R is unbounded.Indeed, consider the sequence of continuous functions:
fn(x).=
n(1− nx) if |x| ≤ 1/n ,
0 otherwise.
Then fn ∈ V and ‖fn‖L1 = 1 for every n ≥ 1, but Φ(fn) = fn(0) = n.
Chapter 3
1. Repeat steps 5 - 6 in the proof of the Stone-Weiertrass theorem.
2. (i) True. For every n ≥ 1, let pn be a polymomial such that maxx∈[−n,n] |f(x) −pn(x)| < 2−n. Such a polynomial exists by the Stone-Weierstrass theorem. The sequence ofpolynomials (pn)n≥1 converges to f uniformly on bounded sets.
18
(ii) False. If p is a polynomial such that supx∈R |p(x)− e−x2 | <∞, then p must be bounded,
hence constant. But there is no way to approximate the function f(x) = e−x2
with constantfunctions, uniformly on R.
3. (i) Extend g to an even function defined on [−π, π], and then by periodicity to an evenfunction defined on the whole real line, periodic of period 2π. By Corollary 3.9 there exists atrigonometric polynomial of the form
p(x) =N∑k=1
ak sin kx+N∑k=0
bk cos kx
such that |p(x)− g(x)| < ε for all x ∈ R. Since g is even, this implies
ε >
∣∣∣∣p(x) + p(−x)
2− g(x) + g(−x)
2
∣∣∣∣ =
∣∣∣∣∣N∑k=0
bk cos kx− g(x)
∣∣∣∣∣ . for all x ∈ R .
(ii) Extend g to an odd function defined on [−π, π], and then by periodicity to an odd func-tion defined on the whole real line, periodic of period 2π. By Corollary 3.9 there exists atrigonometric polynomial of the form
p(x) =N∑k=1
ak sin kx+N∑k=1
bk cos kx
such that |p(x)− g(x)| < ε for all x ∈ R. Since g is odd, this implies
ε >
∣∣∣∣p(x)− p(−x)
2− g(x)− g(−x)
2
∣∣∣∣ =
∣∣∣∣∣N∑k=1
ak sin kx− g(x)
∣∣∣∣∣ . for all x ∈ R .
4. By the Stone-Weierstrass theorem, for any n ≥ 1 there exists a polynomial qn such that|qn(x)− f ′(x)| ≤ 2−n for every x ∈ [a, b]. Define the polynomial
pn(x).= f(a) +
∫ x
aqn(y) dy.
This construction yields
|pn(x)− f(x)| ≤ 2−n(x− a) ≤ 2−n(b− a) , |p′n(x)− f ′(x)| ≤ 2−n
for every n ≥ 1 and x ∈ [a, b].
5. The properties (i) and (ii) are easily checked. However, (iii) fails. For example, thefunction f(θ) = e−iθ cannot be uniformly approximated with functions in the algebra A.Indeed, assume that ∣∣∣∣∣f(θ)−
N∑n=0
cneinθ
∣∣∣∣∣ ≤ 1
2for all θ ∈ [0, 2π].
19
This leads to a contradiction because∣∣∣∣∣∫ 2π
0
(f(θ)−
N∑n=0
cneinθ
)eiθ dθ
∣∣∣∣∣ ≤∫ 2π
0
1
2|eiθ| dθ ≤ π .
On the other hand, since f(θ) = e−iθ, an explicit computation yields∫ 2π
0
(e−iθ −
N∑n=0
cneinθ
)eiθ dθ =
∫ 2π
01 dθ −
N∑n=0
∫ 2π
0cne
i(n+1)θ dθ = 2π .
6. Given f ∈ Lp(Ω), extend f to the entire space Rn by setting f(x) = 0 if x /∈ Ω.
For any ε > 0, using a mollification we obtain a smooth function f = f ∗ Jδ such that‖f − f‖Lp(Rn) < ε.
Since Ω is bounded, its closure Ω is compact. By Corollary 3.6, the continuous function f canbe uniformly approximated by polynomials on the compact set Ω. In particular, there existsa polynomial p(x) = p(x1, . . . , xn) such that
|p(x)− f(x)| ≤ ε for all x ∈ Ω .
The previous inequalities imply
‖f − p‖Lp(Ω) ≤ ‖f − f‖Lp(Ω) + ‖f − p‖Lp(Ω) ≤ ε+
(∫Ωεp dx
)1/p
= ε+ ε(meas(Ω))1/p.
Since ε > 0 was arbitrary, the result is proved.
7. (i) Choose δ > 0 such that |f(x, y) − f(x′, y)| ≤ ε whenever |x − x′| ≤ δ. Construct acontinuous partition of unity g1, g2, . . . , gN on the interval [0, a] such that each gi is supportedon some interval Ii ⊆ [a, b] of length ≤ δ. That means
N∑i=1
gi(x) = 1 for each x ∈ [0, a], gi(x) = 0 for x /∈ Ii .
For each i ∈ 1, . . . , N, choose a point xi ∈ Ii. For every (x, y) ∈ Q = [0, a] × [0, b] we nowhave ∣∣∣∣∣f(x, y)−
N∑i=1
gi(x) f(xi, y)
∣∣∣∣∣ ≤N∑i=1
gi(x)∣∣∣f(x, y)− f(xi, y)
∣∣∣ ≤ ε. (7)
Indeed, the only nonzero terms in the second summation are those for which x ∈ Ii, hence|x− xi| ≤ δ. By construction, this implies |f(x, y)− f(xi, y)| ≤ ε.Defining hi(y)
.= f(xi, y), all requirements are satisfied.
(ii) Let 0 < ε ≤ 1 be given. If f vanishes on the boundary of Q, then in the above constructionwe can choose xi = 0 if 0 ∈ Ii and xi = a if a ∈ Ii. In both cases, f(xi, y) = 0. Therefore, inthe sum (7) such terms gi(x)f(xi, y) can be discarded, because these vanish identically. As aresult, it is not restrictive to assume that in (7) every gi is a continuous function with supportcontained strictly inside the open interval ]0, a[.
20
Since all functions gi vanish at x = 0 and at x = a, and all functions f(xi, y) vanish at y = 0and at y = b, using problem 3.(ii), we can now find coefficients Aim, Bim and an integer Msuch that ∣∣∣∣∣gi(x)−
M∑m=1
Aim sinπmx
a
∣∣∣∣∣ ≤ ε for all x ∈ [0, a] ,
∣∣∣∣∣f(xi, y)−M∑n=1
Bin sinπny
b
∣∣∣∣∣ ≤ ε for all y ∈ [0, b] .
This implies∣∣∣∣∣∣gi(x)f(xi, y)−M∑
m,n=1
AimBin sinπmx
asin
πny
b
∣∣∣∣∣∣≤
∣∣∣∣∣gi(x)−M∑m=1
Aim sinπmx
a
∣∣∣∣∣ |f(xi, y)|+
∣∣∣∣∣M∑m=1
Aim sinπmx
a
∣∣∣∣∣ ·∣∣∣∣∣f(xi, y)−
M∑n=1
Bin sinπny
b
∣∣∣∣∣≤ ε · ‖f‖C0 + (1 + ‖gi‖C0) · ε ≤ (2 + ‖f‖C0) ε .
Therefore, setting cmn.=∑N
i=1AimBin, we obtain∣∣∣∣∣∣N∑i=1
gi(x)f(xi, y)−M∑
m,n=1
cmn sinπmx
asin
πny
b
∣∣∣∣∣∣ ≤ N (2 + ‖f‖C0) ε . (8)
Since ε > 0 was arbitrary, by (7) and (8) the result is proved.
8. Let (fn)n≥1 be a sequence of functions in the Holder space C0,γ(Ω), with ‖fn‖C0,γ ≤ C forsome constant C and every n ≥ 1.
Being Holder continuous, each function fn is uniformly continuous and hence can be uniquelyextended by continuity to the closure Ω, which is a compact set.
Since the functions fn are uniformly bounded and equicontinuous on the compact set Ω, byAscoli’s theorem we can extract a subsequence converging to some continuous function f ,uniformly for x ∈ Ω.
9. (i) f1 ∈ C0,1/3. For any 0 < a < b we have
f1(b)− f1(a) =
∫ b
a
1
3x−2/3 dx ≤
∫ b
a
1
3(x− a)−2/3 dx = (b− a)1/3.
Hence
sup0<a<b<1
|f(b)− f(a)||b− a|1/3
≤ 1 .
On the other hand, choosing an = n−2, bn = n−1, for every γ > 1/3 we have
limn→∞
b1/3n − a1/3
n
|bn − an|1/3 · |bn − an|γ−1/3= lim
n→∞
1
|bn − an|γ−1/3= ∞.
21
Therefore f1 /∈ C0,γ for any γ > 1/3.
(ii) f2 /∈ C0,γ for any γ > 1. Indeed, choose an = 1(n+1)π bn = 1
nπ . observing that sin 1an
=
− sin 1bn
= ±1, we compute
|f2(bn)− f2(an)||bn − an|1/4
=
√1/(n+ 1) +
√1/n
(n+ n2)−1/4≥ 1
2.
This already shows that f2 /∈ C0,γ for any γ > 1/4. The fact that f2 ∈ C0,1/4 is proved byestimating the ratio
|f2(b)− f2(a)||b− a|1/4
separately in two cases:
CASE 1: b ∈[
1nπ ,
1(n−1)π
], a ∈
[1
(n+1)π , b].
CASE 2: b ∈[
1nπ ,
1(n−1)π
], a < 1
(n+1)π .
The first case is handled simply by integrating |f ′2| on the interval [a, b], observing that|f ′2(x)| ≤ x−1/2 + x−3/2. Case 2 follows from Case 1, using the inequality
|f2(b)− f2(a)| ≤ sup
|f2(b)− f2(x)| ; x ∈
[ 1
(n+ 1)π, b]
.
(iii) f3 ∈ C0,γ for every γ ∈ ]0, 1[ . Indeed, for any 0 < a < b < 1 one has
f3(b)− f3(a) =
∫ b
a(| lnx| − 1) dx ≤
∫ b
aCγx
γ−1 dx ≤∫ b
aCγ(x− a)γ−1 dx ≤ Cγ
γ(b− a)γ ,
for some constant Cγ .
10. Ascoli’s theorem guarantees the existence of a subsequence (fnk)k≥1 which converges tosome limit function f ∈ C0,γ([0, 1]), uniformly on [0, 1]. This means ‖fnk−f‖C0 → 0. However,in general it is not true that ‖fnk − f‖C0,γ → 0. In other words, the convergence takes placeonly in the norm of C0, not in the stronger norm of C0,γ .
11. To show that ‖ · ‖ϕ is a norm, we need to check that it satisfies the conditions (N1)–(N3)in the definition. The first two conditions are clear. To prove (N3) we observe that, for anyx, y ∈ Ω, x 6= y, one has
|(f + g)(x)| ≤ |f(x)|+ |g(x)| ,|f(x) + g(x)− f(y)− g(y)|
ϕ(|x− y|)≤ |f(x)− f(y)|
ϕ(|x− y|)+|g(x)− g(y)|ϕ(|x− y|)
.
Taking the supremum over all x, y we obtain ‖f + g‖ϕ ≤ ‖f‖ϕ + ‖g‖ϕ.
It remains to show that the space Cϕ is complete. For this purpose, let (fn)n≥1 be a Cauchysequence w.r.t. the norm ‖ · ‖ϕ. Then this sequence is Cauchy w.r.t. the C0 norm, hence it
22
converges uniformly to a continuous function f : Ω 7→ R. To show that ‖fn−f‖ϕ → 0, observethat, for any ε > 0 there exists N large enough so that
supx 6=y
∣∣∣fn(x)− fm(x)− fn(y) + fm(y)∣∣∣
|x− y|< ε for all m,n ≥ N .
Keeping n fixed and letting m→∞, this implies
supx 6=y
∣∣∣fn(x)− f(x)− fn(y) + f(y)∣∣∣
|x− y|< ε for all n ≥ N .
Hence ‖fn − f‖ϕ → 0.
12. One simply needs to retrace all steps in the proof of Ascoli’s theorem. Since E is compact,given ε > 0 there exists δ > 0 such that
d(x, y) ≤ δ =⇒ |f(x)− f(y)| ≤ ε for all x, y ∈ E and f ∈ F .
Since K is compact, it is precompact. Hence we can choose points α1, . . . , αm ∈ K such thatK ⊆
⋃mj=1B(αj , ε). The remaining steps of the proof can be repeated without change.
Chapter 4
1. (1) The operator Λ is linear and bounded, with ‖Λ‖ = 1, but not compact.
(2) This operator is not linear: (Λ(2f))(x) = sin(2f(x)) 6= 2 sin(f(x)) = 2(Λf)(x).
(3) Λ is linear and bounded, with ‖Λ‖ = 1, but not compact.
(4) Λ is linear, bounded, and compact. Indeed, Λf is always a polynomial of degree ≤ 1.Hence the range of Λ is two-dimensional.
(5) This operator Λ is linear, bounded, and compact. Indeed,
(Λf)(x) =
∫ x
0ey−xf(y) dy ,
Hence, for any x ∈ [0, 1],
|(Λf)(x)| ≤∫ x
0|f(y)| dy ≤ ‖f |C .
To see that Λ is compact, let (fn)n≥1 be a sequence of continuous functions with ‖fn‖C ≤ 1for every n. Then all functions Λ(fn) are Lipschitz continuous with constant 2. Indeed, fromthe differential equation it follows
|y′(x)| ≤ |f(x)|+ |y(x)| ≤ 1 + 1 .
23
By Ascoli’s theorem, the sequence (fn) admits a uniformly convergent subsequence.
2. To show that En is closed, assume ‖fk−f‖L1 → 0 and ‖fk‖2L2 ≤ n for every k. By possiblytaking a subsequence we can assume fk(x) → f(x) for a.e. x ∈ [0, 1]. Fatou’s lemma nowyields ∫ 1
0|f(x)|2dx =
∫ 1
0lim infn→∞
|fk(x)|2dx ≤ lim infn→∞
∫ 1
0|fk(x)|2dx ≤ n .
Hence f ∈ En as well.
To prove that En has empty interior in L1([0, 1]), consider any f ∈ En. For any ε > 0, define
fε(x).=
f(x) + εx−1/2 if f(x) ≥ 0 ,
f(x)− εx−1/2 if f(x) < 0 .
Then ‖fε − f‖L1 = 2ε but |fε(x)| ≥ εx−1/2 for every x ∈ ]0, 1], and hence fε /∈ L2([0, 1]).
3. Let πn : `∞ 7→ R be the projection operator, so that πn(x1, x2, . . .) = xn. If Λ : X 7→ `∞ isbounded, then the composition Λn = πn Λ is bounded as well.
Viceversa, assume that each linear functional Λn is bounded. By assumption, for every x ∈ Xone has
supn≥1|Λn(x)| < ∞ .
The uniform boundedness principle thus implies ‖Λ‖ = supn≥1 ‖Λn‖ <∞.
4. By the closed graph theorem, it suffices to prove that Λ has closed graph. Toward thisgoal, consider a sequence of functions fn ∈ Lp([0, 1]) such that
‖fn − f‖Lp → 0 , ‖Λ(fn)− g‖Lp → 0 ,
for some functions f, g ∈ Lp([0, 1]). We need to show that Λf = g. By choosing a subsequenceand relabeling, we can assume the pointwise convergence
fn(x)→ f(x) , (Λfn)(x)→ g(x) for a.e. x ∈ [0, 1] .
By assumption, this implies (Λfn)(x) → (Λf)(x) for a.e. x. Hence (Λf)(x) = g(x) for a.e.x ∈ [0, 1].
5. Assume that the range Y = K(X) is a closed subspace of X. Then Y itself is complete.The linear operator K is a continuous bijection between X and the Banach space Y . Henceit has a continuous inverse K−1.
Being the composition of a compact operator and a continuous one, the identity map ι(x) =x = K−1K(x) is a compact operator from X into itself. But this is impossible, because X isinfinite dimensional.
6. Let (xn)n≥1 be a bounded sequence of points in X.
CASE 1: Λ1 compact. Then the sequence Λ1(xn) admits a convergent subsequence, sayΛ1(xnk)→ y. Since Λ2 is continuous, this implies the convergence Λ2 Λ1(xnk)→ Λ2(y).
24
CASE 2: Λ2 is compact. Since Λ1 is continuous, the sequence (Λ1(xn))n≥1 is bounded in Y .By the compactness of Λ2, the sequence (Λ2 Λ1(xn))n≥1 admits a convergent subsequence.
