F analysis problems solutions AnhQLe full

Toán cao cấp Bách khoa TPHCM luyện thi Cao học

Functional Analysis Problems with Solutions

ANH QUANG LE, Ph.D.

September 3, 2013

4.1 Linear bounded operators 45

4.2 Linear Functionals 63

2 CONTENTS

Notations:

• B(X, Y ): the space of all bounded (continuous) linear operators from X to Y

• Image (T ) ≡ Ran(T ): the image of a mapping T : X → Y

• x n − → x: x w n converges weakly to x.

• X ∗ : the space of all bounded (continuous) linear functionals on X.

• F or K: the scalar field, which is R or C.

• Re, Im: the real and imaginary parts of a complex number.

For every a ∈ [0, 1] we have

kax + (1 − a)y − x0k = k(x − x0)a + (1 − a)(y − x0)k

4 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

We show that the parallelogram law with respect to the given norm does not hold

for two elements in C[0, 1].

Let f (t) = t, g(t) = 1 − t, t ∈ [0, 1] Then f, g ∈ C[0, 1] and

If the unit sphere of a normed space X contains a line segment [x, y] where x, y ∈

X and x 6= y , then x and y are linearly independent and kx + yk = kxk + kyk

Suppose that the unit sphere contains a line segment [x, y] where x, y ∈ X and

x 6= y Then

kax + (1 − a)yk = 1 for any a ∈ [0, 1].

Choose a = 1/2 then we get k1

2(x + y)k = 1, that is kx + yk = 2 Since x and y belong to the unit sphere, we have kxk = kyk = 1 Hence

kx + yk = kxk + kyk.

Let us show that x, y are linearly independent Assume y = βx for some β ∈ C We

have

1 = kax + (1 − a)βxk = |a + (1 − a)β|.

For a = 0 we get |β| = 1 and for a = 1/2 we get |1 + β| = 2 These imply that

β = 1, and so x = y, which is a contradiction. ¥

Problem 5

Prove that two any norms in a finite dimensional space X are equivalent.

Solution

Since equivalence of norms is an equivalence relation, it suffices to show that an

arbitrary norm k.k on X is equivalent to the Euclidian norm k.k2 Let {e1, , e n } be

a basis for X Every x ∈ X can be written uniquely as x =Pn k=1 c k e k Therefore,

where A = (Pn k=1 |e k |2)1/2 is a non-zero constant This shows that the map x 7→ kxk

is continuous w.r.t the Euclidian norm Now consider S = {x : kxk2 = 1} This is just the unit sphere in (X, k.k2), which is compact The map

6 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

That is

mkxk2 ≤ kxk ≤ Mkxk2.

Hence, the two norms are equivalent ¥

Problem 6

Let X be a normed space.

(a) Find all subspaces of X which are contained in some ball B(a; r) of X (b) Find all subspaces of X which contain some ball B(x0; ρ) of X.

Solution

(a) Let Y be a subspace of X which is contained in some ball B(a; r) of X Note first that the ball B(a; r) must contain the vector zero of X (and so of Y ); otherwise, the question is impossible For any number A > 0 and any x ∈ Y , we have Ax ∈ Y since Y is a linear space By hypothesis Y ⊂ B(a; r), so we have Ax ∈ B(a; r) This implies that kAxk < r + kak Finally

kxk < r + kak

A > 0 being arbitrary, it follows that kxk = 0, so x = 0 Thus, there is only one

subspace of X, namely, Y = {0}, which is contained in some ball B(a; r) of X (b) Let Z be a subspace of X which contain some ball B(x0; ρ) of X Take any

x ∈ B(0; ρ) Then x + x0 ∈ B(x0; ρ) and so x + x0 ∈ Z since Z ⊃ B(x0; ρ) Now, since x0 ∈ Z, x + x0 ∈ Z and Z is a linear space, we must have x ∈ Z Hence B(0; ρ) ⊂ Z.

Now for any nonzero x ∈ X, we have 2kxk ρx ∈ B(0; ρ) ⊂ Z Hence x ∈ Z We can

conclude that Z = X In other words, the only subspace of X which contains some ball B(x0; ρ) of X is X itself. ¥

Problem 7

Prove that any finite dimensional normed space :

(a) is complete (a Banach space),

(b) is reflexive.

Solution

Let X be a finite dimensional normed space Suppose dim X = d.

