Institut für Analysis WS2019/20 Prof. Dr. Dorothee Frey 21.11.2019 M.Sc. Bas Nieraeth

Functional Analysis

Solutions to exercise sheet 6

Exercise 1: Properties of compact operators.

(a) Let X and Y be normed spaces and let T ∈ K(X,Y ). Show that R(T ) is separable.

(b) Let X, Y , and Z be Banach spaces. Let T ∈ K(X,Y ) and S ∈ L(Y,Z), where S is injective. Prove that for every ε > 0 there is a constant C ≥ 0 such that kT xk ≤ εkxk + CkST xk for all x ∈ X.

Solution:

(a) Since bounded sets have a relatively compact image under T , for each n ∈ N the sets T (nBX ) are totally bounded, where nBX := {x ∈ X : kxk ≤ n}. Then it follows from Exercise 1 of Exercise sheet 3 that T (nBX ) is separable. But then ! [ [ T (nBX ) = T nBX = T (X) = R(T ) n∈N n∈N is also separable, as it is a countable union of separable sets. The assertion follows.

(b) First note that by using a homogeneity argument, one finds that this result is true, ifand only it is true for x ∈ X with kxk = 1. Thus, it remains to prove the result for x ∈ X with kxk = 1.

We argue by contradiction. If the result does not hold, then there is an ε > 0 such that for all C ≥ 0 there exists and x ∈ X with kxk = 1 such that kT xk > ε + CkST xk. By choosing C = n for n ∈ N, we find a sequence (xn)n∈N in X with kxnk = 1 such that

kT xnk > ε + nkST xnk. (1)

Since (xn)n∈N is bounded and T is compact, there exists a subsequence (T xnj )j∈N with limit y ∈ Y .

Note that (1) implies that 1 1 kST x k ≤ kT x k ≤ kT k → 0 as n → ∞ n n n n

so that, in particular, ST xnj → 0 as j → ∞ in Z. But, since S is bounded, also ST xnj → Sy as j → ∞ in Z. By uniqueness of limits, we have Sy = 0. Since y is injective, this means that y = 0. But then it follows from (1) that

ε ≤ lim kT xn k = 0, j→∞ j which is absurd. The assertion follows. — Turn the page! — Exercise 2: Examples of compact operators.

(a) Let k be a continuous function k : [0, 1] × [0, 1] → R. We define T : C([0, 1]) → C([0, 1]) by Z 1 T f(t) := k(t, s)f(s) ds. 0 Show that T is a compact operator.

(b) Show that the inclusion mapping ι : C1([0, 1]) → C([0, 1]), ι(f) := f is a compact operator.

(c) Let m ∈ C([0, 1]) with m(x) 6= 0 for all x ∈ [0, 1]. Show that the operator T : C([0, 1]) → C([0, 1]) given by T f := mf is not compact.

(d) Let X := {f ∈ C([0, 1]) : f(0) = 0}. Then (X, k · k∞) is a Banach space. For each t ≥ 0 we define ( f(x − t) if x − t ∈ [0, 1]; Ttf(x) = 0 otherwise. Show that

(i) Tt ∈ L(X) with kTtk ≤ 1 for all t ≥ 0,

(ii) TsTt = Ts+t for all s, t ≥ 0,

(iii) Tt is compact if and only if t ≥ 1.

Solution:

(a) Note that this is the operator from Exercise 3 of Exercise sheet 4. To prove compactness of T , we show that for any bounded sequence (fn)n∈N in C([0, 1]), the sequence (T fn)n∈N has a uniformly convergent subsequence. For this we use the Arzelà-Ascoli Theorem.

It remains to check that (T fn)n∈N is uniformly bounded and uniformly equicontinuous. For the boundedness, note that since we assumed that (fn)n∈N is bounded, we have

kT fnk∞ ≤ kT kkfnk∞ ≤ kT k sup kfnk∞, n∈N

proving that (T fn)n∈N is also bounded.

For the uniform equicontinuity, we repeat the proof of Exercise 3(a) of Exercise sheet 4,

but replace the quantity kfk1 in that proof by supn∈N kfnk∞. Then, analogously to that proof, for each ε > 0 we find a δ > 0 (that does not depend on n now) such that Z 1 0 ε |T fn(t) − T fn(t )| ≤ |fn(s)| ds 1 + supn∈N kfnk∞ 0 εkf k ≤ n ∞ < ε 1 + supn∈N kfnk∞ whenever |t − t0| < δ. This proves uniform equicontinuity. Thus, by the Arzelà-Ascoli Theorem, the sequence (T fn)n∈N has a uniformly convergent subsequence. The assertion follows.

(b) This follows immediately from Exercise 3(b) of Exercise sheet 3. 1 (c) Note that since m is nowhere vanishing, we also have m ∈ C([0, 1]). This implies that T 1 has a bounded inverse C([0, 1]) → C([0, 1]), g 7→ m g. Since compact operators are not invertible on infinite dimensional spaces (otherwise Id = TT −1 would also be compact, which is only true on finite dimensional spaces), T can not be compact.