7. If Y.= Range(Λ) is an infinite dimensional normed space, then the unit ball in Y is not
precompact. As shown in the proof of Theorem 2.22, there exist a sequence of points yn ∈ Ysuch that ‖yn‖ ≤ 1, ‖yn − ym‖ ≥ 1/2 for all m,n ≥ 1 m 6= n. Hence from this sequence(yn)n≥1 one cannot extract any convergent subsequence.
This leads a contradiction, because the assumptions imply that Λ is compact and Λ(y) = yfor every y ∈ Y .
8. First of all, observe that the assumptions imply 0 ∈ U . Call V.= −U ∩ U . Since V ⊆ U ,
it suffices to prove that V contains a neighborhood of the origin. Observe that
(i) V is closed, convex.
(ii) V = −V , 0 ∈ V .
(iii)⋃n≥1 nV = X.
Indeed, (i)-(ii) are clear. To prove (iii), consider any x ∈ X. Then there exist integers n1, n2
large enough so that x ∈ n1U , −x ∈ n2U . Calling n.= maxn1, n2 we have x
n ∈ U , −xn ∈ U .
By convexity, this implies
A.= co
xn, −x
n
=θx ; |θ| ≤ 1
n
⊆ U .
Since A = −A, we conclude that A ⊆ V . Hence x ∈ nA ⊆ nV .
Since X is a Banach space, X =⋃n≥1 nV , and each set nV is closed, by Baire’s category
theorem at least one of the sets nV must have non-empty interior. By homogeneity, V itselfhas non-empty interior. If B(x, r) ⊆ V , then (ii) implies B(−x, r) ⊆ V . By convexity,
B(0, r) =1
2B(x, r) +
1
2B(−x, r) ⊆ V.
9. Assume that K is surjective. By the open mapping theorem, K is an open map. Inparticular, the image of the unit ball B1
.= x ∈ X ; ‖x‖ < 1 contains a neighborhood of
the origin. But this is impossible, because the closure K(B1) is compact, and cannot containany open set in the infinite dimensional space Y .
In the particular example, the point y =(
1, 14 ,
19 , . . . ,
1n2 , . . .
)lies in `1 but not in the image
K(`1).
10. Fix any x ∈ R and consider a sequence xn → x. Since f is bounded, there exists theupper and lower limits
m.= lim inf
y→xf(y) ≤ lim sup
y→xf(y)
.= M .
We claim that m = M = f(x). Indeed, assume m 6= f(x). Then we can find a sequencexn → x such that f(xn)→ m. Hence the point (x,m) lies in the closure of the graph of f . If
25
f(x) 6= m, this graph is not closed and we reach a contradiction. The proof that f(x) = M isentirely similar.
To show that the result fails without the boundedness assumption, consider the functionf(x) = x−1 if x 6= 0, f(0) = 0. Then f has closed graph but is not continuous at x = 0.
11. Let f : X 7→ Y be continuous. Assume that xn → x and f(xn) → y. By continuity,f(x) = limn→∞ f(xn) = y, hence the graph of f is closed.
To construct the example, choose any function 0 6= f ∈ L1(R). Define the map g : t 7→ gt(·)from R into L1(R) by setting
gt(x).=
0 if t ≤ 0, x ∈ R ,
f(x− 1
t
)if t > 0, x ∈ R .
Then g is not continuous at t = 0 because, for t > 0, ‖gt − g0‖L1 = ‖f‖L1 and does notapproach zero as t→ 0+. On the other hand, the map g has closed graph. To see this, simplyobserve that, for any sequence tk → 0+, the sequence f tk has no limit.
12. Assume that limk→∞ λk → 0. For each n ≥ 1 consider the truncated operator
Λn(x1, x2, . . .) = (λ1x1, λ2x2, . . . , λnxn, 0, 0, . . .). (9)
Each Λn is continuous and has finite dimensional range, hence is compact. Observing that‖Λ− Λn‖ = supj>n |λj | → 0 as n→∞, by Theorem 4.10 we conclude that Λ is compact.
On the other hand, assume that lim supn→∞ |λn| > 0. Then there exists ε > 0 and a subse-quence (λnk)k≥1 such that |λnk | ≥ ε for every k ≥ 1. Consider the unit vectors
e1 = (1, 0, 0, 0, . . .), e2 = (0, 1, 0, 0, . . .), e3 = (0, 0, 1, 0, . . .), . . .
Then the sequence (enk)k≥1 is bounded, but the sequence Λ(enk) = λnkenk does not admitany convergent subsequence. Indeed,
‖Λenj − Λenk‖ = ‖λnjenj − λnkenk‖ ≥ ‖λnkenk‖ ≥ ε .
Hence Λ is not compact.
13. Let B be continuous at the origin. Then there exists δ > 0 such that |B(x, y)| ≤ 1whenever ‖x‖ ≤ δ and ‖y‖ ≤ δ.
For any (x, y) ∈ X × Y we now have
|B(x, y)| =1
δ2· ‖x‖ ‖y‖ ·B
(δx
‖x‖,δy
‖y‖
)≤ C · ‖x‖ ‖y‖ ,
with C = δ−2.
14. Given any polynomial p, we have∣∣∣∣∫ 1
0p(t) q(t) dt
∣∣∣∣ ≤ maxx∈[0,1]
|p(x)| ·∫ 1
0|q(t)| dt.
26
Hence the map Λp : q 7→ B(p, q) is a bounded linear operator from X into itself, with norm‖Λp‖ = ‖p‖C([0,1]).
To prove that the bilinear map B is not continuous on the product space X×X, we proceed asfollows. Choose a sequence of numbers xn > 0 such that | lnxn| > n2. Define the continuousfunctions
fn(x).=
1√nx
if x ∈ [xn, 1] ,
1√nxn
if x ∈ [0, xn] .
Observe that ∫ 1
0fn(x) dx <
1√n
∫ 1
0
√x dx =
2√n.∫ 1
0f2n(x) dx >
∫ 1
xn
1
nxdx >
1
n| lnxn| ≥ n .
By the Stone-Weierstrass theorem, each function fn can be uniformly approximated by apolynomial. We can thus construct a sequence of polynomials (pn)n≥1 such that∫ 1
0pn(x) dx <
2√n,
∫ 1
0p2n(x) dx > n
for every n ≥ 1. We thus have (pn, pn)→ 0 in X ×X, but B(pn, pn)→∞.
15. Since S is bounded, there exists M such that ‖x‖ ≤ M for all x ∈ S. Let ε ∈ ]0, 1] begiven. Let Yε be a finite dimensional space such that d(x, Yε) ≤ ε for all x ∈ S.
In the finite dimensional space Yε, the closed ball B.= y ∈ Yε ; ‖y‖ ≤ 1 + M centered at
the origin with radius 1 +M is compact. Hence it can be covered with finitely many balls ofradius ε, say B(yn, ε), n = 1, . . . , N . We claim that
S ⊆N⋃n=1
B(yn, 2ε) .
Indeed, for every x ∈ S there exists yx ∈ Yε such that ‖x − yx‖ ≤ ε. Clearly, we must have‖yx‖ ≤M + ε. Hence yx ∈ B(yn, ε) for some n. This implies x ∈ B(yn, 2ε).
The previous analysis shows that S is precompact. Namely, for every ε > 0 is can be coveredby finitely many balls of radius ε. By assumption, S is a closed subset of a Banach space,hence it is complete. We thus conclude that S is compact.
16. If A 6= λI, consider two cases.
CASE 1: There exists a vector f ∈ X such that g = Af is not a scalar multiple of f .
Clearly, the two vectors f, g must be linearly independent.
Consider the two-dimensional subspace V = spanf, g. Let ϕ : V 7→ R be the linear func-tional such that ϕ(f) = 0, ϕ(g) = 1. Otherwise stated,
ϕ(αf + βg).= β for all α, β ∈ K .
27
Using the Hahn-Banach extension theorem, this functional can be extended to a boundedlinear functional, still called ϕ, defined on the entire space X.
We then define the operator K : X 7→ X by setting K(x).= ϕ(x) g. Notice that this implies
K(f) = ϕ(f) g = 0, K(g) = ϕ(g) g = g .
Moreover, K has one-dimensional range, hence it is compact.
We now check that A(Kf) = 0 while K(Af) = Kg = g. Hence AK 6= KA.
CASE 2: There exists two nonzero vectors f, g ∈ X such that Af = λf , Ag = λ′g with λ 6= λ′.Then A(f + g) = λf + λ′g is not a scalar multiple of f + g, and Case 1 again applies.
17. Since V is an infinite dimensional normed space, we can find a sequence of points (xn)n≥1
in V such that ‖xn‖ ≤ 1, ‖xm − xn‖ ≥ 1/2 for all m 6= n. Then the sequence (Λxn)n≥1 doesnot admit any convergent subsequence. Indeed, by assumption
‖Λxm − Λxn‖ ≥ ε‖xm − xn‖ ≥ε
2for all m 6= n.
Therefore, the operator Λ cannot be compact.
18. Consider the family ϕn ; n ≥ 1 of bounded linear functionals on X∗, defined by
ϕn(x∗) =n∑k=1
〈x∗, xk〉.
By assumption,limn→∞
ϕn(x∗) = ϕ(x∗)
exists for every x∗ ∈ X∗. Hence supn≥1 |ϕn(x∗)| < ∞ for every x∗ ∈ X∗.
By the uniform boundedness principle, M.= supn≥1 ‖ϕn‖ <∞. Therefore
‖ϕ‖ .= sup
‖x∗‖≤1|ϕ(x∗)| ≤ sup
‖x∗‖≤1supn≥1|ϕn(x∗)| ≤ sup
n≥1‖ϕn‖ = M,
showing that ϕ is bounded.
19. For every continuous function f : [0, 1] 7→ R, the mean value theorem implies
limt→0
1
t
∫ t
0f(s) ds = f(0) .
Hence the function Λf is continuous as well.
(i) It is clear that the operator Λ is linear. Observing that the average value of f over theinterval [0, t] satisfies ∣∣∣∣1t
∫ t
0f(s) ds
∣∣∣∣ ≤ maxs∈[0,t]
|f(s)| ≤ ‖f‖C([0,1] ,
28
we conclude that Λ is a bounded operator, with norm ‖Λ‖ ≤ 1.
(ii) If (Λf)(x) = 0 for every x ∈ [0, 1], then∫ t
0 f(s) ds = 0 for every t > 0 and hence f(s) = 0for every x ∈ [0, 1]. This proves that Λ is one-to-one.
If g = Λf for some continuous function f , taking derivatives we obtain
g′(t) =f(t)− g(t)
tfor all t ∈ ]0, 1[ .
Hence g is continuously differentiable. Any continuous function g : [0, 1] 7→ R which is notcontinuously differentiable cannot be in the range of Λ.
(iii) To prove that Λ is not compact, consider the sequence of functions
fn(x).=
sin 2nx if x ∈ [0, 21−nπ] ,
0 otherwise.
Observe that Λfn(x) = 0 for x ≥ 21−nπ. Hence, for m < n a direct computation yields
‖Λfm − Λfn‖C0 ≥∣∣∣Λfm(2−mπ)− Λfn(2−mπ)
∣∣∣ =∣∣∣ 2π− 0∣∣∣ =
2
π.
As a consequence, the sequence (Λfn)n≥1 does not admit any uniformly convergent subse-quence.
20. (i) The multiplication operator Mf is one-to-one if and only if g(x) 6= 0 for a.e. x ∈ Ω.
(ii) In general, the range of Mg is not closed. For example, let Ω = ]0, 1[ and g(x) = x. Thenthe function h(x) ≡ 1 is not in the range of Mg, because the function f(x) = h(x)/g(x) = 1/xdoes not lie in Lp(Ω). However, if 1 ≤ p <∞, then the characteristic functions fn
.= χ
[1/n , 1]
lie in the range of Mg. Moreover, ‖fn − f‖Lp → 0 whenever 1 ≤ p <∞.
To see a positive result, assume that there exists ε > 0 such that the domain Ω can bedecomposed as Ω = Ω0 ∪ Ωε, with Ω0, Ωε disjoint measurable sets, and
g(x) = 0 for a.e. x ∈ Ω0 , |g(x)| ≥ ε for a.e. x ∈ Ωε .
Then Λ has closed range. Indeed,
Range(Λ) =f ∈ Lp(Ω) ; f(x) = 0 for a.e. x ∈ Ω0
.
(iii) If g does not coincide a.e. with the zero function, then the operator Mg is not compact.
To see this, assume that, for some ε > 0, the set Aε.= x ∈ Ω ; |g(x)| ≥ ε has strictly
positive measure. Then we can find countably many disjoint measurable sets An such that
An ⊂ Aε , 0 < meas(An) < 1 for all n ≥ 1 .
Define the sequence of functions
fn(x).=
cn if x ∈ An ,0 otherwise,
29
choosing the constants cn so that ‖fn‖Lp = 1. By construction, the sequence (fn)n≥1 isbounded. However, the sequence (Mgfn)n≥1 does not admit any convergent subsequence.Indeed, for any m 6= n, since fn and fm have disjoint support, one has
‖fng − fmg‖Lp ≥ ‖fng‖Lp ≥ ε ‖fn‖Lp = ε.
21. In the space `1, consider the compact operator
Λ(x1, x2, x3, . . .) =(x1
1,x2
2,x3
3, . . .
).
Then Λ is one-to-one and its range is dense in `1. However, given any ε > 0, choose n > ε−1
and consider the truncated operator
Λn(x1, x2, x3, . . .) =(x1
1,x2
2,x3
3, . . . ,
xnn, 0, 0, . . .
).
Then ‖Λn − Λ‖ < ε but Λn is not one-to-one and its range is not dense in `1.
22. (i) By the open mapping theorem, the functional Λ has a continuous inverse. Hence thereexists a constant C > 0 such that ‖Λ−1y‖ ≤ C ‖y‖ for every y ∈ Y . Taking β = C−1, thisimplies β‖x‖ ≤ ‖Λx‖ for all x ∈ X.
(ii) We check that, if γ.= ‖Ψ‖ < 1/β, then the map u 7→ Λ−1(f −Ψu) is a strict contraction.
Indeed, for any u, v ∈ X one has
‖Λ−1(f−Ψu)−Λ−1(f−Ψv)‖ ≤ ‖Λ−1(Ψv)−Λ−1(Ψu)‖ ≤ ‖Λ−1‖·‖Ψ‖ ‖u−v‖ ≤ β · γ‖u−v‖ .
Since β · γ < 1, this is a strict contraction and therefore has a unique fixed point.
(iii) Let Λ : X 7→ Y be a continuous bijection, and let β > 0 be such that ‖Λx‖ ≥ β‖x‖ for allx ∈ X. We claim that, if Ψ : X 7→ Y is any bounded linear operator with ‖Ψ‖ < β−1, thenthe sum Λ + Ψ is still a bijection. Indeed, by (ii) for any f ∈ Y there exists a unique elementu ∈ X such that u = Λ−1(f −Ψu). This implies Λu = f −Ψu, hence (Λ + Ψ)u = f . showingthat the operator ΛΨ is a bijection.
23. (i) From the definition it follows
|yn| =
∣∣∣∣∣ 1nn∑i=1
xi
∣∣∣∣∣ ≤ max1≤i≤n
|xi| ≤ ‖x‖`∞ .
Hence this averaging operator has norm ‖Λ‖ = 1.
To prove that Λ is not compact, consider the sequences xk = (xk1, xk2, xk3, . . .) ∈ `∞ definedby
xki.=
1 if 2k−1 < i ≤ 2k ,0 otherwise.
Clearly ‖xk‖`∞ = 1 for every k. However, this bounded sequence does not admit any subse-quence (xk`)`≥1 such that Λxk` converges. Indeed, if j < k, then
‖Λxj − Λxk‖`∞ ≥ |xj,2j − xk,2j | =∣∣∣12− 0∣∣∣ =
1
2.
30
(ii) Λ is not a bounded operator from `1 into `1. For example, if x = (1, 0, 0, . . .) ∈ `1, then
Λx =(
1, 12 ,
13 , . . .
)/∈ `1.
24. (i) If Y = X, let ι : Y 7→ X be the identity map. This is a bijective, continuous, linearmap from Y onto X. By Corollary 4.5, the inverse map ι−1 : X 7→ Y is continuous as well.Hence there exists a constant C ′ such that ‖x‖Y ≤ C ′‖x‖X for all x ∈ X.
(ii) Assume Y 6= X. Consider the sets Sn.= y ∈ Y ; ‖y‖Y ≤ n, and let Sn be the closure
of Sn in the space X. We claim that
(C) Each Sn has empty interior.
If (C) holds, then
Y =⋃n≥1
Sn ⊆⋃n≥1
Sn .
Hence Y is of first category, being contained in the union of countably many closed sets withempty interior.