(a) By Problem 5, it suffices to consider the Euclidian norm in X Let {e1, , e d }

be a basis for X For x ∈ X there exist numbers c1, , c d such that

where α k = f (e k ) Let us define f k ∈ X ] by the relation f k (x) = c k , k = 1, , d.

For any x ∈ X and f ∈ X ], we get

Hence, α k = 0 for all k = 1, , d and thus, dim X ] = d For the space X ∗ we have

X ∗ ⊂ X ] , so dim X ∗ = n ≤ d and dim(X ∗)] = n From the relation X ⊂ (X ∗)∗ ⊂

(X ∗)] we conclude that d ≤ n Thus, n = d, and so X = (X ∗)∗ ¥

8 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

Problem 8 ( Reed-Simon II.4)

(a) Prove that the inner product in a normed space X can be recovered from the polarization identity:

h

(kx + yk2 − kx − yk2) − i(kx + iyk2− kx − iyk2)

i

= 12

Now suppose that the norm satisfies the parallelogram law Assume the field is

C, and define the inner product via the polarization identity from part (a) If

Combining these results we have

hx, αyi = αhx, yi for α ∈ Q + iQ.

10 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

Now, if α ∈ C, by the density of Q + iQ in C, there exists a sequence (α n ) in Q + iQ converging to α It follows that

hx, αyi = αhx, yi for α ∈ C.

Thus the h., i is linear.

Since ki(x − iy)k = kx − iyk, we have

So this shows that the norm is induced by h., i and that it is also positive definite,

and thus it is an inner product ¥

Problem 9 (Least square approximation Reed-Simon II.5)

Let X be an inner product space and let {x1, , x N } be an orthonormal set.

hx n , xix n + z, for some z ∈ X (∗)

We observe that for all n = 1, , N ,

Review: Quotient normed space.

• Let X be a vector space, and let M be a subspace of X We define an equivalence relation on

Then X/M is a vector space.

If the subspace M is closed, then we can define a norm on X/M by

Let X be a normed space and M a closed subspace of X Let π : X → X/M

be the canonical map Show that the topology induced by the standard norm on X/M is the usual quotient topology, i.e that O ⊂ X/M is open in X/M if and only if π −1 (O) is open in X.

12 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

Solution

• If O is open in X/M, then π −1 (O) is open in X since π is continuous.

• Now suppose that O ⊂ X/M and that π −1 (O) is open in X We show that O is open in X/M Consider an open ball B(0; r), r > 0 in X Let x ∈ B(0; r) Then

so ϕ is an isometry (and thus injective and continuous) Finally, constants are in

X = C[0, 1], so ϕ is surjective and thus an isometric isomorphism ¥

Problem 12

If 0 < p < 1 then ` p is a vector space but kxk p = (Pn |x n | p)1/p is not a norm for

` p

≤ kx n − xk(ky n − yk + kyk) + kxkky n − yk

≤ kx n − xkky n − yk + kx n − xkkyk + kxkky n − yk.

Since kxn − xk → 0 and ky n − yk → 0 as n → ∞, we see that

hx n , y n i → hx, yi as n → ∞. ¥

14 CHAPTER 1 NORMED AND INNER PRODUCT SPACES

Problem 14

Prove that if M is a closed subspace and N is a finite dimensional subspace of a normed space X, then M + N := {m + n : m ∈ M, n ∈ N} is closed.

Solution

Assume dim N = 1, say N = Span{x} The case x ∈ M is trivial Suppose x / ∈ M.

Consider the sequence zk := αk x + m k, where mk ∈ M, α k ∈ C, and suppose

z k ∈ M + N → y as k → ∞ We want to show y ∈ M + N The sequence (α k)

is bounded; otherwise, there exists a subsequence (α k 0 ) such that 0 < |α k 0 | → ∞ as

so x must be 0, which is in M This is a contradiction Consequently, (α k) is bounded

and therefore it has a subsequence (αk 0 ) which is converging to some α ∈ C Thus

m k 0 = z k 0 − α k 0 x → y − αx as k 0 → ∞.

Hence, y − αx is in M since M is closed Thus y ∈ M + N.

The solution now follows by induction ¥

Suppose that X is complete Let (a n ) be a sequence in X such thatP∞ n=1 ka n k < ∞.

Let S n=Pn i=1 a i be the partial sum Then for m > n,

By hypothesis, the series P∞ n=1 ka n k converges, so Pm i=n+1 ka i k → 0 as n → ∞.