One can also give a direct proof. Choose any bounded sequence (fn)n∈N in C([0, 1]) which 1 does not have a uniformly convergent subsequence, e.g., fn(x) = x n . Then the sequence 1 1 ( m fn)n∈N is also bounded, but (T ( m fn))n∈N = (fn)n∈N has no uniformly convergent sub- sequence. Thus, T is not compact, as asserted.

(d) We only show (iii). Note that if t ≥ 1, then Ttf = 0 for all f ∈ X. Since the 0 map is compact (the set {0} is compact), we conclude that Tt is compact for t ≥ 1.

Conversely, we will show that if t ∈ [0, 1), then Tt is not compact. Indeed, consider the 1 sequence fn : [0, 1] → R given by fn(x) := x n . Then fn(0) = 0 and kfnk∞ = 1 for all n ∈ N, so this sequence is bounded in X. We will show that (Ttfn)n∈N does not have a convergent subsequence in X.

We argue by contradiction. Suppose there is a subsequence (Ttfnj )j∈N with limit g ∈ X.

Since uniform convergence implies pointwise convergence, we find that Ttfnj also converges pointwise to g. But since ( 1 (x − t) n if x − t ∈ [0, 1]; Ttfn(x) = 0 otherwise. converges pointwise to the function ( 1 if x ∈ (t, 1]; g˜(x) = 0 if x ∈ [0, t], it follows from uniqueness of limits that g =g ˜. But this is a contradiction, since g is continuous, while g˜ is not. This proves the assertion.

Exercise 3: Operators on `p given by infinite matrices. 0 1 1 Let p ∈ (1, ∞) and define p ∈ (1, ∞) through the relation p + p0 = 1. Suppose for each j, k ∈ N we are given an aj,k ∈ K with

p 1   p ∞ ∞ ! p0 X X p0 γ :=  |aj,k|  < ∞. j=1 k=1

p P∞ p (a) Let x ∈ ` . Show that the sequence y = (yj)j∈N given by yj := k=1 aj,kxk is again in ` .

(b) Prove that the operator T : `p → `p, T x := y with y defined as in part (a) is bounded with kT k ≤ γ.

(c) Show that T is a compact operator.

Solution:

(a) By Hölder’s inequality we have

1 ∞ ∞ ! p0 X X p0 |yj| ≤ |aj,kxk| ≤ |aj,k| kxkp k=1 k=1 — Turn the page! — for each j ∈ N, so that p ∞ ∞ ∞ ! p0 X p X X p0 p p p |yj| ≤ |aj,k| kxkp = γ kxkp < ∞. j=1 j=1 k=1

p Hence, y ∈ ` with kykp ≤ γkxkp.

(b) Boundedness of T follows from the estimate kT xkp = kykp ≤ γkxkp we did in part (a). Moreover, this estimate implies that kT k ≤ γ, as asserted.

(c) There are several ways of doing this. For example, to prove that T is compact we need to show that T (B`p ) is totally bounded. For this we can check the totally boundedness p criterion from Exercise 4 of Exercise sheet 3. Indeed, let y ∈ T (B`p ) and pick x ∈ ` with kxkp ≤ 1 such that T x = y. Then

p ∞ ∞ ∞ ! p0 X p X X p0 |yj| ≤ |aj,k| j=J+1 j=J+1 k=1

by a calculation analogous to the one in part (a). Hence,

p ∞ ∞ ∞ ! p0 X p X X p0 sup |yj| ≤ |aj,k| → 0 as J → ∞, y∈T (B`p ) j=J+1 j=J+1 k=1 since the tail of a convergent series converges to 0. Thus, by Exercise 4 of Exercise sheet 3, T (B`p ) is totally bounded, and we conclude that T is compact.

Alternatively, we can find an explicit sequence of finite rank operators that converge to T .

Indeed, fix n ∈ N, and let Tn be the operator given by the matrix (aj,k)j∈{1,...,n}, k∈N, i.e, P∞ Tnx := y with yj := k=1 aj,kxk for j ∈ {1, . . . , n} and yj = 0 for j > n. Note that the range of Tn is at most n-dimensional, and thus Tn is a finite rank operator.

p For each x ∈ ` with kxkp = 1 we have, using Hölder’s inequality,

p p ∞ ∞ n ∞ ! p0 0 p X X X X p kT x − Tnxkp = aj,kxk ≤ |aj,k| j=1 k=n+1 j=1 k=n+1 Hence, taking a supremum over all such x, we find that

p ∞ ∞ ! p0 ∞ p X X p0 X kT − Tnk ≤ |aj,k| =: bj,n. (2) j=1 k=n+1 j=1

We wish to apply the Dominated Convergence Theorem with the sequence (bj,n) to take a limit under the series to conclude that the limit in (2) as n → ∞ tends to 0. Note that p P∞ p0  p0 |bj,n| ≤ k=1 |aj,k| =: bj, and

∞ X p |bj| = γ < ∞. j=1

Thus, we may apply the Dominated Convergence Theorem to (2). Noting that for a fixed j, the sequence bj,n is the tail of a convergent series, we have limn→∞ bj,n = 0. Thus, by (2) we have ∞ ∞ X X kT − Tnk ≤ lim bj,n = lim bj,n = 0. n→∞ n→∞ j=1 j=1 We conclude that T is the operator norm limit of finite rank operators. Since finite rank operators are compact, and because the set of compact operators is closed with respect to operator norm convergence, we conclude that T is also a compact operator, as asserted.