To prove (C), we argue by contradiction. If some Sn has nonempty interior, by positivehomogeneity the set S1 has nonempty interior as well. To fix the ideas, assume that the openball B(x, r) ⊂ S1 for some x ∈ S1 and r > 0. Since S1 = −S1, we also have B(−x, r) ⊂ S1.By convexity, this implies
B(0, r) =1
2B(x, r) +
1
2B(−x, r) ⊂ S1 .
In turn, by positive homogeneity this yields
B(0, 2−nr) ⊂ S2−n.= y ∈ Y ; ‖y‖Y ≤ 2−n .
Repeating the argument in step 3. of the proof of Theorem 4.4 (the Open Mapping Theorem),we conclude that B(0, r/2) ⊂ S1.
For any x ∈ X we thus have
x =3 ‖x‖Xr
· rx
3‖x‖X∈ 3 ‖x‖X
r·B(0, r/2) ⊂ 3 ‖x‖X
r· S1 ⊂ Y.
This shows that X = Y , against the assumption.
Chapter 5
1. Using the properties of inner products one obtains
(x, y + z) = (y + z, x) = (y, x) + (z, x) = (x, y) + (x, z).
31
(x, αy) = (αy, x) = α(y, x) = α(x, y) .
Saying that the vectors x, y are mutually orthogonal means that (x, y) = 0. This implies
‖x+ y‖2 = (x+ y, x+ y) = (x, x) + (x, y) + (y, x) + (y, y) = ‖x‖2 + 0 + 0 + ‖y‖2.
2. Let x, y ∈ H and ε > 0 be given. Let M.= max‖x‖, ‖y‖. Choose 0 < δ < 1, such that
δ = ε/(1 + 2M). If ‖x′ − x‖ ≤ δ and ‖y′ − y‖ ≤ δ, then
|(x′, y′)− (x, y)| =∣∣∣(x′ − x, y′ − y) + (x′ − x, y) + (x, y′ − y)
∣∣∣≤ ‖x′ − x‖ ‖y′ − y‖+ ‖x′ − x‖ ‖y‖+ ‖x‖ ‖y′ − y‖ ≤ δ2 + 2Mδ ≤ ε .
3. (i) The parallelogram identity is proved by writing
‖x+y‖2+‖x−y‖2 = (x+y, x+y)+(x−y, x− y) = 2(x, x)+2(y, y) = 2‖x‖2+2‖y‖2 . (10)
(ii) Assume that the norm ‖ · ‖ satisfies the parallelogram identity. We claim that the assign-ment
(x, y).=
1
2
(‖x+ y‖2 − ‖x‖2 − ‖y‖2
)=
1
4
(‖x+ y‖2 − ‖x− y‖2
)(11)
satisfies all properties of an inner product. Note that the equality between the last two termsin (11) is an easy consequence of (10).
By (11), the identity (x, y) = (y, x) is obvious.
To check that (x+ x′, y) = (x, y) + (x′, y), using the parallelogram identity we obtain
‖x+ x′ + y‖2 = 2‖x+ y‖2 + 2‖x′‖2 − ‖x+ y − x′‖2
= 2‖x′ + y‖2 + 2‖x‖2 − ‖x′ + y − x‖2
= ‖x‖2 + ‖x′‖2 + ‖x+ y‖2 + ‖x′ + y‖2 − 1
2‖x+ y − x′‖2 − 1
2‖x′ + y − x‖2 .
This implies
(x+ x′, y) =1
4
(‖x+ x′ + y‖2 − ‖x+ x′ − y‖2
)=
1
4
(‖x‖2 + ‖x′‖2 + ‖x+ y‖2 + ‖x′ + y‖2 − 1
2‖x+ y − x′‖2 − 1
2‖x′ + y − x‖2
)
−1
4
(‖x‖2 + ‖x′‖2 + ‖x− y‖2 + ‖x′ − y‖2 − 1
2‖x− y − x′‖2 − 1
2‖x′ − y − x‖2
)
=1
4
(‖x+ y‖2 − ‖x− y‖2
)+
1
4
(‖x′ + y‖2 − ‖x′ − y‖2
)= (x, y) + (x′, y) .
32
In particular, from the above identity by induction it follows
(nx, y) = (x+ x+ · · ·+ x, y) = n(x, y) .
In turn, this implies(
1nx, y
)= 1
n(x, y), and hence
(λx, y) = λ(x, y) (12)
for every rational number λ = mn . By continuity, we conclude that (12) remains valid for every
λ ∈ R.
Finally, we check that if x 6= 0, then
(x, x) =1
4
(‖x+ x‖2 − ‖x− x‖2
)= ‖x‖2 > 0.
Hence the inner product is positive definite, and√
(x, x) = ‖x‖.
4. (i) Choose x = (1, 0), y = (0, 1). Then ‖x‖p = ‖y‖p = 1, ‖x + y‖p = ‖x − y‖p = 21/p. Ifp 6= 2, then
‖x + y‖2 + ‖x− y‖2 = 22/p + 22/p 6= 2 + 2 = 2‖x‖2 + 2‖y‖2
(ii) Choose f(x) = sinπx, g(x) = 1 − sinπx. Then ‖f‖ = ‖g‖ = ‖f + g‖ = ‖f − g‖ = 1 andthe parallelogram identity fails.
5. Since the monomials 1, x are mutually orthogonal in L1([−1, 1]), the first two polynomialsin an orthonormal basis are
p0(x) =1
‖1‖L2
=1√2, p1(x) =
x
‖x‖L2
=
√3
2x .
To find p2, we compute the inner product
(p0, x2)L2 =
∫ 1
−1p0(x)x2 dx =
∫ 1
−1
x2
√2dx =
√2
3.
Hence q(x) = x2 − (p0, x2)L2 · p0(x) = x2 − 1
3 is perpendicular to p0, and to p1 as well. Wecompute
‖q‖2L2 =
∫ 1
−1
(x2 − 1
3
)2dx = 2
∫ 1
0
(x4 − 2
3x2 +
1
9
)dx = 2
(1
5− 2
9+
1
9
)=
8
45.
Therefore
p2(x) =q(x)
‖q‖L2
=
√45
8
(x2 − 1
3
).
6. For every x ∈ H, Bessel’s inequality (5.10) implies∑∞
n=1 |(x, en)|2 ≤ ‖x‖2. Hencelimn→∞ |(x, en)| = 0. By definition, this means that en 0.
33
7. Since H is infinite dimensional, for a given vector x ∈ H we can find an orthonormalsequence (ek)k≥1 such that (x, ek) = 0 for every k. Choose λ
.=√
1− ‖x‖2. Then the sequencexn
.= x+λen satisfies all requirements. Indeed, by Pythagora’s theorem ‖xn‖2 = ‖x‖2+λ2 = 1.
Moreover, for any y ∈ H by the previous problem 6. we have
limn→∞
(y, xn) = (y, x) + limn→∞
λ(y, en) = (y, x) .
8. (i) By assumption,∑
k ‖αkvk‖ <∞, hence the series is absolutely convergent. Since H iscomplete, the series converges.
(ii) Let (vk)k≥1 be an orthonormal sequence. Then (assuming m ≤ n)
lim supm,n→∞
∣∣∣∣∣∣∑
m<k≤nαkvk
∣∣∣∣∣∣ = lim supm,n→∞
∑m<k≤n
|αk|2 .
This shows that the sequence of partial sums is Cauchy if and only if∑
k |αk|2 <∞.
9. For any f ∈ L2(Rn), performing the change of variable y = φ(x) and using the fact thatdetDφ ≡ 1, we compute∫
Rn|(Λf)(x)|2 dx =
∫Rn|f(φ(x))|2 dx =
∫Rn|f(y)|2 dy.
Hence ‖Λf‖L2 = ‖f‖L2 for every f ∈ L2(Rn), proving that ‖Λ‖ = 1.
The adjoint operator Λ∗ is defined by the identity∫Rnf(y)·(Λ∗g)(y) dy =
∫Rn
(Λf)(x)·g(x) dx =
∫Rnf(φ(x))·g(x) dx =
∫Rnf(y)·g(φ−1(y)) dy .
Hence (Λ∗g)(y) = g(φ−1(y), showing that Λ∗ = Λ−1.
10. The adjoint operator is defined by the identity
(Λx, y) =∞∑k=1
xi+1yi =∞∑j=2
xjyj−1 = (x,Λ∗y).
Hence Λ∗(y1, y2, y3, . . .) = (0, y1, y2, . . .). Clearly, Λ is onto but not one-to-one, while Λ∗ isone-to-one but not onto.
11. (i) Assume x = θx1 +(1−θ)x2, for some x1 6= x2, with ‖x1‖ ≤ 1, ‖x2‖ ≤ 1 and 0 < θ < 1.
By the parallelogram’ identity
‖x1 + x2‖2 = ‖x1‖2 + ‖x2‖2 + 2Re(x1, x2) = 2‖x1‖2 + 2‖x2‖2 − ‖x1 − x2‖2.
If x1 6= x2, then we have the strict inequality
2Re(x1, x2) < ‖x1‖2 + ‖x2‖2.
34
Therefore‖x‖2 =
(θx1 + (1− θ)x2 , θx1 + (1− θ)x2
)= θ2‖x1‖2 + (1− θ)2‖x2‖2 + 2θ(1− θ)Re(x1, x2)
< θ2‖x1‖2 + (1− θ)2‖x2‖2 + θ(1− θ)(‖x1‖2 + ‖x2‖2)
≤ θ2 + (1− θ)2 + 2θ(1− θ) = 1.
This proves that every point x in the closed unit ball of a Hilbert space which is not extremalmust have norm ‖x‖ < 1. Hence all points with ‖x‖ = 1 are extremal.
(ii) Next, assume f ∈ L1([0, 1]) with ‖f‖L1 = 1. Then we can find a set A ⊂ [0, 1] withpositive measure such that |f(x)| ≥ 1/2 for every x ∈ A. To fix the ideas, assume f(x) ≥ 1/2for all x ∈ A, with meas(A) > 0. Choose two disjoint measurable subsets A1, A2 ⊂ [0, 1] withA1 ∪A2 = A, meas(A1) = meas(A2). Consider the function
g(x) =
1/2 if x ∈ A1 ,−1/2 if x ∈ A2 ,
0 if x /∈ A1 ∪A2 .
Then
‖f + g‖1L = ‖f − g‖L1 = 1 , f =f + g
2+f − g
2.
Hence f is not an extreme point of the unit ball in L1, being a convex combination of the twofunctions f + g and f − g.
12. By taking a subsequence and relabeling, one can assume ‖xn‖ → C as n→∞. Since thissequence is bounded, by Theorem 5.14 it admits a weakly convergent sequence, say xnj xfor some x ∈ H. We now compute
‖x‖2 = (x, x) = limj→∞
(x, xnj ) ≤ lim supj→∞
‖x‖ ‖xnj‖ ≤ C‖x‖ .
Dividing by ‖x‖ we conclude ‖x‖ ≤ C.
13. By assumption, H admits a countable, everywhere dense set x1, x2, x3, . . .. Usingthe Gram-Schmidt procedure, from this set we construct a countable orthonormal basise1, e2, e3, . . ..
DefiningΛx
.= a = (a1, a2, a3, . . .) with ak
.= (x, ek),
we obtain the desired bijection. Indeed,
‖Λx‖`2 =∞∑k=1
|ak|2 =
∞∑k=1
|(x, ek)|2 = ‖x‖ .
14. (i) Assume x ∈ V .= spanv1, . . . , vn, so that
x =
n∑k=1
θkvk
35
v
v2
v1
h
1
v2
v3
h
2
3
Figure 1: Computing the area of a parallelogram and the volume of a parallelepiped. Here h2 =d(v2 , spanv1) while h3 = d(v3 , spanv1, v2).
for some coefficients θk. To actually compute these coefficients, we observe that, for everyj = 1, . . . , n, one must have
(x, vj) =n∑k=1
θk(vk, vj) .
Therefore the numbers θ1, . . . , θn are obtained by solving the system of n linear equations(v1, v1) · · · (vn, v1)...
. . ....
(v1, vn) · · · (vn, vn)
θ1
...θn
=
(x, v1)...
(x, vn)
. (13)
Take x to be the zero vector in (13). Then that the vectors v1, . . . , vn are linearly independentif and only if θ1 = θ2 = · · · = θn = 0 is the only solution of (13) when x = 0. This is the caseif and only if the determinant G((v1, . . . , vn) of the n×n symmetric matrix in (13) is not zero.
(ii) If x ∈ H and y ∈ V then G(x, v1, v2, . . . , vn) = G(x + y, v1, v2, . . . , vn). In particular,taking y = PV (x) and z = x− y = PV ⊥(x), we obtain
d(x, V ) = ‖z‖ , G(x, v1, v2, . . . , vn) = G(z, v1, v2, . . . , vn) = (z, z)G(v1, v2, . . . , vn).
Therefore
d(x, V ) = ‖z‖ = (z, z)1/2 =
√G(x, v1, v2, . . . , vn)
G(v1, v2, . . . , vn).
(iii) The volume of the parallelepiped with edges v1, . . . , vn can be computed as a product:
‖vn‖ · d(vn−1 ; spanvn
)· d(vn−2 ; spanvn−1, vn
)· · · d
(v1 ; spanv2, . . . , vn
)√G(vn) ·
√G(vn−1, vn)
G(vn)· · ·
√G(v1, v2, . . . , vn)
G(v2, . . . , vn)=√G(v1, v2, . . . , vn),
where each factor was computed using (ii).
15. Let U ⊂ L2(R) be the subspace of all even functions. This is a closed subspace, hence forevery f ∈ L2 the perpendicular projection g0 = πUf is well defined. We claim that
g0(x) =f(x) + f(−x)
2.
36
Indeed, g0(x) = g0(−x) for every x ∈ R, hence g0 ∈ U . It remains to check that f − g0 isperpendicular to every function g ∈ U . This is indeed the case because, if g is even,∫
g(x)
(f(x)− f(x) + f(−x)
2
)dx =
∫g(x)
f(x)
2dx−
∫g(x)
f(−x)
2dx
=
∫g(x)
f(x)
2dx−
∫g(−x)
f(x)
2dx = 0 .
16. To prove that (i)=⇒(ii), assume that K is compact. Let B1 = x ∈ X ; ‖x‖ ≤ 1 be theclosed unit ball in X. Given any ε > 0, by assumption the image K(B1) is precompact, henceit can be covered with finitely many balls of radius ε. Say, K(B1) ⊆
⋃ni=1B(yi, ε).
Call Y.= spany1, . . . , yn and set
Kε(x).= πY K(x) .
Clearly, Kε has finite dimensional range. Moreover,
‖K −Kε‖ = sup‖x‖≤1
‖K(x)−Kε(x)‖ = sup‖x‖≤1
‖K(x)− πYK(x)‖ ≤ sup‖x‖≤1
d(K(x), Y ) ≤ ε .
Indeed,d(K(x), Y ) ≤ min
1≤i≤n‖K(x)− yi‖ ≤ ε .
The converse implication (ii)=⇒ (i) is an immediate consequence of Theorem 4.10.
17. Assume that∑
n ‖un − vn‖ < 1. Let (un)n≥1 be complete. If (vn)n≥1 is not complete,then there exists a unit vector x ∈ H which is perpendicular to every vn. A contradiction isobtained by writing
1 = ‖x‖2 =
∞∑n=1
(x, un)2 =
∞∑n=1
(x, un − vn)2 ≤∞∑n=1
‖x‖2 ‖un − vn‖2 ≤∞∑n=1
‖un − vn‖ .
18. (i) Let (un)n≥1 be mutually orthogonal. Assume θ0y+ θ1u1 + · · ·+ θNuN = 0, for somecoefficients θj not all zero. Since the vectors u1, . . . , uN , uN+1 are linearly independent, wemust have θ0 6= 0 and uj 6= 0 for every j. This yields a contradiction because
0 = (0, uN+1) = (θ0y + θ1u1 + · · ·+ θNuN , uN+1) = θ0(y, uN+1) = θ02−N−1‖uN+1‖2 .
(ii) Let (un)n≥2 be an orthonormal set, and define v.=∑∞
n=2 2−nun, u1.= v/‖v‖. Observe
that in this case the two vectors y, u1 are parallel.
19. Assume that the strong convergence ‖xn − x‖ → 0 holds. Then for every y ∈ H one has
|(y, xn)− (y, x)| ≤ ‖y‖ · ‖xn − x‖ → 0 as n→∞.
37
Hence one has the weak convergence xn x as well. Moreover, if ‖xn‖ ≥ ‖x‖, then∣∣∣‖xn‖ − ‖x‖∣∣∣ ≤ ‖xn − x‖ → 0 as n→∞.