Therefore, (S n ) is a Cauchy sequence in the Banach space X Thus, (S n) converges,that is, the seriesP∞ n=1 a n converges

Conversely, suppose P∞ n=1 a n converges in X whenever P∞ n=1 ka n k < ∞ We show

that X is complete Let (y n ) be a Cauchy sequence in X Then

16 CHAPTER 2 BANACH SPACES

Let X be a Banach space Prove that the closed unit ball B(0; 1) ⊂ X is compact

if and only if X is finite dimensional.

Solution

• Suppose dim X = n Then X is isomorphic to R n (with the standard topology).The result then follows from the Heine-Borel theorem

• Suppose that X is not finite dimensional We want to show that B(0; 1) is not

compact To do this, we construct a sequence in B(0; 1) which have no convergent

subsequence

We will use the following fact usually known as Riesz’s Lemma: (See the proof below)

Let M be a closed subspace of a Banach space X Given any r ∈ (0, 1), there exists an x ∈ X such

that

kxk = 1 and d(x, M ) ≥ r.

Pick x1 ∈ X such that kx1k = 1 Let S1 := Span {x1} Then S1 is closed According

to Riesz’s Lemma, there exists x2 ∈ X such that

kx2k = 1 and d(x2, S1) ≥ 1

2.

Now consider the subspace S2generated by {x1, x2} Since X is infinite dimensional,

S2 is a proper closed subspace of X, and we can apply the Riesz’s Lemma to find

an x3 ∈ X such that

kx3k = 1 and d(x3, S2) ≥ 1

2.

If we continue to proceed this way, we will have a sequence (x n) and a sequence of

closed subspaces (S n ) such that for all n ∈ N

Proof of Riesz’s Lemma:

Take x1 ∈ M Put d = d(x / 1, M ) = inf m∈M km − x1k Then d > 0 since M is closed For any

ε > 0, by definition of the infimum, there exists m1∈ M such that

18 CHAPTER 2 BANACH SPACES

sequence in X/M such that P∞ n=1 k[x n ]k < ∞ We show that

∃[x] ∈ X/M :

k

X

n=1 [x n ] → [x] as k → ∞.

For each n, k[x n ]k = inf z∈M kx n + zk, and therefore there is z n ∈ M such that

Since kx (i) − x (j) k p ≥ |x (i) k − x (j) k | for every k, it follows that

|x (i) k − x (j) k | → 0 for every k as i, j → ∞.

This tells us that the sequence ¡x (i) k ¢ is a Cauchy sequence in F, which is complete,

so that¡x (i) k ¢ converges to x k ∈ F as i → ∞ for each k Set x = (x1, , x k , ) We

will show that

20 CHAPTER 2 BANACH SPACES

Letting M → ∞, since the last sum is finite, we see that

(a) Show that ` ∞ equipped with the norm k.k ∞ is a Banach space.

(b) Let c0 be the space of sequences converging to 0 Show that c0 is a closed subspace of ` ∞

Solution

(a) We need to show that ` ∞ is complete Assume that the sequence (x (n)) is Cauchy

in ` ∞ That is, for all ² > 0, there exists N ∈ N such that

This shows that kxk ∞ < ∞ and so x ∈ ` ∞.

(b) Of course c0 ⊂ ` ∞ Assume (x (n) ) ∈ c0 that converges in ` ∞ to x We have to show that x ∈ c0 Let ² > 0 be arbitrary Since x (n) → x in ` ∞, we can choose

(a) Let c00 be the space of sequences such that if x = (x n)n∈N ∈ c00 then x n = 0

for all n ≥ n0, where n0 is some integer number Show that c00 with the norm k.k ∞ is NOT a Banach space.

(b) What is the closure of c00 in ` ∞ ?

So the sequence x (n) is a Cauchy sequence in c00 Evidently, it is also a Cauchy

sequence in ` ∞ Now consider the sequence

22 CHAPTER 2 BANACH SPACES

We see that x (n) ∈ c / 00, and

This tells us that there is a Cauchy sequence in c00 which does not converge to

something in c00 Therefore, c00 equipped with the k.k ∞ norm is not a Banachspace

(b) We claim that c00 = c0 (closure taken in ` ∞)

According to Problem 19, c0 is closed, so we have c00 ⊂ c0 We show the inverse

inclusion Take an arbitrary sequence x = (x1, x2, ) ∈ c0 We build a sequence a (n)