Exercise 4: Isolated points in metric spaces. Let (X, d) be a non-empty complete metric space. A point x ∈ X is called an isolated point if {x} is an open set in X.

(a) Show that if X has no isolated points, then X is uncountable.

(b) Show that if X is countable, then the set of all isolated points in X is dense in X.

Hint: use the Baire Category Theorem.

Solution:

(a) If X has no isolated points, then each set {x} is nowhere dense. Indeed, the closure of {x} is {x} itself. Since x is not an isolated point, {x} is not open, and thus x is not an interior point of {x}. Hence, the interior of the closure of {x} is empty, proving that {x} is a nowhere dense set.

Since X is a complete metric space, it follows from the Baire Category Theorem that X is a Baire second category set and is thus not the union of countably many nowhere dense set. Since X 6= ∅, [ X = {x}, x∈X is a way to write X as a union of nowhere dense sets. It follows that X can not be countable. Hence, X is uncountable.

(b) Denote the set of all isolated points in X by S. If x∈ / S, then X\{x} is an open and dense set in X (If it were not dense, then its closure would have to coincide with X\{x}, which would imply that {x} is open and thus x ∈ S). Then \ S = X\{x} x∈X\S

is also dense in X by the Baire Category Theorem, as it is the intersection of countably many open dense sets in X. The assertion follows.

(Bonus) Exercise 5: Proper inclusions in `p spaces. Let p ∈ (1, ∞).

(a) Show that for any 1 < q < p, the set `q is a Baire first category subset of `p.

S q p (b) Show that q∈(1,p) ` 6= ` .

Solution:

(a) In Exercise 4 of Exercise sheet 5 we have shown that `q ⊆ `p whenever q < p, which we will use throughout this exercise. — Turn the page! — p q q Let N ∈ N and define AN := {x ∈ ` : x ∈ ` and kxkq ≤ N}. Note that then ` = S A `q `p N∈N N . Thus, to conclude that is a Baire first category subset of , it suffices to p show that the sets (AN )N∈N are nowhere dense in ` .

(n) Fix N ∈ N. We first prove that AN is closed. Let (x )n∈N be a sequence in An that converges in `p with limit x ∈ `p.

(n) (n) Note that then for each j ∈ N we have |xj − xj | ≤ kx − x kp → 0 as n → ∞, so that (n) xj → xj in K as n → ∞. But then

J J X q X (n) q (n) q q |xj| = lim |xj | ≤ lim sup kx kq ≤ N . n→∞ n→∞ j=1 j=1

q We conclude that x ∈ ` with kxkq ≤ N and thus x ∈ AN . We conclude that AN is closed.

p Next, we prove that AN has an empty interior in ` . Pick any x ∈ AN . We will show that p q x is not an interior point of AN . By Exercise 4 of Exercise sheet 5, we can pick a y ∈ ` \` . r > 0 x˜ := r y + x x˜ ∈ `p `p Let and set 2kykp . Note that as the sum of two elements of , q q 2kykp q q but x˜ ∈ / ` . Indeed, if x˜ ∈ ` , then also y = r (˜x − x) ∈ ` , contradicting y∈ / ` . We conclude that x˜ ∈ `p\`q.

Note that r kx˜ − xk = < r. p 2 q However, since x˜ ∈ / ` , we do not have x ∈ AN . Thus, the ball B(x, r) is not contained in AN . As r > 0 was arbitrary, we conclude that x is not an interior point of AN and thus, p AN has an empty interior. Thus, we have shown that AN is a nowhere dense set in ` , proving the desired result.

1 (b) Set qn := p − n . Then we claim that [ [ `q = `qn . q∈(1,p) n∈N

Indeed, the inclusion “⊇” is clear. For the converse inclusion, note that for any q ∈ (1, p) q qn we can find an n ∈ N such that q < qn. Hence ` ⊆ ` , proving the inclusion. The claim follows.

Since `p is a complete metric space, it is a Baire second category set. Thus, it can not be written as a countable union of nowhere dense sets. Since by part (a) every `qn space S `qn can be written as a countable union of nowhere dense sets, the union n∈N can also be written a countable union of nowhere dense sets. Hence, it can not be equal to `p, proving the result.

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