Viceversa, assume xn x and ‖xn‖ → ‖x‖. Then
lim supn→∞
‖xn − x‖2 = lim supn→∞
(xn − x , xn − x)
= limn→∞
((xn, xn)− (xn, x)− (x, xn) + (x, x)
)= ‖x‖2 − ‖x‖2 − ‖x‖2 + ‖x‖2 = 0.
20. The orthogonal subspace is U⊥ = W =
w ∈ L2(Q) ;
∫ 1
0w(x, y) dx = 0 for a.e. y ∈ [0, 1]
.
Given f ∈ L2(Q), we have the perpendicular decomposition f = u+w with u ∈ U , w ∈W =U⊥. Here
u(x, y) = ϕ(y) =
∫ 1
0f(s, y) ds , w(x, y) = f(x, y)−
∫ 1
0f(s, y) ds .
Observe that u and w are perpendicular because∫Qu(x, y)w(x, y) dxdy =
∫ 1
0ϕ(y)
(∫ 1
0w(x, y) dx
)dy =
∫ 1
0ϕ(y) · 0 dy = 0 .
In this case, the function g ∈ U which has minimum distance from f is the perpendicularprojection: g = πU (f) = u.
21. Repeat the arguments in the proof of Theorem 4.2.
22. (i) The closure and convexity of Ω ⊂ L2(R) are straightforward. If ‖fn − f‖L2 → 0, wecan find a subsequence such that fn(x) → f(x) for a.e. x ∈ R. If fn(x) ≤ ex for every n ≥ 1and a.e. x ∈ R, then also f(x) ≤ ex for a.e. x. Hence Ω is a closed subset of L2(R).
Moreover, if f(x) ≤ ex and g(x) ≤ ex for a.e. x ∈ R, the same is true for any convexcombination: θf(x) + (1− θ)g(x) ≤ ex for every θ ∈ [0, 1] and a.e. x.
(ii) Clearly the function g(x) = minf(x), ex lies in the convex set Ω. By the result provedin problem 21, it suffices to check that∫
(ω(x)− g(x)) · (g(x)− f(x)) dx ≥ 0 for all ω ∈ Ω . (14)
By assumption, ω(x) ≤ ex for a.e. x. For each x ∈ R we consider two cases.
Case 1: f(x) > ex. Then g(x) = ex and therefore
ω(x)− g(x) ≤ 0 , g(x)− f(x) < 0 .
Case 2: f(x) ≤ ex. Then g(x) = f(x) and therefore g(x)− f(x) = 0.
38
In both cases, (ω(x)− g(x)) · (g(x)− f(x)) ≥ 0. Hence (14) holds.
23. (i) The operator Λ has norm ‖Λ‖ =√
2. Indeed,
‖Λf‖L2 =
(∫ 0
−∞|f(−x)|2 dx+
∫ ∞0|f(x)|2dx
)1/2
=
(2
∫ ∞0|f(x)|2dx
)1/2
≤(
2‖f‖2L2
)1/2=√
2 · ‖f‖L2 ,
with equality holding whenever f is supported on the positive half axis.
(ii) Ker(Λ) = f ∈ L2 ; f(x) = 0 for a.e. x > 0,Range(Λ) = f ∈ L2 ; f(x) = f(−x) for a.e. x ∈ R.
(iii) The adjoint operator Λ∗ is defined by the identity∫f(x)(Λ∗g)(x) dx =
∫(Λf)(x)g(x) =
∫ 0
−∞f(−x)g(x) dx+
∫ ∞0
f(x)g(x)dx
=
∫ ∞0
f(x)(g(x) + g(−x)
)dx .
This implies
(Λ∗g)(x) =
0 if x < 0 ,
g(−x) + g(x) if x > 0 .
24. (i) For any f ∈ L2([0,∞[), the substitution y = ex yields∫ ∞0|f(ex)|2 dx =
∫ ∞1
|f(y)|2
ydy ≤
∫ ∞0|f(y)|2dy .
Hence ‖Λf‖L2 ≤ ‖f‖L2 and ‖Λ‖ ≤ 1. To prove the converse inequality consider the sequenceof functions fn =
√n · χ[1,1+1/n]. Then
‖fn‖L2 = 1 , ‖Λfn‖2 =
∫ 1+1/n
1
n
ydy = n ln
(1 +
1
n
)→ 1 as n→∞.
Therefore ‖Λ‖ = 1.
(ii) Ker(Λ) =f ∈ L2([0,∞[) ; f(x) = 0 for a.e. x ∈ [0, 1]
.
Range(Λ) =g ∈ L2([0,∞[) ;
∫ ∞1
|g(ln y)|2
ydy < ∞
.
(iii) The adjoint operator Λ∗ is characterized by the identity∫ ∞0
f(x)(Λ∗g)(x) dx =
∫ ∞0
(Λf)(x)g(x) dx =
∫ ∞0
f(ex)g(x) dx =
∫ ∞1
f(y)g(ln y)
ydy .
Therefore
(Λ∗g)(y) =
0 if y ∈ [0, 1] ,
g(ln y)
yif y > 1 .
39
25. Assume fn f . Then for any b ∈ [0, T ], taking g = χ[0,b] ∈ L2([0, T ]) we have
limn→∞
∫ b
0fn(x) dx = lim
n→∞
∫ T
0fn(x)g(x) dx =
∫ T
0f(x)g(x) dx =
∫ b
0f(x) dx .
Viceversa, assume that ‖f‖L2 ≤M , ‖fn‖L2 ≤M for all n ≥ 1, and that
limn→∞
∫ b
0fn(x) dx =
∫ b
0f(x) dx for every b ∈ [0, T ] .
If ϕ is a piecewise constant function of the form
ϕ =N∑k=1
ck · χ[0,bk] (15)
for some constants ck, bk, then by linearity we still have
limn→∞
∫ T
0fn(x)ϕ(x) dx =
N∑k=1
limn→∞
∫ bk
0ckfn(x) dx =
N∑k=1
∫ bk
0ckf(x) dx =
∫ T
0f(x)ϕ(x) dx .
Now consider an arbitrary function g ∈ L2([0, T ]). Given any ε > 0, we can find a piecewiseconstant function ϕ of the form (15) such that ‖g − ϕ‖L2 < ε. This yields the estimate
lim supn→∞
∣∣∣∣∫ T
0(fn − f)g dx
∣∣∣∣ ≤ lim supn→∞
∣∣∣∣∫ T
0(fn − f)ϕdx
∣∣∣∣+ supn≥1
∣∣∣∣∫ T
0(fn − f)(ϕ− g) dx
∣∣∣∣≤ 0 + ‖fn − f‖L2‖ϕ− g‖L2 ≤ 2M ε .
Since ε > 0 is arbitrary, this implies∫ T
0 fng dx →∫ T
0 fg dx, for every g ∈ L2([0, T ], provingthe weak convergence fn f .
26. Consider first the case fn(x) =√n · cosnx . Then for any b ∈ [0, 1] we have∫ b
0fn(x) dx =
1√n
sinnb → 0 as n→∞ .
By taking linear combinations, it is clear that
limn→∞
∫ 1
0fn(x)g(x) dx = 0 (16)
for any piecewise constant function g. In spite of the fact that piecewise constant functionsare dense in L2([0, 1]), the weak convergence fn 0 FAILS, because the sequence (fn)n≥1 isnot bounded. Indeed,∫ 1
0|fn(x)|2 dx = n
∫ 1
0cos2 nx dx → ∞ as n→∞ .
Next, consider the case
fn(x) =
n2/3 if x ∈ [0, n−1] ,
0 if x > n−1 .
40
For each fixed 0 < b ≤ 1, and all n > b−1 we have∫ b
0fn(x) dx =
∫ 1/n
0n2/3 dx = n−1/3 → 0 as n→∞ .
Hence, also in this case (16) holds for every piecewise constant function g. However,∫ 1
0|fn(x)|2 dx =
∫ 1/n
0|n2/3|2 dx = n1/3 → ∞ as n→∞ .
Also in this case, the sequence (fn)n≥1 is not bounded in L2([0, 1]), hence it cannot convergeweakly. Notice that in this case the result proved in problem 25 cannot be used.
For example, take g(x) = x−3/7. Then g ∈ L2([0, 1]) but∫ 1
0fn(x)g(x) dx =
∫ 1/n
0n2/3x−3/7 dx = n2/3 · 7
4n−4/7 → ∞ as n→∞ .
27. We only need to prove the implication (ii) =⇒ (i). By linearity, it is clear that theconvergence (xn, y) → (x, y) must hold for every y ∈ span(S). By assumption, span(S) is asubspace whose closure has non-empty interior. Hence span(S) = H.
Let y ∈ H be given. For any ε > 0 we can find a point y ∈ span(S) with ‖y − y‖ < ε. Byassumption, there exists a constant M such that ‖x‖ ≤M and ‖xn‖ ≤M for all n ≥ 1. Thisyields the estimate
lim supn→∞
∣∣∣(xn − x, y)∣∣∣ ≤ lim sup
n→∞
∣∣∣(xn − x, y − y)∣∣∣+ lim sup
n→∞
∣∣∣(xn − x, y)∣∣∣
≤ supn≥1‖xn − x‖ · ‖y − y‖+ 0 ≤ 2M · ε .
Since ε > 0 was arbitrary, this proves the weak convergence xn x.
28. Use the result in problem 27, choosing S as the set of all characteristic functions χQ,where Q = [a1, b1]× · · · × [aN , bN ]. Observe that the subspace spanned by all these functionsis dense on L2(Ω).
29. If y /∈ S = coxn ; n ≥ 1, then by Theorem 2.33 there exists a linear functional φ : H 7→R that strictly separates the compact, closed convex set y from the closed convex set S. Bythe Riesz’ representation theorem, there exists an element z ∈ H and constants c1 < c2 suchthat
(y, z) = φ(y) ≤ c1 < c2 ≤ φ(xn) = (xn, z) for all n ≥ 1 .
This leads to a contradiction, because the weak convergence xn y implies (xn, z)→ (y, z).
Chapter 6
41
1. (i) Define
(Λ1f)(x).= f(x+ 1) , (Λ2f)(x) =
f(x− 1) if x > 1 ,
0 if x ∈ [0, 1] .
Then
(Λ1 Λ2f)(x) = f(x) , (Λ2 Λ1f)(x)
f(x) if x > 1 ,
0 if x ∈ [0, 1] .
(ii) If Λ(I−K) = I, then Ker(I−K) = 0 hence the Fredholm operator I−K is one-to-one.By Theorem 6.1, I−K is onto. We conclude that I−K is a bijection, and hence Λ = (I−K)−1
commutes with I −K.
On the other hand, if (I −K)Λ = I, then I −K must be onto. Hence by Theorem 6.1 theFredholm operator I − K is one-to-one. We conclude that I − K is a bijection, and henceΛ = (I −K)−1 commutes with I −K.
2. (i) Λ is a bounded operator, with norm ‖Λ‖ = 2. Indeed,
‖Λf‖2L2([0,∞[) =
∫ ∞0|(Λf)(x)|2 dx =
∫ ∞0|2f(x+1)|2 dx = 4
∫ ∞1|f(y)|2 dy ≤ 4‖f‖L2([0,∞[) ,
with equality holding if f(x) = 0 for a.e. x ∈ [0, 1].
Λ is not compact. Indeed, consider the sequence of characteristic functions fn = χ[n, n+1]
.
Then‖fn‖L2 = 1 , ‖Λfn − Λfm‖L2 = 2
√2 for all m 6= n .
Therefore, from the sequence (Λfn)n≥1 one cannot extract any convergent subsequence.
(ii) To compute the adjoint operator we write∫ ∞0
f(y)(Λ∗g)(y) dy =
∫ ∞0
(Λf)(x)g(x) dx =
∫ ∞0
2f(x+1)g(x) dx =
∫ ∞1
f(y) 2g(y−1) dy.
Therefore
(Λ∗g)(y) =
2g(y − 1) if y > 1 ,
0 if y ∈ [0, 1] .
(iii) One has KerΛ =f ∈ L2([0,∞[ ; f(x) = 0 for a.e. x ∈ [0, 1]
, while KerΛ∗ = 0.
Of course, this implies that Λ is not a Fredholm operator. In particular, it cannot be writtenas λI +K, with λ ∈ R and K compact.
3. Assume η > M . Then the operator ηI − Λ is strictly positive definite. Indeed,
(ηx− Λx , x) ≥ η‖x‖2 − ‖Λx‖‖x‖ ≥ (η − ‖Λ‖) ‖x‖2.
An application of Theorem 5.12, with β = η−‖Λ‖, yields the existence of a continuous inverse(ηI − Λ)−1, having norm ‖(ηI − Λ)−1‖ ≤ 1/β. Therefore, η ∈ σ(Λ).
The case η < −M is entirely similar.
42
4. (i) Let M.= max(x,y)∈[0,1]×[0,1] |K(x, y)| . Then, by Holder’s inequality,
|Λf(x)| =
∣∣∣∣∫ 1
0K(x, y)f(y) dy
∣∣∣∣ ≤ M
∫ 1
0|f(y)| dy ≤ M ‖f‖L2([0,1]) .
This proves that Λ is bounded, with ‖Λ‖ ≤ 1.
Since K(x, y) = K(y, x), the fact that Λ is self-adjoint follows from
(Λf, g)L2 =
∫ 1
0
(∫ 1
0K(x, y)f(y) dy
)g(x) dx =
∫ 1
0
∫ 1
0K(x, y)f(y)g(x) dxdy
=
∫ 1
0
(∫ 1
0K(y, x)g(x)dx
)f(y)dy = (f, Λg)L2 .
To show that Λ is compact, follow the proof of Theorem 4.12. Observe that, if a sequenceof continuous functions gn = Λfn converges uniformly on [0, 1], then it also converges inL2([0, 1]).
(ii) Consider the special case where
u(x) = (Λf)(x) =
∫ x
0(1− x)y f(y) dy +
∫ 1
x(1− y)xf(y) dy .
Differentiating twice w.r.t. x one obtains
u′(x) =
∫ x
0(−y)f(y) dy +
∫ 1
x(1− y)f(y) dy ,
u′′(x) = − f(x) .
5. (i) The fact that Λ is self-adjoint is checked by writing∫ π
0f(x) (Λ∗g)(x) dx =
∫ π
0(Λf)(x) g(x) dx =
∫ π
0f(x) sinx g(x) dx .
Hence (Λ∗g)(x) = sinx g(x).
(ii) For every f ∈ L2([0, π]) one has
‖Λf‖2L2 =
∫ π
0
∣∣∣ sinx f(x)∣∣∣2 dx ≤ ∫ π
0
∣∣∣f(x)∣∣∣2 dx .
Hence ‖Λ‖ ≤ 1. To prove the converse inequality, consider the set An =x ∈ [0, π] ; sinx >
1− 1n
and define fn = χAn . Then ‖Λfn‖L2 ≥
(1− 1
n
)‖fn‖L2 . We thus conclude that ‖Λ‖ = 1.
However, there is no function f ∈ L2 such that Λf = f . Hence 1 /∈ σp(Λ).
(iii) According to problem 20 in Chapter 4, the multiplication operator is not compact.
6. (i) Assume that u ∈ L2([0, 1]), say with ‖u‖L2 = M . Then, for any 0 ≤ t1 < t2 ≤ 1, usingHolder’s inequality we obtain∣∣∣Λu(t2)− Λu(t1)
∣∣∣ ≤ ∫ t2
t1
1 · |u(s)| ds ≤ ‖1‖L2([t1,t2]) · ‖u‖L2 =√t2 − t1 · ‖u‖L2 .
43
Hence Λu ∈ C0,1/2.
(ii) Consider any sequence (fn)n≥1 with ‖fn‖L2 ≤ 1 for every n. Then the corresponding func-tions Λfn are uniformly bounded and Holder continuous, hence equicontinuous. By Ascoli’stheorem, one can extract a subsequence (fnk)k≥1 that converges uniformly on [0, 1] to somecontinuous function f . In turn, this implies ‖fnk − f‖L2 → 0, proving that the operator Λ iscompact.
(iii) Writing∫ 1
0u(x) (Λ∗v)(x) dx =
∫ 1
0(Λu)(x) v(x) dx =
∫ 1
0
(∫ x
0u(s) ds
)v(x) dx
=
(∫ 1
0u(s) ds
)(∫ 1
0v(s) ds
)−∫ 1
0u(x)
(∫ x
0v(s) ds
)dx ,
we conclude
(Λ∗v)(x) =
∫ 1
xv(s) ds .
(iv) The operator (I−K) is a Fredholm operator on L2([0, 1]). If u = Ku, then u(x) =∫ x
0 u(s) ds .hence u is an absolutely continuous solution of the Cauchy problem
d
dxu(x) = u(x) , u(0) = 0 .