(a) Let x = (x1, x2, ) ∈ ` p Then, for n large enough, we have |x n | < 1 and hence

|x n | q ≤ |x n | p since 1 ≤ p < q < ∞ That implies x ∈ ` q Now we want to show

(b) Let x = (x1, x2, ) ∈ ` p0 for some p0 Clearly, kxk p ≥ max n |x n | = kxk ∞ for any

finite p On the other hand,

(a) First we show that E := {x ∈ ` p : x n = 0, n ≥ N for some N} is dense

in ` p Indeed, if x ∈ ` p , x = P∞ k=1 x k e k, where ek is the sequence such that the

k-component is 1 and the others are zero, then

But Pn k=1 x k e k ∈ E, so E is dense in ` p Now let A ⊂ E consisting of elements

x = (x1, x2, , x n , 0, 0, ) ∈ E such that x k = a k + ib k , a k , b k ∈ Q Since Q is dense

in R, A is dense in E Hence A is dense in ` p Since A is countable, ` p is separable

(b) We now show that it is not the case for ` ∞

Let F := {x ∈ ` ∞ : ∀k ≥ 1, x k = 0 or x k = 1} Then F is uncountable Note that for x ∈ F, kxk ∞ = 1 Moreover, x, y ∈ F, x 6= y ⇒ kx − yk ∞ = 1 Assume

that ` ∞ is separable Then there is a set A = {a1, a2, } dense in ` ∞ So, for all

x ∈ F , there exists k ∈ N such that kx − a k k ∞ ≤ 1

3 Let F be the family of closed balls B(x;1

3), x ∈ F If B 6= B 0 then B ∩ B 0 = ∅ This allows us to construct an injection f : F → A which maps each B ∈ F with an element a ∈ B ∩ A This is impossible since F is uncountable and A is countable ¥

∗ ∗ ∗∗

24 CHAPTER 2 BANACH SPACES

Problem 23 (The space C[0, 1])

Let C[0, 1] be the space of all continuous functions on [0, 1].

(a) Prove that if C[0, 1] is equipped with the uniform norm

kf k = max

x∈[0,1] |f (x)|, f ∈ C[0, 1]

then C[0, 1] is a Banach space.

(b) Give an example to show that C[0, 1] equipped with the L1-norm

This shows that for every x ∈ [0, 1], the sequence ¡f n (x)¢ is a Cauchy sequence of

numbers and therefore converges to a number which depends on x, say, f (x) In (∗), fix n and let m → ∞, we have

One can check that the sequence (f n) is a Cauchy sequence with respect to the

L1-norm, but it converges to the function

26 CHAPTER 2 BANACH SPACES

28 CHAPTER 3 HILBERT SPACES

which implies v = 0 Hence, x = u ∈ M.

(b) Assume that there are two linearly independent vectors x, y ∈ M ⊥ Recall that

M is the zero vector of the linear space X/M Consider the cosets [x], [y] Assume α[x] + β[y] = M, for some scalars α, β Then αx + βy ∈ M and, since αx + βy ∈ M ⊥

as well, we conclude that αx + βy = 0 Hence, α = β = 0 and therefore [x], [y] are linearly independent This contradicts the hypothesis codim M = 1 ¥

kT (x + y)k2 = kx + yk2 for every x, y ∈ H1.

By opening up the norm using the inner product, we obtain

RehT x, T yi = Rehx, yi.

Similarly, kT (x + iy)k2 = kx + iyk2 gives that

ImhT x, T yi = Imhx, yi.

• If η = 0, then, by definition of infimum, there is a sequence (z j ) in C such that

kz j k → 0 as j → ∞ Therefore, z j → 0 as j → ∞ Since C is closed, 0 ∈ C, and 0

is the unique element of minimal norm

• Suppose η > 0 First we show that C contains an element of minimal norm.

Take (z j ) in C such that kz j k → η as j → ∞ The convexity of C implies that

z ∈ C The norm function is continuous, so kzk = η This shows that C contains

an element of minimal norm

Assume that there are two elements a1, a2 ∈ C such that ka1k = ka2k = η By the

30 CHAPTER 3 HILBERT SPACES

This inner product gives rise to the norm

kxk2 =phx, xi =

à ∞X

i=1

|x i |2

!1

.

According to Problem 18, the normed space `2 is complete So `2 is a Hilbert space

• Consider the general case where 1 ≤ p < ∞ Assume that ` p with the

corre-sponding inner product is a Hilbert space Consider two elements e n , e m ∈ ` p with

(a) Show that M is closed in H Find M ⊥

(b) Calculate the distance from y to M for y(t) = t2.

