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Functional Analysis Solutions to Exercise Sheet 6

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Functional Analysis Solutions to Exercise Sheet 6 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 2 K(X; Y ). Show that R(T ) is separable. (b) Let X, Y , and Z be Banach spaces. Let T 2 K(X; Y ) and S 2 L(Y; Z), where S is injective. Prove that for every " > 0 there is a constant C ≥ 0 such that kT xk ≤ "kxk + CkST xk for all x 2 X. Solution: (a) Since bounded sets have a relatively compact image under T , for each n 2 N the sets T (nBX ) are totally bounded, where nBX := fx 2 X : kxk ≤ ng. 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 ) n2N n2N 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 2 X with kxk = 1. Thus, it remains to prove the result for x 2 X with kxk = 1. We argue by contradiction. If the result does not hold, then there is an " > 0 such that for all C ≥ 0 there exists and x 2 X with kxk = 1 such that kT xk > " + CkST xk. By choosing C = n for n 2 N, we find a sequence (xn)n2N in X with kxnk = 1 such that kT xnk > " + nkST xnk: (1) Since (xn)n2N is bounded and T is compact, there exists a subsequence (T xnj )j2N with limit y 2 Y . Note that (1) implies that 1 1 kST x k ≤ kT x k ≤ kT k ! 0 as n ! 1 n n n n so that, in particular, ST xnj ! 0 as j ! 1 in Z. But, since S is bounded, also ST xnj ! Sy as j ! 1 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 " ≤ lim kT xn k = 0; j!1 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 2 C([0; 1]) with m(x) 6= 0 for all x 2 [0; 1]. Show that the operator T : C([0; 1]) ! C([0; 1]) given by T f := mf is not compact. (d) Let X := ff 2 C([0; 1]) : f(0) = 0g. Then (X; k · k1) is a Banach space. For each t ≥ 0 we define ( f(x − t) if x − t 2 [0; 1]; Ttf(x) = 0 otherwise: Show that (i) Tt 2 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)n2N in C([0; 1]), the sequence (T fn)n2N has a uniformly convergent subsequence. For this we use the Arzelà-Ascoli Theorem. It remains to check that (T fn)n2N is uniformly bounded and uniformly equicontinuous. For the boundedness, note that since we assumed that (fn)n2N is bounded, we have kT fnk1 ≤ kT kkfnk1 ≤ kT k sup kfnk1; n2N proving that (T fn)n2N 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 supn2N kfnk1. Then, analogously to that proof, for each " > 0 we find a δ > 0 (that does not depend on n now) such that Z 1 0 " jT fn(t) − T fn(t )j ≤ jfn(s)j ds 1 + supn2N kfnk1 0 "kf k ≤ n 1 < " 1 + supn2N kfnk1 whenever jt − t0j < δ. This proves uniform equicontinuity. Thus, by the Arzelà-Ascoli Theorem, the sequence (T fn)n2N 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 2 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)n2N 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)n2N is also bounded, but (T ( m fn))n2N = (fn)n2N 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 2 X. Since the 0 map is compact (the set f0g is compact), we conclude that Tt is compact for t ≥ 1. Conversely, we will show that if t 2 [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 kfnk1 = 1 for all n 2 N, so this sequence is bounded in X. We will show that (Ttfn)n2N does not have a convergent subsequence in X. We argue by contradiction. Suppose there is a subsequence (Ttfnj )j2N with limit g 2 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 2 [0; 1]; Ttfn(x) = 0 otherwise: converges pointwise to the function ( 1 if x 2 (t; 1]; g~(x) = 0 if x 2 [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 2 (1; 1) and define p 2 (1; 1) through the relation p + p0 = 1. Suppose for each j; k 2 N we are given an aj;k 2 K with p 1 0 1 p 1 1 ! p0 X X p0 γ := @ jaj;kj A < 1: j=1 k=1 p P1 p (a) Let x 2 ` . Show that the sequence y = (yj)j2N 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 1 1 ! p0 X X p0 jyjj ≤ jaj;kxkj ≤ jaj;kj kxkp k=1 k=1 — Turn the page! — for each j 2 N, so that p 1 1 1 ! p0 X p X X p0 p p p jyjj ≤ jaj;kj kxkp = γ kxkp < 1: j=1 j=1 k=1 p Hence, y 2 ` 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 2 T (B`p ) and pick x 2 ` with kxkp ≤ 1 such that T x = y. Then p 1 1 1 ! p0 X p X X p0 jyjj ≤ jaj;kj j=J+1 j=J+1 k=1 by a calculation analogous to the one in part (a). Hence, p 1 1 1 ! p0 X p X X p0 sup jyjj ≤ jaj;kj ! 0 as J ! 1, y2T (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 2 N, and let Tn be the operator given by the matrix (aj;k)j2f1;:::;ng; k2N, i.e, P1 Tnx := y with yj := k=1 aj;kxk for j 2 f1; : : : ; ng 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 2 ` with kxkp = 1 we have, using Hölder’s inequality, p p 1 1 n 1 ! p0 0 p X X X X p kT x − Tnxkp = aj;kxk ≤ jaj;kj j=1 k=n+1 j=1 k=n+1 Hence, taking a supremum over all such x, we find that p 1 1 ! p0 1 p X X p0 X kT − Tnk ≤ jaj;kj =: 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 ! 1 tends to 0.
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