By Gronwall’s inequality, the only solution is u(x) = 0 for every x. We conclude that theoperator I − K is one-to-one. By Fredholm’s theorem, I − K is onto. Hence for everyg ∈ L2([0, 1]) there exists one and only one function u such that u − Ku = g. The inversemap g 7→ u = (I −K)−1 is also a bounded linear operator.
If g is continuously differentiable, then u is an absolutely continuous function that satisfiesthe Cauchy problem
u′(x) = g′(x) + u(x) , u(0) = g(0) .
To compute the solution u(·), we set v = u− g. The previous ODE yields
v′(x) = v + g , v(0) = 0 ,
v(x) =
∫ x
0ex−yg(y) dy .
This provides an explicit formula for the inverse operator:
u(x) =(
(I −K)−1g)
(x) = g(x) +
∫ x
0ex−yg(y) dy .
7. To fix the ideas, let w1, . . . ,wN be the eigenvectors with corresponding eigenvalues λ1 =· · · = λN = 1, while λk 6= 1 for k > N . Then the equation u −Ku = f admits a solution ifand only if
(f,wk) = 0 for every k ∈ 1, . . . , N. (17)
44
If (17) holds, then u is a solution if and only if
u =N∑k=1
ckwk +∑k>N
(f,wk)
1− λwk ,
for arbitrary constants c1, . . . , cN .
8. Assume A = A∗, B = B∗. Then for every x, y ∈ H we have
(ABx, y) = (Bx, Ay) = (x, BAy).
Hence AB is self adjoint if and only AB = BA.
9. In the case where H is finite dimensional, the result is trivial: every operator is compactand (ii) is satisfied simply because the set of orthonormal sequences is empty. In the followingwe thus assume that H is infinite dimensional.
(i) =⇒ (ii). Assume that Λ : H 7→ H is compact, and let v1, v2, . . . be an orthonormalsequence of vectors in H.
If the sequence Λvn does not converge to zero, by taking a subsequence we can assume ‖Λvn‖ ≥ε > 0 for all n ≥ 1. A contradiction is obtained as follows.
By compactness, there exists a subsequence such that Λvnk → w for some vector w. Ourprevious assumption implies ‖w‖ ≥ ε.
By choosing a further subsequence, we can assume
‖Λvnk − w‖ <‖w‖
2for all k ≥ 1 . (18)
Consider the vector v =∑∞
k=1 vnk/k. Since the unit vectors vnk are mutually orthogonal, theseries is convergent and ‖v‖2 =
∑∞k=1 k
−2 < ∞. Since Λ is a bounded operator, we shouldhave
N∑k=1
Λvnkk
= Λ
(N∑k=1
vnkk
)→ Λ(v) as N →∞.
But this is impossible, because by (18)∥∥∥∥∥N∑k=1
Λvnkk
∥∥∥∥∥ ≥∥∥∥∥∥N∑k=1
w
k
∥∥∥∥∥−N∑k=1
∥∥∥∥w − Λvnkk
∥∥∥∥ ≥ N∑k=1
1
k
(‖w‖ − ‖w‖
2
)and the right hand side approaches infinity as n→∞.
(ii) =⇒ (i). Assume that Λ is not compact. Then the set S.= Λ(B1) defined as the closure
of the image of the unit ball, is not compact. By the result proved in problem 15, Chapter 4,there exists ε > 0 such that
(Pε) S is not contained in the ε-neighborhood of any finite-dimensional subspace of H.
45
We now inductively construct an orthonormal sequence v1, v2, . . ., as follows.
• By (Pε), there exists a vector v1 such that ‖Λv1‖ ≥ ε.
• Assume that the orthonormal set v1, . . . , vn has been constructed. Again by (Pε),the set S is not contained in the ε-neighborhood of the finite dimensional space Vn
.=
spanΛv1, . . . ,Λvn. Hence there exists a vector w such that ‖w‖ ≤ 1 and d(Λw, Vn) ≥ ε.Define the vectors
wn+1.= w −
n∑k=1
(w, vn)vn , vn+1.=
wn+1
‖wn+1‖.
Observe that wn+1 is the perpendicular projection of w on spanv1, . . . , vn⊥. Hence0 < ‖wn+1‖ ≤ ‖w‖ ≤ 1. Moreover, observing that Λwn+1 − Λw ∈ Vn, we obtain
‖Λvn+1‖ ≥ ‖Λwn+1‖ ≥ d(Λwn+1, Vn) = d(Λw, Vn) ≥ ε .
By induction, we thus obtain an orthonormal sequence vn ; n ≥ 1 such that ‖Λvn‖ ≥ ε > 0for every n. Hence (ii) fails.
To prove the last statement, consider the orthonormal sequence
wn = 2−n/2∑
2n<k≤2n+1
ek .
Then define the operator Λ by setting
Λv.=
∞∑n=1
(wn, v)en .
This is a bounded linear operator with norm ‖Λ‖ = 1. It is not compact, because theimage of the unit ball contains the entire orthonormal sequence en ; n ≥ 1. However,limk→∞ Λek = 0. Indeed, for 2n < k ≤ 2n+1 we have
‖Λek‖ = |(wn, ek)| = 2−n/2.
10. Let V ⊆ H be a closed subspace. Consider the orthogonal decomposition V = V1 + V2,where
V1.= V ∩Ker(I −K) , V2
.= V ∩ V ⊥1 .
Then (I − K)(V ) = (I − K)(V2). By possibly replacing V with the closed subspace V2, wecan thus assume that V ∩Ker(I −K) = 0, i.e. I −K is one-to-one restricted to V .
We claim that there exists a constant β > 0 such that
‖u−Ku‖ ≥ β ‖u‖ for all u ∈ V. (19)
Otherwise, we could find a sequence (un)n≥1 such that
‖un‖ = 1 , yn.= un −Kun → 0 .
46
Since K is compact, there exists a subsequence such that Kunj → z for some z ∈ H. Thisyields
limn→∞
un = limn→∞
(yn +Kun) = z ∈ V .
Hence(I −K)z = lim
n→∞(I −K)un = 0 ,
in contradiction with the assumption that (I −K) is one-to-one restricted to V .
Now assume vn ∈ V for every n ≥ 1, and
yn = (I −K)vn → y.
We need to prove that y ∈ (I −K)(V ).
Using (19) we obtain
0 = lim supm,n→∞
‖ym − yn‖ ≥ β lim supm,n→∞
‖vm − vn‖ .
Therefore the sequence (vn)n≥1 is Cauchy, and converges to some limit v ∈ V . By continuity,y = (I −K)v.
11. Let w1,w2, . . . be an orthonormal basis of a real Hilbert space H, consisting of eigen-vectors of a linear, compact, self-adjoint operator K. Let λ1, λ2, . . . be the correspondingeigenvalues.
Assume that t 7→ u(t) provides a solution to the Cauchy problem
d
dtu(t) = Ku(t) , u(0) = f , (20)
for some f ∈ H. Taking the inner product of u(t) with wk we obtain
d
dt(u(t), wk) = (Ku(t), wk) = (u(t), Kwk) = λk(u(t), wk) , (u(0), wk) = (f, wk) .
(21)Therefore, the solution can be written as
u(t) =∞∑k=1
ck(t)wk , (22)
where each coefficient ck(·) is obtained by solving the linear scalar Cauchy problem
c′k(t) = λkck(t) , ck(0) = (f,wk) .
This yields ck(t) = eλkt(f,wk), and hence
u(t) =∑k≥1
eλkt(f,wk)wk .
12. In the more general case of a non-homogenous equation, the coefficients ck(·) satisfy theequations
c′k(t) = λkck(t) + (g(t), wk) , ck(0) = (f,wk) .
47
Hence the formula (22) remains valid, with
ck(t) = eλkt(f,wk) +
∫ t
0eλk(t−s)(g(t),wk) ds .
13. Taking the inner product of the solution with wk, in this case we find that the functionck(t)
.= (u(t), wk) satisfies the second order initial value problem
c′′k(t) = λkck(t) , ck(0) = (f,wk) , c′k(0) = (g,wk) .
14. Assume that ηI − Λ is a continuous bijection. Then there exists β > 0 such that, forevery bounded linear operator Ψ with norm ‖Ψ‖ < β, the operator ηI−Λ + Ψ is a continuousbijection. Hence every η ∈ R with |η − η| < β lies in the resolvent set ρ(Λ).
Chapter 7
1. (i) The norm of the diagonal operator ASt is computed by
‖ASt‖ = supk
∣∣∣λkeλkt∣∣∣ .Using the assumption
λk = αk + iβk = ω − rk(cos θk + i sin θk)
for some rk ≥ 0 and |θk| ≤ θ < π/2, we obtain
|λkeλkt| ≤ (|ω|+ rk)e(ω−rk cos θk)t ≤ (|ω|+ rk)e
(ω−ηrk)t η.= cos θ > 0 .
Hence, for a fixed t > 0,
‖ASt‖ ≤ supr≥0
(|ω|+ r)e(ω−ηr)t < ∞ .
(ii) Similarly,
‖AnSt‖ = supk
∣∣∣λnkeλkt∣∣∣ ≤ supr≥0
(|ω|+ r)ne(ω−ηr)t < ∞ .
2. This is an immediate consequence of the formula (7.31). Namely, if S is a semigroup oftype ω, then
‖etAλ‖ ≤ e2ωt
for all t ≥ 0 and λ ≥ 2ω. If S is contractive semigroup, then S is of type ω = 0.
48
3. The computation of the various integrals is an exercise in basic Calculus.
From the identities∫ ∞0
Wn(t) dt = 1 ,
∫ ∞0
tWn(t) = T ,
∫ ∞0
(t− T )2Wn(t) dt =T 2
n
we deduce ∫|t−T |>ε
Wn(t) dt ≤√T 2
n→ 0 as n→∞.
This proves (7.39).
4. The formula
(I − hA)−nu =
∫ ∞0
wn(t)Stu dt (23)
is proved by induction on n. The case n = 1 is already known. Now assume (23) is valid.Using the variable τ = t+ s we obtain
(I − hA)−(n+1)u = (I − hA)−1(I − hA)−nu =
∫ ∞0
w(t)St
((I − hA)−nu
)dt
=
∫ ∞0
w(t)St
(∫ ∞0
wn(s)Ssu ds
)dt =
∫ ∞0
(w ∗ wn)(τ)Sru dτ .
Hence the same formula (23) is valid with n replaced by n+ 1.
The convergence(I − T
nA
)−nu =
∫ ∞0
Wn(t)Stu dt → STu as n→∞. (24)
is proved by using (7.39). Indeed, let ε > 0 By the continuity of the map t 7→ Stu there existsδ > 0 such that
‖Stu− STu‖ ≤ ε for all t ∈ [T − δ, T + δ] .
By (7.39), for all n large enough we have∫|t−T |>δ
Wn(t) dt < ε .
Recalling that ‖Stu‖ ≤ ‖u‖ for every t ≥ 0, we thus obtain∥∥∥∥∥STu−(I − T
nA
)−nu
∥∥∥∥∥ ≤∥∥∥∥STu− ∫ ∞
0Wn(t)Stu dt
∥∥∥∥≤∥∥∥∥STu− ∫ ∞
0Wn(t)STu dt
∥∥∥∥+
∫|t−T |≤δ
Wn(t)‖Stu− STu‖ dt+
∫|t−T |>δ
Wn(t)‖Stu− STu‖ dt
≤ 0 + ε+ ε 2‖u‖ .
Since ε > 0 was arbitrary, this proves (24).
49
5. (i) =⇒ (ii). Assume that Stu ∈ Ω for every u ∈ Ω and t ≥ 0. Then
(I − hA)−1u =
∫ ∞0
e−t/h
hStu dt .
This shows that (I−hA)−1u is an integral average of points Stu ∈ Ω, with weight w(t) = e−t/h
hsatisfying
∫∞0 w(t) dt = 1. Approximating the integral with a finite sum, we see that (I −
hA)−1u can be approximated by convex combinations of elements Stiu ∈ Ω. Since Ω is closedand convex, we conclude (I − hA)−1u ∈ Ω.
(ii) =⇒ (i). Using the result of the previous problem 4,(I − T
nA
)−n=
∫ ∞0
Wn(t)Stu dt , (25)
where Wn is a smooth averaging kernel, with∫∞
0 Wn(t) dt = 1. Since Ω is closed and convex,the right hand side of (25) lies in Ω. Letting n→∞, by the previous problem 4 the left handside of (25) converges to STu. Since Ω is closed, we conclude STu ∈ Ω.
6. Let St ; t ≥ 0 be a semigroup of type ω, with generator A. Then
‖eγtStu‖ = eγt‖Stu‖ ≤ eγteωt‖u‖ .
Hence eγtSt ; t ≥ 0 is a semigroup of type γ + ω. Moreover, if u ∈ Dom(A), then
limh→0+
eγhShu− uh
= limh→0+
eγhShu− Shuh
+ limh→0+
Shu− uh
= γu+Au .
Calling Aγ the generator of the semigroup eγtSt ; t ≥ 0, this proves that Dom(Aγ) ⊇Dom(A) and Aγ(u) = Au+ γu.
Inverting the roles of A,Aγ , we see that Dom(A) ⊇ Dom(Aγ) and Au = Aγu − γu. Thisconcludes the proof.
7. (i) The semigroup properties are easily checked:
(S0f)(x) = e−2·0f(x+ 0) = f(x) ,
(St+sf)(x) = e−2(t+s)f(x+ t+ s) = e−2s(e−2tf((x+ t) + s)
)= (Ss(Stf))(x) .
In the following, we first assume 1 ≤ p <∞. By definition, the function f ∈ Lp(R) lies in thedomain of the generator A if and only if the following limit exists in Lp(R):
Au.= lim
h→0+
e−2hf(x+ h)− f(x)
h= lim
h→0+
e−2hf(x+ h)− f(x+ h)
h+ limh→0+
f(x+ h)− f(x)
h
= − 2f(x) + limh→0+
f(x+ h)− f(x)
h.
(26)
50
Hence
Dom(A) =u ∈ Lp(R) ; u is absolutely continuous and ux ∈ Lp(R)
Moreover, Au = −2u+ ux for every u ∈ Dom(A).
(ii) On the space L∞(R) the above semigroup is not strongly continuous. For example, considerthe function
u(x) =
1 if x ∈ [0, 1], ,0 otherwise .
Then
Stu(x) =
e−2t if x ∈ [−t, 1− t] ,
0 otherwise .
As t→ 0+ we then have the convergence ‖Stu− u‖Lp(R) → 0 for every 1 ≤ p <∞. However
limt→0+
‖Stu− u‖L∞(R) = 1 .
Hence the map t 7→ Stu is not continuous from [0,∞[ into the space L∞(R).
8. The generator A of the semigroup must be a linear operator on Rn with dense domain.The only dense subspace of Rn is Rn itself. Hence Dom(A) = Rn, and A must be describedby an n×n matrix. In particular, the linear operator A is continuous, hence for every u ∈ Rnand t ≥ 0 we must have
d
dtStu = AStu, S0u = 0 .
Since t 7→ Stu is the solution to the above Cauchy problem, we conclude that Stu = etAu.
9. (i) On the space of all bounded continuous functions w : [0, T ] 7→ X, consider the Picardoperator defined as
Φ(w)(t).= Stu+
∫ t
0St−sf(s, w(s)) ds .
Following the proof of Theorem 7.1, one checks that Φ is a strict contraction w.r.t. the equiv-alent norm
‖w‖†.= max
t∈[0,T ]e−2Lt‖w(t)‖ .
Hence Ψ has a unique fixed point, which by definition provides a mild solution of (7.40).
(ii) Assume that the generator A is a bounded linear operator. Then we can differentiate(7.41) w.r.t. time, and obtain
d+
dtu(t) = lim
h→0+
St+hu− Stuh
+ limh→0+
1
h
∫ t+h
tSt−sf(s, u(s)) ds
+ limh→0+
∫ t
0
St+h−sf(s, u(s))− St−sf(s, u(s))
hds
= AStu+ f(t, u(t)) +∫ t
0 ASt−sf(s, u(s)) ds
= Au(t) + f(t, u(t)) .
51
Observing that the forward derivative (obtained by taking the limit as h→ 0+) is a continuousfunction of time, we conclude that it coincides with the backward derivative (obtained byletting h→ 0−). This completes the proof.
10. For u ∈ L1([0, 1]), define
(Stu)(x) =
0 if x ≤ t/T ,
u(x− t
T
)if x > t/T .
(27)
This corresponds to solving the PDE ut + ux = 0 with boundary condition u(t, 0) = 0.
11. Assume Sτ is compact. If t > τ then St = St−τSτ is the composition of the continuousoperator St − τ with the compact operator Sτ , hence it is compact.