Solution

(a) Let 1 ∈ H be the function 1(t) = 1, ∀t ∈ [−1, 1] Define the map T : H → C

by

x 7→ h1, xi.

Then T is linear We show that T is bounded, so continuous.

(a) Show that E is closed in H Find E ⊥

(b) Calculate the distance from h to E for h(t) = e t

= kf k2.

32 CHAPTER 3 HILBERT SPACES

In fact, S is an isometry It follows that I − S is continuous By definition, E = Ker(I − S), so E is closed.

By definition, E consists of all even functions, so E ⊥ is the linear subspace of allodd functions In fact, we have

(b) The distance from h ∈ H to E is the length of the projection vector of h on E ⊥

By the above expression,

Let H be a Hilbert space and M be a closed subspace of H Denoting by P :

H → M the orthogonal projection of H onto M, prove that, for any x, y ∈ H,

• For all u ∈ H, there exist unique u M ∈ M and u M ⊥ ∈ M ⊥ such that

34 CHAPTER 3 HILBERT SPACES

n∈N

G n

!

.

(a) Prove that d(x, G) = lim n→∞ d(x, G n ), ∀x ∈ X.

(b) Prove that P G (x) = lim n→∞ P G n (x), ∀x ∈ X Note: P G is the orthogonal

Then there are n1, , n k ∈ N such that x1 ∈ G n1, , x k ∈ G n k Let n = max{n1, , n k }.

By hypothesis, the sequence (G n ) is increasing , so we have G n1, , G n k ⊂ G n, so

x1, , x k ∈ G n And since G n is a linear space,

36 CHAPTER 3 HILBERT SPACES

Let H be a Hilbert space.

(a) Prove that for any two subspaces M, N of H we have

(M + N) ⊥ = M ⊥ ∩ N ⊥ (b) Prove that for any two closed subspaces E, F of H we have

(E ∩ F ) ⊥ = E ⊥ + F ⊥

Solution

(a) If x ∈ (M + N) ⊥ , then for every m ∈ M and n ∈ N we have

hm + n, xi = 0

since m + n ∈ M + N For n = 0 we have hm, xi = 0 This holds for all m ∈ M, so

x ∈ M ⊥ Similarly x ∈ N ⊥ Thus x ∈ M ⊥ ∩ N ⊥ , and hence (M + N) ⊥ ⊂ M ⊥ ∩ N ⊥

If x ∈ M ⊥ ∩ N ⊥, then we have

hm, xi = 0 and hn, xi = 0, ∀m ∈ M, n ∈ N.

hm + n, xi = 0.

This means that x ∈ (M + N) ⊥ Hence M ⊥ ∩ N ⊥ ⊂ (M + N) ⊥

(b) From part (a) it follows that

A system {x i } i∈N in a normed space X is called a complete system if

Span {Pn i=1 α i x i : ∀n ∈ N, α i ∈ F} is dense on X.

If {x i } i∈N is a complete system in a Hilbert space H and x⊥x i for every i, show that x = 0.

Solution

Given x ∈ X, if x⊥x i for every i, then x⊥ Span {Pn i=1 α i x i } Let D := Span {Pn i=1 α i x i }.

By definition, D = X Then there exists a sequence (x n ) in D converging to x and

x⊥x n for every n Hence

0 = hx, x n i → hx, xi = kxk2 as n → ∞.

Thus x = 0. ¥

Problem 35

Let H be a Hilbert space and {ϕ i } ∞

i=1 an orthonormal system in H Show that

38 CHAPTER 3 HILBERT SPACES

i=1

|hx, e i i|2

!1

2 Ã ∞X

i=1

|hx, e i i|2

!1 2

Let H be a Hilbert space and A and B be any two subsets of H Show that

(a) A ⊥ is a closed subspace of H.

Hence αx + βy ∈ A ⊥ , so A ⊥ is a subspace of H.

We show now that A ⊥ is closed Suppose that the sequence (x n ) in A ⊥ converges

to some x ∈ H Since hx n , ai = 0 for all a ∈ A, by continuity of the inner product,

(b) For any x ∈ A, we have x⊥A ⊥ This implies that x ∈ (A ⊥)⊥ Thus

That gives that αx + βy ∈ M Hence M is a linear subspace of `2

Let us prove that it is closed Take x ∈ M There exists a sequence x (k) = (x (k) n ) ∈ M converging to x as k → ∞ Since x (k) 2n = 0, we obtain

x 2n = lim

k→∞ x (k) 2n = 0,