To see that the converse may not hold in general, observe that the semigroup (27) is triviallycompact for t ≥ T but not compact for 0 ≤ t < T .
12. (i) (Stu)(x) = u(x + t). Not that this is well defined also for t < 0. This is a group ofisometries: ‖Stu‖L1 = ‖u‖L1 .
(ii) If v = E−h u.= (I−hA)−1u, then v = u+hvx. We are thus looking for a function v ∈ L1(R)
such that v − hv′ = u, hence
v′(x) =v(x)
h− u(x)
hx ∈ R .
The explicit solution is provided by
v(x) =
∫ ∞x
e(x−y)/h
hu(y) dy.
Setting t = y − x, it is interesting to observe that the above formula is equivalent to
v(x) =
∫ ∞0
e−t/h
hu(x+ t) dt =
∫ ∞0
e−t/h
h(Stu)(x) dt .
(iii) If u ∈ C∞c with support contained in the interval [a, b], then the same is true of thederivative ux. Hence (E+
h u)(x) = u(x) + hux(h) is a smooth function supported inside [a, b].By induction on n we see that the same holds for (E+
h )nu.
(iv) Take T ≥ b− a. If u vanishes outside [a, b], then STu is supported inside [a− T, b− T ].In particular, (STu)(x) = u(x+ T ) = 0 for every x ∈ [a,∞[ .
On the other hand, every forward Euler approximation vanishes outside the interval [a, b],hence the same must be true for the limit (if it exists). We conclude that, if the function u isnot identically zero, the forward Euler approximations cannot converge to STu.
Chapter 8
52
1. (i) This is a distribution of infinite order.
(ii)-(iii) These linear functionals are not distributions on R. They are both distributions onthe open half line ]0,∞[ .
(iv) This is a distribution of order zero on Ω = ]0,∞[ .
2. By assumption, f has a weak derivative Df ∈ Lp(]a, b[) ⊂ L1(]a, b[). By Corollary 8.17, fcoincides a.e. with an absolutely continuous function.
If f is continuously differentiable, for any a < x < y < b, using Holder’s inequality on theinterval [x, y] we obtain
|f(x)− f(y)| ≤∫ y
x1 · |f ′(s)| ds ≤ ‖1‖Lq([x,y]) · ‖f ′‖Lp([x,y])
1
q+
1
p= 1 .
‖1‖Lq([x,y]) = |x−y|1/q = |x−y|1−1p , ‖f ′‖Lp([x,y]) ≤ ‖f ′‖Lp([x,y]) ≤ ‖f‖W 1,p(]a,b[) .
By approximation, the same inequality holds for every f ∈W 1,p.
3. Fix any δ > 0 and consider the smaller square
Qδ.=
(x1, x2) ; δ < x1 < 1− δ, δ < x2 < 1− δ.
For any ε ∈ ]0, δ], the mollified functions fε.= Jε ∗ f are well defined on Qδ. Moreover
Dx1fε = Jε ∗ Dx1f = 0. Hence there exist a smooth function gε : [ε, 1 − ε] 7→ R such thatfε(x1, x2) = gε(x2) for all (x1, x2) ∈ Qδ.
Letting εn → 0, we obtain ‖fε − f‖L1(Qδ) → 0. Hence, for a suitable sequence εn → 0 weachieve the pointwise convergence
f(x1, x2) = limε→0
fε(x1, x2) = limε→0
gε(x2) for a.e. (x1, x2) ∈ Qδ .
Since δ > 0 is arbitrary, this proves the result.
4. For every test function ϕ ∈ C∞c (Ω′) we have∫Ω′g ϕ dx = −
∫Ω′f Dx1ϕdx =
∫Ω′
∂f
∂x1ϕdx .
By the uniqueness of the weak derivative, proved in Lemma 8.12, this implies g(x) = ∂f∂x1
(x)for a.e. x ∈ Ω′.
5. (i) Consider the mollifications uε.= Jε ∗ u. The assumption u ∈W 1,∞(Ω) implies that
|∇uε(x)| ≤ ‖u‖W 1,∞ for all x ∈ Ωε.= x ∈ Ω ; B(x, ε) ⊂ Ω .
Since Ωε is convex, for any x, y ∈ Ωε we have
|uε(x)−uε(y)| ≤∫ 1
0|∇uε(θx+(1−θ)y)·(x−y)| dθ ≤
(maxz∈Ωε
|∇uε(z)|)|x−y| ≤ ‖u‖W 1,∞ ·|x−y| .
53
Therefore, each uε is Lipschitz continuous with Lipschitz constant L = ‖u‖W 1,∞ . Taking thelimit as ε→ 0 we obtain the result.
(ii) Consider the open set
Ω.= (x1, x2) ; 1 < x2
1 + x22 < 4 \ (x1, x2) ; x2 = 0 , x1 > 0.
Let u be the angle function, in polar coordinates. In other words,
u(x1, x2).= θ , if (x1, x2) = r(cos θ, sin θ) , r =
√x2
1 + x22, 0 < θ < 2π .
6. (i) Consider first the case where Ω is a bounded, open, convex set. Let u be Lipschitzcontinuous with constant C. For every fixed (n− 1)-tuple (x1, x2, . . . , xi−1, xi+1, . . . , xn), thefunction
s 7→ u(x1, x2, . . . , xi−1, s, xi+1, . . . , xn)
is defined for s in some open interval ]a, b[ (possibly empty). Moreover, it is Lipschitz continu-ous of the same constant C. Being absolutely continuous, it is differentiable almost everywhere.We thus conclude that the partial derivative
uxi(x) = limh→0
u(x+ hei)− u(x)
h.
exists for a.e. x ∈ Ω. An integration by parts shows that this function provides the weakderivative Dxiu. Since u is bounded and |uxi(x)| ≤ C at every point x where the partialderivative exists, it is clear that u ∈W 1,∞(Ω).
The result can be easily extended to a general open set Ω, observing that
‖u‖W 1,∞(Ω) = supB(x,r)⊂Ω
‖u‖W 1,∞(B(x,r)) ,
where the supremum is taken over all open balls contained in Ω.
(ii) Consider any bounded open subset Ω′ ⊂ Ω. Then W 1,∞(Ω′) ⊂ W 1,n+1(Ω′). Hence byTheorem 8.41 the function u is differentiable a.e. on Ω′. By varying the set Ω′ we concludethat u is differentiable a.e. on Ω.
7. A direct computation shows that
‖f‖nLn(Ω) = cn ·∫ 1
0rn−1
[ln ln
(1 +
1
r
)]ndr < ∞
because the integrand on the right hand side is bounded. Moreover, for n ≥ 2 we have
‖∇f‖nLn(Ω) = cn ·∫ 1
0rn−1
1
ln(
1 + 1r
) · 1
1 + 1r
· 1
r2dr
n dr≤ cn ·
∫ 1
0
1
r lnn(
1 + 1r
) dr < ∞ .
54
8. (i) Let f be continuously differentiable. Then
|f(0)| ≤ |f(x)|+∫
[0,x]|f ′(y)| dy.
integrating both sides over the interval [−1, 1] we obtain
2|f(0)| ≤∫ 1
−1|f(x)| dx+
∫ 1
−1(1− |x|)|f ′(x)| dx ≤ ‖f‖W 1,1 .
Since C1(Ω) is dense in W 1,1(Ω), this functional can be extended by continuity to a boundedlinear functional on the whole space W 1,1(Ω).
(ii) Next, assume Ω = B(0, 1) ⊂ R2. For p > 2, setting γ = 1− 2p , Morrey’s inequality yields
|f(0)| ≤ ‖f‖C0,γ(Ω) ≤ C · ‖f‖W 1,p(Ω) ,
for some constant C and every f ∈ W 1,p ∩ C1. Hence the map T : f 7→ f(0) can be uniquelyextended to a bounded linear functional on the entire space W 1,p(Ω).
To see that this functional is not continuous on W 1,p(Ω) for p ∈ [1, 2], we prove that
supf∈W 1,2(Ω)
|f(0)|‖f‖W 1,2
= ∞ . (28)
Observing that, for every p ∈ [1, 2], one has ‖f‖W 1,p ≤ Cp‖f‖W 1,2 , from (28) we conclude thatthe functional f 7→ f(0) is unbounded in the space ‖f‖W 1,p as well.
To prove (28), consider the function
f(x).= ln ln
(1 +
1
|x|
)and the decreasing sequence of continuous functions
fn(x).= min
1 ,
f(x)
n
.
As shown in Problem 7, one has f ∈W 1,2(Ω). Hence
|fn(0)| = 1 , ‖fn‖W 1,2 ≤1
n‖f‖W 1,2 → 0.
9. Assume p > 2 and set γ = 1− p2 . Let f ∈ C∞0 (R3). Then for every t ∈ R the restriction of
f to the planeΣt
.= (t, x2, x3) ; x2, x3 ∈ R
lies in C∞c (Σt). Morrey’s inequality yields
|f(t, 0, 0)| ≤ ‖f‖C0,γ(Σt) ≤ C ‖f‖W 1,p(Σt) .
we have ∫R2
(|f(x1, x2, x3)|p + |∇f(x1, x2, x3)|p
)dx2dx3 < ∞ .
55
If p is suitably large, then f coincides a.e. with a Holder continuous function on the plane(x1, x2, x3) ; x1 = x1 .
10. Construct a sequence of smooth approximations uν ∈ C∞c (Rn) such that ‖uν−u‖W 1,p(Rn) ≤ 2−ν−1
for every ν ≥ 1. Setting u0 = 0, vν.= uν+1 − uν , we have
u = w0 +∞∑ν=1
wν , ‖wν‖W 1,p ≤ 2−ν forall ν ≥ 1 .
We shall use the notation x = v + y, with v ∈ V , y ∈ V ⊥. Observing that∫ n
R
∞∑ν=0
(|wν |p + |∇wν |p) dx =
∫V ⊥
∞∑ν=0
(∫y+V
(|wν |p + |∇wν |p) dv)dy < ∞ ,
by Fubini’s theorem we conclude that∫y+V
∞∑ν=0
(|wν |p + |∇wν |p) dv < ∞ (29)
for a.e. z ∈ V ⊥. If (29) holds, then the sequence of partial sums is absolutely convergent inW 1,p(y + V ), hence also in C0,γ(y + V ), by Morrey’s embedding theorem, with γ = 1 − m
p .Hence, restricted to the affine subspace y + V , the function u coincides a.e. with a Holdercontinuous function.
⊥
v
y
x = v+y
V
y+V
0
V
Figure 2: Each affine subspace y+V has dimension m < p, hence Morrey’s inequality can be applied.
(ii) For any smooth function w ∈ C∞c (Rn) with compact support, one has∫V ⊥|w(y)|pdy ≤
∫V ⊥‖w‖pC0,γ(y+V )
dy ≤ C
∫V ⊥‖w‖p
W 1,p(y+V )dy = C‖w‖p
W 1,p(Rn).
By approximation, we conclude that ‖u‖Lp(V ⊥) ≤ C‖w‖pW 1,p(Rn)
for every u ∈W 1,p(Rn).
11. Since the inequality
‖g‖C0,γ(Ω) ≤ C ‖g‖W 1,p(Ω) , γ = 1− n
p
is valid for every g ∈ C∞c (Ω), by approximation the same holds for every f ∈W 1,p0 (Ω), after a
modification on a set of measure zero.
56
12. Consider any open set Ω ⊂ Rn. Consider the open subsets
Ω1/n.=x ∈ Ω ; |x| < n , B(x, 1/n) ⊂ Ω
. (30)
By the dominated convergence theorem, for any f ∈ Lp(Ω), 1 ≤ p <∞, the approximations
fn(x).=
f(x) if x ∈ Ω1/n
0 otherwise
converge to f in Lp. In turn, for ε < 1/n the mollifications Jε ∗ fn lie in C∞c (Ω) and convergeto fn as ε → 0. Hence f can be approximated in Lp(Ω) by smooth functions with compactsupport.
Next, consider any sequence of functions fn ∈ C∞c (Ω). and assume ‖fn − f‖L∞ → 0. Thisimplies that f is the uniform limit of a sequence of continuous functions which vanish on theboundary of Ω. Hence f is a bounded continuous function which satisfies
limn→∞
sup |f(x)| ; x ∈ Ω \ Ω1/n = 0 . (31)
Viceversa, every bounded continuous function f satisfying (31) lies in W 0,∞0 (Ω). Indeed, if
this holds, then we can approximate f with a sequence of functions fn,εn ∈ C∞c (Ω), choosing
fn,εn = Jεn ∗ fn = Jεn ∗ (f · χΩ1/n
) , εn <<1
n.
13. Let fk(x).= f(x)ϕ(k − |x|). Then fk is supported on the set where |x| ≤ k and coincides
with f for |x| ≤ k − 1. By the dominated convergence theorem
‖fk − f‖pW 1,p ≤ C
∫|x|>k−1
(|f |p +
∑i
∣∣∣∣ ∂f∂xi∣∣∣∣p)dx → 0 as k →∞.
Performing the mollifications fk,εk = Jεk ∗ fk with εk << 1/k we obtain a sequence of smoothfunctions with compact support which converge to f in W 1,p(Rn).
14. If u ∈ W 2,p(R+) then u is absolutely continuous, hence the limit u(0) =.= limx→0+ u(x)
is well defined. The even extension Eu(x) = u(|x|) is absolutely continuous, with derivative(Eu)x(x) = (sign x)ux(|x|) for a.e. x ∈ R. Our assumptions imply∫ ∞
−∞
(|Eu|p + |(Eu)x|p
)dx = 2
∫ ∞0
(u|p + |ux|p
)dx < ∞.
Hence Eu ∈W 1,p(R).
In general, Eu /∈ W 2,p(R). For example, take u(x) = x/2. Then Eu(x) = |x|/2 is a functionwhose first derivative is (Eu)x = 1
2sign x. However, the function x 7→ 12sign x has a jump at
x = 0. Its distributional derivative is a Dirac distribution, concentrating a unit mass at x = 0.Hence Eu does not have a weak second derivative.
57
15. (i) A direct computation yields
‖uλ‖Lq =
(∫|u(λx)|q dx
)1/q
=
(∫|u(y)|q λ−ndy
)1/q
= λ−n/q‖u‖Lq .
Hence the conclusion holds with α = −n/q.
(ii) A similar computation yields
‖∇uλ‖Lp =
(∫|λ∇u(λx)|p dx
)1/p
=
(∫|u(y)|p λp−ndy
)1/p
= λ(p−n)/p‖u‖Lp .
Hence the conclusion holds with β = (p− n)/p.
(iii) We have α = β if −nq = p−n
p . This is the case if 1q = 1
p −1n , i.e. if q = p∗ is the Sobolev
conjugate exponent to p.
16. Since H1(Ω) is a Hilbert space and the sequence (um)m≥1 is bounded, we can extracta convergent subsequence umk u in H1(Ω). Moreover, since the embedding H1(Ω) ⊂L2(Ω) is compact, by possibly extracting a further subsequence we obtain the convergence‖umk − u‖L2 → 0. This clearly implies u = u, hence u ∈ H1(Ω).
To prove a bound on ‖u‖H1 , set
M.= lim inf
m→∞‖um‖H1 .
By possibly extracting a subsequence and relabeling, we can assume
M = limm→∞
‖um‖H1 .
Writing0 ≤ (um − u, um − u)H1 = (um, um)H1 + (u, u)H1 − 2(um, u)H1 ,
and letting m→∞, we obtain
M2 + ‖u‖2H1 − 2‖u‖2H1 ≥ 0 .
Hence ‖u‖H1 ≤M .
17. (i) Consider the functions
fn(x).=
1− n+1
n |x| if |x| ≤ nn+1 ,
0 otherwise.
Show that each fn has compact support in Ω. Moreover, ‖fn−f‖W 1,p → 0. Choose a sequenceεn → 0 so that the mollified functions Jεn ∗fn have compact support and converge to f in H1.
(ii) To show that f ∈ W 1,20 (Ω0), let ϕ(x) = log log
(1 + 1
|x|
). Notice that ϕ ∈ W 1,2(Ω) as
claimed in problem 6. For each integer n, define the function
gn(x) = max
0 , min
1− n+ 1
n|x| , n− ϕ(x)
.
58
Show that each gn has compact support in Ω0. Moreover, ‖gn − f‖W 1,p → 0. Choose asequence εn → 0 so that the mollified functions Jεn ∗ fn have compact support contained inΩ0 and converge to f in H1.
(iii) If the functions ϕn ∈ C∞c (Ω0) form a Cauchy sequence w.r.t. the W 1,p norm, with p > 2,then by Morrey’s inequality the sequence (ϕn)n≥1 converges to some continuous function g,uniformly on the closed disc Ω. In particular, g = 0 at the origin. Derive a contradiction,showing that f cannot coincide a.e. with a continuous function that vanishes at the origin.
18. (i) Since Ω is connected, by Corollary 8.16 the assumption implies that u coincides a.e. witha constant. Since the set where u = 0 has positive measure, we conclude that u(x) = 0 fora.e. x ∈ Ω.
(ii) If (8.82) fails, consider a sequence of functions un such that
‖un‖L2 = 1 , ‖∇un‖L2 <1
n,
and meas(x ∈ Ω ; un(x) = 0
)≥ α for every n.
Since the embedding H1(Ω) ⊂ L2(Ω) is compact, we can extract a subsequence (unk)k≥1 suchthat ‖unk − u‖L2 → 0 for some limit function u ∈ L2(Ω), and moreover unk u in H1(Ω).By taking a further subsequence, we can assume the pointwise convergence unk(x)→ u(x) fora.e. x ∈ Ω.
By assumption, the weak gradients ∇unk converge to zero in L2(Ω). Hence ∇u ≡ 0 andu(x) = c for some constant c and a.e. x ∈ Ω. We claim that c = 0, because otherwise
‖unk − u‖2L2 =
∫Ω|unk(x)− c|2 dx ≥
∫x ; unk (x)=0
c2 dx ≥ αc2
for every k.
On the other hand, the strong convergence in L2 implies ‖u‖L2 = limk→∞ ‖unk‖L2 = 1,providing a contradiction.
19. Fix i ∈ 1, . . . , n We claim that the function v = uxi , defined a.e. as the classicalderivative of u, provides the weak derivative of u on Rn.
In the following, without loss of generality we assume i = 1 and use the notation x = (x1, x′).
Moreover, we consider the set
N .=x′ = (x2, . . . , xn) ∈ Rn−1 ; (t, x′) ∈ K for some t ∈ R
.
By assumption, this set has zero (n− 1)-dimensional measure.
For any test function ϕ ∈ C∞c (Rn), we now have∫Rnuϕx1 dx =
∫Rn−1
∫ ∞−∞
uϕx1 dx1 dx′ =
∫Rn−1\N
(−∫ ∞−∞
ux1ϕdx1
)dx′ = −
∫Rnux1ϕdx .
Hence ux1 = Dx1u, proving our claim.
59
In turn, the assumptions u,∇u ∈ Lp(Ω) imply u ∈W 1,p(Ω).
20. (i) Take f(x) = g(x) = |x|−1/2
(iii) Take f(x) = g(x) = |x|32−n.
21. Assume 1 ≤ p < ∞. If the result were not true, we could find a sequence un ∈ W 1,p(Ω)such that
‖un‖Lp(Ω) = 1 , ‖un‖Lp(Ω′) → 0 , ‖∇un‖Lp(Ω) → 0 .
By the compact embedding W 1,p(Ω) ⊂⊂ Lp(Ω) there exists a subsequence unk such that‖unk − u‖Lp → 0 for some u ∈ Lp(Ω)
The above conditions imply ‖u‖Lp(Ω′) = 0, ‖∇u‖Lp(Ω) = 0. Since Ω is connected, we concludethat u ≡ 0. However, this is in contradiction with ‖u‖Lp(Ω) = limn→∞ ‖u‖Lp(Ω) = 1.
22. The set S can be represented as
S =f : ]0, 1[ 7→ R ; ‖f‖C1 ≤M , f ′ is Lipschitz continuous with constant M
.
Assume ‖fn − f‖C0 → 0, with fn ∈ S for every n. By assumption, all functions fn, ∂xfn areuniformly bounded and Lipschitz continuous with constant M . Hence they can be extendedby continuity to the closed interval [0, 1].
Since fn(x) → f(x) uniformly on [0, 1], this implies that f is also Lipschitz continuous withconstant M . By Ascoli’s theorem, by taking a subsequence we can assume the uniform con-vergence ∂xfnk → v for some Lipschitz continuous function v. Together, the limits
fnk(u) → f(x) , ∂xfnk(x) → v(x) uniformly for x ∈ ]0, 1[
imply that v(x) = f ′(x). Since this limit does not depend on the particular subsequence, weconclude that the entire sequence ∂xfn converges to ∂xf(x), uniformly on ]0, 1[.
Notice that this implies that ∂xf is also Lipschitz continuous with constant M . The aboveshows that f ∈ S, hence S is a closed subset of C0(]0, 1[). The above arguments prove both(ii) and (i).
23. Let Eu ∈ W 1,p(Rn) be an extension of u to the entire space Rn, with ‖Eu‖W 1,p(Rn) ≤C‖u‖W 1,pΩ). Consider the mollifications uk
.= J1/k ∗ Eu.
24. The assumption implies that f is absolutely continuous on any bounded interval. Hencef differentiable at a.e. point. Writing
f(x) = f(a) +
∫ x
ag(s) ds,
we have
gn(x) =1
n
∫ x+1/n
xg(s) ds .
60
This implies the pointwise convergence gn(x) → g(x) at every Lebesgue point x of g, hencealmost everywhere.
To prove that ‖gn − g‖L1([a,b]) → 0, let ε > 0 be given. Choose a continuous function ϕ suchthat ‖g − ϕ‖L1([a, b+1]) < ε and define
ϕn(x) =1
n
∫ x+1/n
xϕ(s) ds .
Then
lim supn→∞
‖gn − g‖L1([a,b])
≤ lim supn→∞
‖gn − ϕn‖L1([a,b]) + lim supn→∞
‖ϕn − ϕ|L1([a,b]) + lim supn→∞
‖ϕ− g‖L1([a,b])
≤ ε+ 0 + ε .
Since ε > 0 was arbitrary, this proves the result.
25. The net smoothness of a function u ∈ W 2,2(R3) is k − np = 2 − 3
2 = 12 . Therefore,
the Sobolev embedding theorem implies that, after a modification on a set of measure zero,un ∈ C0,1/2, with ‖un‖C0,1/2 ≤ C, for some constant C and every n ≥ 1. Being the pointwiselimit of a sequence of uniformly Holder continuous functions, u is Holder continuous as well.
Chapter 9
1. The PDE can be rewritten as Lu = −f , where Lu = −(
(ux)x + (xux)y + (uy)y
). To
prove that the operator L is uniformly elliptic, it suffices to check that the quadratic form(ξ1, ξ2) 7→ ξ2
1 + xξ1ξ2 + ξ22 is strictly positive definite for (x, y) ∈ Ω. The conclusion is then
obtained by applying Theorem 9.8.
2. (i) The fact that B[u, v] = 〈u, v〉♦ is a continuous bilinear map on H10 is clear. We need to
show that it is strictly positive definite. For |y| ≤ 1 one has
a2 + 2b2 + 2yab ≥ (2−√
2)(a2 + 2b2) +√
2(a2 + 2b2) + 2yab ≥ (2−√
2)(a2 + 2b2) .
Therefore, by Poincare’s inequality,
B[u, u] =
∫Ω
(u2x + 2u2
y + 2yuxuy) dxdy ≥ (2−√
2)
∫Ω
(u2x + u2
y) dxdy ≥ β ‖u‖2H1
for some β > 0 and all u ∈ H10 (Ω).
(ii) We can now use the Lax-Milgram theorem and conclude that for every f ∈ L2(Ω) thereexists a unique u ∈ H1
0 (Ω) such that B[u, v] = (f, v)L2 for every v ∈ H10 (Ω). By a formal
61
integration by parts, we find that this function u provides a weak solution to the ellipticboundary value problem
−uxx − 2uyy − (2 + y)uxy = f on Ω ,u = 0 on ∂Ω .
3. The operator L is elliptic in a neighborhood of the point (x, y) ∈ R2 if and only if the
coefficient matrix
(x 11 y
)is strictly positive definite. This is the case if and only if
(x, y) ∈ Ω+ .= (x, y) ∈ R2 ; x > 0, y > 0, xy > 1 .
The operator L is uniformly elliptic on a bounded domain Ω if and only if the closure Ω isentirely contained in the open set Ω+.
4. If φ ∈ H10 is a weak solution, then∫
Ω∇φ · ∇v dx = µ
∫Ωφ v dx for all v ∈ H1
0 (Ω).
In particular, taking v = φ we obtain
‖φ‖2L2(Ω) =
∫Ω|∇φ|2 dx = µ
∫Ω|φ|2 dx = µ‖φ‖2L2(Ω) .
By assumption,
β2 = supu∈H1
0 ,u6=0
‖φ‖2L2(Ω)
‖∇φ‖2L2(Ω)
,
hence µ ≥ 1/β2.
(ii) Using the representation formula (9.57), the solution of the parabolic equation with initialdata u(0) = g is provided by
u(t) =
∞∑k=1
e−µkt(g, φk)L2φk , (32)
where φk ∈ H10 (Ω), the set φk ; k ≥ 1 is an orthonormal basis of L2(Ω) consisting of
eigenfunctions of the Laplace operator, and µk are the corresponding eigenvalues. Since µk ≥β−2 for every k ≥ 1, from (32) it follows
‖u(t)‖2L2 =
∞∑k=1
∣∣∣e−µkt(g, φk)L2
∣∣∣2 ≤ ∞∑k=1
∣∣∣e−t/β2(g, φk)L2
∣∣∣2 = e−2t/β2 ‖g‖2L2 .
Taking square roots of both sides we obtain the result.
5. As in Theorem 9.9, let φk ; k ≥ 1 be an orthonormal basis of L2(Ω), consisting ofeigenfunctions of the operator −∆. For every v ∈ C∞c (Ω) we have
B[v, v].=
∫Ω|∇v|2 dx = −
∫Ωv ·∆v dx =
∞∑k=1
µk(v, φk)2 ≥ µ1 ‖v‖2L2 .
62
By continuity, the same inequality remains valid for every v ∈ H10 (Ω). On the other hand, the
eigenfunction φ1 ∈ H10 (Ω) satisfies
B[φ1, φ1] =
∫Ω|∇φ1|2 dx = −
∫Ωφ1 ·∆φ1 dx = µ1 = µ1 ‖φ1‖2L2 .
We thus obtain a representation for the first eigenvalue:
µ1 = inf06=v∈H1
0 (Ω)
B[v, v]
‖v‖2L2
= inf0 6=v∈H1
0 (Ω)
∫Ω |∇v|
2 dx∫Ω |v|2 dx
. (33)
Similarly,
µ1 = inf0 6=v∈H1
0 (Ω)
B[v, v]
‖v‖2L2
= inf0 6=v∈H1
0 (Ω)
∫Ω|∇v|2 dx∫
Ω|v|2 dx
.
The inclusion Ω ⊆ Ω implies H10 (Ω) ⊂ H1
0 (Ω). Hence the above representation formulas yieldµ1 ≤ µ1.
6. Choose γ.= maxt∈[0,T ] q(t) and consider the operator
Lγu.= − (p(t)u′)′ + (γ − q(t))u .
Since p(t) ≥ θ > 0, the operator Lγ is uniformly elliptic. We claim that the bilinear form
Bγ [u, v] =
∫ T
0
(p(t)u′(t)v′(t) + (γ − q(t))u(t)v(t)
)dt
is strictly positive definite on H10 (]0, T [). Indeed,
Bγ [u, u] ≥∫ T
0p(t)|u′(t)|2 dt ≥ β‖u‖H1
for some β > 0 and all u ∈ H10 (]0, T [).
As in the Theorem 9.9, the inverse operator L−1γ is a linear, compact self-adjoint operator from
L2(Ω) into itself. By the Hilbert-Schmidt theorem, the space L2(Ω) admits an orthonormalbasis φk ; k ≥ 1 consisting of eigenfunctions of L−1
γ . The corresponding eigenvalues satisfyλk > 0, λk → 0 as k →∞. This implies(
1
λk− γ)φk = Lφk .
The above analysis shows that the eigenvalues of the Sturm-Liuville problem (9.85) are givenby µk = γ − 1
λk. Hence limk→∞ µk = −∞.
7. By assumption, for each k ≥ 1 we have∫Ω∇φk · ∇v dx = µk
∫Ωφkv dx
for some eigenvalue µk and all v ∈ H10 (Ω). Assume j 6= k and choose v = φj . Then∫
Ω∇φk · ∇φj dx = µk
∫Ωφkφj dx = 0 .
63
Hence (φk, φj)H1 = 0.
8. (i) Let ϕk ; k ≥ 1 be an orthonormal basis of L2(Ω) consisting of eigenfunctions ofthe Laplace operator ∆. We claim that, for every f ∈ L2(Ω) and m ≥ 1, the linear system ofalgebraic equations
B[um, ϕj ] =
m∑k=1
ckB[ϕk, ϕj ] = (f, ϕj)L2 j = 1, . . . ,m
has a unique solution c1, . . . , cm.
For this purpose, it suffices to show that the m × m matrix of coefficients A = (aij), withaij = B[ϕk, ϕj ], is invertible. Assume, on the contrary, that there exists a nonzero vectorξ = (ξ1, . . . , ξn) such that
n∑j=1
B[ϕk, ϕj ]ξj = 0 k = 1, . . . ,m .
Then, setting v =∑
j ξjϕj 6= 0 we obtain
B[v, v] =m∑
j,k=1
B[ϕk, ϕj ]ξjξk = 0,
contradicting the assumption that the bilinear form B[·, ·] is strictly positive definite.
(ii) If the identityB[um, v] = (f, v)L2
holds for v = ϕ1, . . . , ϕm, then by linearity it remains valid for every v ∈ spanϕ1, . . . , ϕm.In particular, since B is strictly positive definite, this implies
β‖um‖2H1 ≤ B[um, um] = (f, um)L2 ≤ ‖f‖L2 ‖um‖L2 ≤ ‖f‖L2 ‖um‖H1 ,
for some constant β > 0. Therefore
‖um‖H1 ≤ β−1‖f‖L2 .
Thanks to the uniform bound, we can extract a subsequence (unj )j≥1 and achieve the weakconvergence unj u in H1
0 (Ω). We now have
B[u, ϕk] = limm→∞
B[um, ϕk] = (f, ϕk).
By linearityB[u, v] = lim
m→∞B[um, v] = (f, v) (34)
for every v ∈ spanϕk ; k ≥ 1. By approximation, (34) remains valid for every v ∈spanϕk ; k ≥ 1 = H1
0 (Ω). This proves that u is a weak solution to the elliptic problem.
9. According to (9.36), the solution u of the boundary value problem satisfies an abstractequation of the form u − Ku = h, where K : L2(Ω) 7→ L2(Ω) is a compact operator and
64
h = L−1γ f , where L−1
γ is another compact operator. Moreover, we are assuming that theoperator I −K is a continuous bijection. By the open mapping theorem, the inverse operator(I −K)−1 is continuous. Being the composition of a compact operator and a continuous one,the map f 7→ u = (I −K)−1L−1
γ f is compact operator.
10. (i) By Problem 9 in Chapter 2, every function ϕ ∈ C∞c (Q) can be uniformly approximatedby a finite linear combination of the functions φm,n. Since C∞c (Q) is dense in L2(Q), weconclude that spanφm,n ; m,n ≥ 1 is dense in L2(Q).
A straightforward computation shows that
−∆φm,n = µm,n φm,n , µm,n =(mπa
)2+(nπb
)2.
(ii) If µ1 is the smallest eigenvalue of −∆ on Ω, then by the result proved in problem 5. the
assumption Ω ⊆ Q implies µ1 ≥ µ1,1 = π2
a2+ π2
b2. The conclusion now follows from (33).
11. This is a special case of problem 8. Consider the 2 × 2 matrix A = (aij) and the vectorb = (b1, b2) with entries
aij = B[ϕi, ϕj ] =
∫ 3
0ϕ′i(x)ϕ′j(x) dx , bi =
∫ 3
01 · ϕi dx .
An explicit calculation yields
A =
(2 11 2
), b =
(11
).
Solving the system of two algebraic equations(2 11 2
)(ξ1
ξ2
)=
(11
),
we find ξ1 = ξ2 = 1/3. The Galerkin approximation thus yields
u2(x) =ϕ1(x) + ϕ2(x)
3=
x/3 if x ∈ [0, 1] ,1/3 if x ∈ [1, 2] ,
(3− x)/3 if x ∈ [2, 3] .
The exact solution of the boundary value problem is u(x) = 32x−
12x
2.
12. (i) Here L is the symmetric operator defined by
Lu = −(2ux)x −1
2(yuy)x −
1
2(yux)y − (3uy)y .
The corresponding symmetric bilinear form is
B[u, v] =
∫Ω
(2uxvx +
y
2(uxvy + uyvx) + 3uyvy
)dx .
65
On the open domain Ω = (x, y) ; x2 + y2 < 1, the operator L is uniformly elliptic. Indeed,for |y| ≤ 1 we have
2ξ21 + 2yξ1ξ2 + 3ξ2
2 ≥ ξ21 + ξ2
2 for every ξ = (ξ1, ξ2) ∈ R2.
(ii) Define the energy as
E(t) =1
2‖ut‖2L2(Ω) +
1
2B[u, u] =
1
2
∫Ωu2t dx+
1
2
∫Ω
(2u2
x + yuxuy + 3u2y
)dx .
Differentiating w.r.t. time, and using the boundary condition u = ut = 0 on ∂Ω, performingan integration by parts we obtain
d
dtE(t) = (ut, utt)L2 +B[u, ut]
=
∫Ωututt dx+
∫Ω
(2uxuxt +
y
2(uyuxt + uxuyt) + 3uyuyt
)dx
=
∫Ωututt dx−
∫Ωut
(2uxx + yuxy + 2uyy +
ux2
)dx = 0 .
13. (i) The fact that on H10 (Ω) the norm ‖·‖♦ is equivalent to the H1 norm is a straightforward
consequence of the fact that the bilinear form B[·, ·] is strictly positive definite.
(ii) Repeat the arguments used in the proof of Lemma 9.7.
14. For any g ∈ L2(Ω) the formula
Stg =
∞∑k=1
e−µkt(g, φk)L2φk . (35)
defines a trajectory t 7→ Stg ∈ L2(Ω). Here φk ; k ≥ 1 is an orthonormal basis of L2(Ω).We need to show that this map is n times continuously differentiable, for every n ≥ 1. Towardthis goal we observe that, for a given N ≥ 1, the partial sum is continuously differentiable:
dn
dtn
(N∑k=1
e−µkt(g, φk)L2φk
)=
N∑k=1
(−µk)ne−µkt(g, φk)L2φk .
We claim that, as N → ∞, the above sum converges to a well defined limit, uniformly for tin compact subsets of ]0,∞[ . Since the φk form an orthonormal sequence, it suffices to provethat
limN→∞
∑n>N
∣∣∣(−µk)ne−µkt(g, φk)L2
∣∣∣2 = 0
uniformly for t in compact subsets of ]0,∞[ .
Consider any interval [a, b], with 0 < a < b <∞. Then
M.= sup
µ≥0,t∈[a,b]µne−µt < ∞ .
66
Given any ε > 0, choose N so large that
M∑n>N
|(g, φk)|2 < ε .
This is certainly possible because∑
n≥1 |(g, φk)|2 = ‖g‖2L2 <∞. Then, for every t ∈ [a, b],
∑n>N
∣∣∣(−µk)ne−µkt(g, φk)L2
∣∣∣2 ≤ M∑n>N
|(g, φk)|2 < ε .
15. For γ > 0, the bilinear form
Bγ [u, v] =
∫Ω∇u · ∇v + γuv dx
satisfiesBγ [u, u] ≥ β ‖u‖2H1 , β
.= min1, γ.
Hence for every f ∈ L2(Ω) there exists a unique u ∈ H1(Ω) such that
Bγ [u, v] = (f, v)L2 for every v ∈ H1(Ω).
The map f 7→ u = L−1γ f is a self-adjoint, compact linear operator from L2(Ω) into itself.
A function u ∈ L2(Ω) is a weak solution to the Neumann problem if and only if
u− L−1γ γu = L−1
γ f .
This can be written in the abstract form
(I −K)u = h , K.= γL−1
γ , h.= L−1
γ f . (36)
The equation (36) has a solution if and only if
h ∈ Range(I −K) = [Ker(I −K)]⊥.
This is the case if and only if (h, v)L2 = 0 for every solution v of v −Kv = 0, i.e. for everyweak solution of the homogeneous Neumann problem
−∆u = 0 on Ω ,
∂u
∂ν= 0 on ∂Ω .
(37)
By definition, v ∈ H1(Ω) is a weak solution of the above problem if∫Ω∇v · ∇w dx = 0 for every w ∈ H1(Ω) .
Choosing w = v, this implies ∫Ω|∇v|2 dx = 0 .
67
Since the open set Ω is connected, we conclude that the solutions of (37) are precisely theconstant functions.
Going back to the original problem, by the previous analysis the Neumann boundary valueproblem has a solution if and only if
∫Ω f · v dx = 0 for every constant function v. Of course,
this holds if and only if∫
Ω f dx = 0.
16. On the space H20 (Ω) consider the continuous bilinear form
B[u, v].=
∫Ω
∆u∆v dx .
We claim that there exists β > 0 such that B[u, u] ≥ β ‖u‖2H2 , namely∫Ω
(∑i
uxixi
)(∑j
uxjxj
)dx ≥
∫Ω
∑ij
|uxixj |2 dx+
∫Ω
∑i
|uxi |2 dx+
∫Ω|u|2 dx . (38)
If u ∈ C∞c (Ω), integrating twice by parts we obtain∫Ωuxixiuxjxj dx = −
∫Ωuxixixjuxj dx =
∫Ωuxixjuxixj dx .
Hence ∫Ω
(∑i
uxixi
)(∑j
uxjxj
)dx =
∫Ω
∑ij
|uxixj |2 dx . (39)
Applying Poincare’s inequality to each function uxi we obtain∫Ω|uxi |2 dx ≤ C ·
∫Ω|∇uxi |2 dx = C ·
∫Ω
∑j
|uxixj |2 dx (40)
for some constant C. Similarly,∫Ω|u|2 dx ≤ C ·
∫Ω
∑j
|uxi |2 dx . (41)
Together, (39)–(41) yield (38), whenever u ∈ C∞c (Ω). By an approximation argument, thesame estimate is valid for every u ∈ H2
0 (Ω).
Since B[·, ·] is strictly positive definite, the Lax-Milgram theorem yields the existence of aunique u ∈ H2
0 (Ω) such that
B[u, v] = (f, v)L2 for every v ∈ H20 (Ω) .
By definition, u is the unique weak solution of the boundary value problem∆2u = f x ∈ Ω ,
u =∂u
∂ν= 0 x ∈ ∂Ω ,
Chapter 10
68
1. Since K is compact, it can be covered by finitely many of the sets Ai, say K ⊆ A1 ∪A2 ∪· · ·∪An. Consider the functions ρi(x) = d(x,Aci ), where Aci = K \Ai denotes the complementof the set Ai. These functions are all Lipschitz continuous. Namely |ρi(x)− ρi(y)| ≤ d(x, y).Hence the function r(x) = maxi ρi(x) is also Lipschitz continuous.
We now observe that, if x ∈ Ai then ρ(x) ≥ ρi(x) > 0. Hence the minimum of the continuousfunction r(·) on the compact set K is strictly positive. By construction, it is now clear thatthe constant ρ
.= minx∈K r(x) > 0 satisfies the requirement.
2. (i) If xn → x then every subsequence converges to x as well. To prove the converseimplication, assume that the sequence does not converge to x. Hence there exists ε > 0such that for every N one can find n > N such that d(xn, x) > ε. If this holds, then wecan construct a subsequence (xnk)k≥1 such that d(xnk , x) > ε for every k. Clearly, from thissubsequence one cannot extract any further subsequence converging to x.
(ii) If a convergent subsequence (xnk)k≥1 exists, then it would be Cauchy. In particular,limj,k→∞ d(xnj , xnk) = 0. but this is impossible if d(xnj , xnk) ≥ δ for j 6= k.
(iii) The assumption implies that, given ε > 0, from any subsequence (xn)n∈I one can extracta further subsequence (xn)n∈I′ such that
d(xm, xn) < 2ε for all m,n ∈ I ′.
Here I ′ ⊂ I ⊂ N are infinite sets of natural numbers.
We argue by induction on k = 1, 2, . . . Let I1 ⊂ N be a infinite set of indices such that
d(xm, xn) < 2−1 for all m,n ∈ I1 .
After Ik−1 ⊂ N has been constructed, we choose an infinite set Ik ⊂ Ik−1 such that
d(xm, xn) < 2−k for all m,n ∈ Ik .
After all the sets Ik have been constructed, we choose a sequence n1 < n2 < n3 < · · · suchthat nk ∈ Ik for every k. We claim that the sequence (xnk)k≥1 is Cauchy. Indeed, if m < n,then d(xm, xn) ≤ 2−m. Since the space E is complete, this subsequence has a limit.
3. This is an elementary integral:
‖f‖L1 =
∫ 1/2
0
1
x(lnx)2dx =
∫ − ln 2
−∞
1
y2dy =
1
ln 2.
Next, observe that the mollifier Jε is supported on the interval −ε, ε] and strictly positive onthe subinterval [−ε/2, ε/2]. Namely, it satisfies
Jε(x) ≥ φε(x).=
δ
εif 0 < x <
ε
2,
0 otherwise ,
69
for some constant δ > 0. For 0 < x < 1/6, choosing ε = 2x we obtain
F (x) ≥ (J2x∗f)(x) ≥ (φ2x∗f)(x) =
∫ 2x
0
δ
2x
1
s(ln s)2ds =
δ
2x
∫ ln 2x
−∞
1
y2dy =
δ
2x| ln(2x)|.
Hence F /∈ L1(R).
4. All functions fn are uniformly bounded, because
|fn(x)| ≤ |fn(0)|+ ‖g‖L1 for all n ≥ 1, x ∈ R
and by assumption the values fn(0) range in a bounded set.
Using the previous Problem 2 (iii), given ε > 0 it suffices to prove that from any subsequenceone can extract a further subsequence (fnk)k≥1 such that
lim supj,k→∞
‖fnj − fnk‖C(R) ≤ ε . (42)
Toward this goal, choose a constant M large enough so that∫ −M−∞
g(x) dx+
∫ ∞M
g(x) dx <ε
2.
We claim that, on the compact interval [−M,M ], the functions fn are equicontinuous. Indeed,
Fix x ∈ [−M,M ] and ε > 0. Choose δ > 0 such that∫ x+δx−δ g(s) ds < ε. Then |fn(y)−fn(x)| < ε
for all y ∈ [x− δ , x+ δ] and n ≥ 1.
Given any subsequence, we can thus apply Ascoli’s theorem and extract a further subsequence(fnk)k≥1 which converges uniformly on the interval [−M,M ]. We now estimate (42) by writing
lim supj,k→∞
‖fnj − fnk‖C(R) = max
lim supj,k→∞
‖fnj − fnk‖C(]−∞,−M ])
+ lim supj,k→∞
‖fnj − fnk‖C([−M,M ]) + lim supj,k→∞
‖fnj − fnk‖C([−M,M ])
(43)
We now havelim supj,k→∞
‖fnj − fnk‖C([−M,M ]) = 0 .
Moreover
lim supj,k→∞
‖fnj − fnk‖C(]−∞,−M ]) ≤ lim supj,k→∞
|fnj (−M)− fnk(−M)|+ 2
∫ −M−∞
g(x) dx ≤ 0 + ε ,
lim supj,k→∞
‖fnj − fnk‖C([M,∞[) ≤ lim supj,k→∞
|fnj (M)− fnk(M)|+ 2
∫ ∞M
g(x) dx ≤ 0 + ε .
Hence (42) holds.
5. Set gn(x).= arctan fn(x) and consider the functions
a(x).= lim inf
n→∞gn(x) ≤ lim sup
n→∞gn(x)
.= b(x) .
70
The functions a, b are Lebesgue measurable, and so is the set A.= x ∈ R ; − π < a(x) =
b(x) < π. By replacing each function fn with fn.= χA · fn, it is not restrictive to assume
that the sequence (fn)n≥1 converges pointwise at every point x ∈ R. By Fatou’s lemma,
‖f‖L1(R) =
∫ (limn→∞
|fn(x)|)dx ≤ lim inf
n→∞
∫|fn(x)| dx ≤ C .
6. Let f : [a, b] 7→ R be absolutely continuous, and let A ⊂ [a, b] be a set with measure zero.Fix any ε > 0. We need to show that meas(f(A)) ≤ ε.
Toward this goal, let δ > 0 be such that∑N
i=1 |f(bi) − f(ai)| < ε for any finite family ofdisjoint intervals [ai, bi] ⊂ [a, b], i = 1, . . . , N , with total length
∑ni=1(bi − ai) < δ. Since
meas(A) = 0, there exists an open set V ⊇ A such that meas(V ) < δ. Since V ⊂ R is open,we can write V as a countable union of its connected components, which are disjoint openintervals: V =
⋃k≥1 ]ck, dk[ . We now have
meas(f(A)) ≤ meas(f(V )) ≤ supn
n∑k=1
meas(f([ck, dk])
)= sup
n
n∑k=1
∣∣∣f(Mk)−f(mk)∣∣∣. (44)
Here mk,Mk ∈ [ck, dk] are points where the continuous function f attains respectively itsminimum and maximum, restricted to the interval [ck, dk]. Observing that
n∑k=1
|mk −Mk| ≤n∑k=1
(dk − ck) < δ for all n ≥ 1 ,
we conclude that the right hand side of (44) is ≤ ε.
7. (i) Choose a subsequence such that ‖fnk‖L1 ≤ 2−k for every k ≥ 1. Then for any N ≥ 1we have ∫ 1
0
(supn>N|fn(x)|
)dx ≤
∫ 1
0
∑n>N
|fn(x)| dx ≤ 2−N .
Calling
Aε.=x ∈ [0, 1] ; lim sup
n→∞|fn(x)| ≥ ε
,
for every integer N we have
meas(Aε) ≤ meas(x ∈ [0, 1] ; sup
n>N|fn(x)| ≥ ε
)≤∫ 1
0
(supn>N|fn(x)|
)dx ≤ 2−N
ε.
Since N is arbitrary, this yields meas(Aε) = 0. The above shows that lim supn→∞ |fn(x)| = 0for a.e. x ∈ [0, 1]. Hence limn→∞ fn(x) = 0 for a.e. x.
(ii) For 2k < n ≤ 2k+1, define
fn(x) =
1 if x ∈
[n− 2k − 1
2k,n− 2k
2k
],
0 otherwise .
71
8. Let B(xj , rj), j ≥ 1 be a countable family of disjoint open balls contained in Ω. For eachj, let gj be the characteristic function of the set B(xj , rj). Clearly, all these functions lie inLp(Ω) and are linearly independent, hence the space Lp(Ω) is infinite dimensional.
Moreover, ‖gj‖L∞ = 1, ‖gi − gj‖L∞ = 1 for i 6= j.
To cover the case 1 ≤ p <∞, define
fj(x).= cjgj(x) , cj =
(1
meas(B(xj , rj))
)1/p
.
Then‖fj‖Lp(Ω) = 1 , ‖fi − fj‖Lp(Ω) = 21/p for all i 6= j .
9. It is clear that the set S is totally ordered: if (t1, f(t1)) and (t2, f(t2)) lie in S, with t1 ≤ t2,then (t1, f(t1)) (t2, f(t2)). To prove that S is maximal, assume (x, y) /∈ S. To fix the ideas,assume f(x) < y, the other case being similar. Then there exists t > x such that f(t) < y.The set S ∪ (x, y) is not totally ordered, because the two relations
(x, y) (t, f(t)), (t, f(t)) (x, y)
are both false.
A maximal totally ordered subset of different kind is
S.= (−1, y) ; y ≤ 0 ∪ (x, 0) ; x ∈ [−1, 1] ∪ (1, y) ; y ≥ 0 .
10. The proof is by induction. When m = 2, this is the standard Holder’s inequality. Assumethat the result is true for m = 2, 3, . . . , N − 1. Let fk ∈ Lpk(Ω), with 1
p1+ 1
p2+ . . .+ 1
pN= 1.
If pN =∞, then by the inductive hypothesis we immediately get∫Ω|f1 · · · fN−1 · fN | dx ≤
∫Ω|f1 · · · fN−1| dx · ‖fN‖L∞ ≤
N∏k=1
‖fk‖Lpk .
If 1 ≤ pN <∞, choose q so that
1
q
.= 1− 1
pN=
1
p1+
1
p2+ . . .+
1
pN−1.
The standard Holder inequality yields∫Ω|f1 · · · fN−1 · fN | dx ≤ ‖f1 · · · fN−1‖Lq‖fN‖LpN . (45)
We now observe that qp1
+ · · · + qpN
= 1, while |fk|q ∈ Lpk/q(Ω) for k = 1, . . . , N − 1. Theinductive assumption yields
‖f1 · · · fN−1‖Lq =
(∫Ω|f1|q · · · |fN−1|q dx
)1/q
≤
(N−1∏k=1
‖f qk‖Lpk/q
)1/q
=N−1∏k=1
‖fk‖Lpk .
(46)Together, (45)-(46) yield the result for m = N . By induction, the generalized Holder inequalityis valid for every m ≥ 2.
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