Functional Analysis (2006) Homework assignment 2

Transcription

1 Functional Analysis (26) Homework assignment 2 All students should solve the following problems: 1. Define T : C[, 1] C[, 1] by (T x)(t) = t x(s) ds. Prove that this is a bounded linear operator, and compute T. Also prove that the inverse T 1 : R(T ) C[, 1] exists but is not bounded. 2. Let M = { x L 2 [, 1] : x(t) dt =, tx(t) dt =, t 2 x(t) dt = }. Given x L 2 [, 1], find a formula for the vector in M which lies closest to x (in the L 2 [, 1]-norm). 3. (Problem 3.9: 4). Let H 1 and H 2 be two Hilbert spaces and let T : H 1 H 2 be a bounded linear operator. Suppose that we are given subsets M 1 H 1 and M 2 H 2 such that T (M 1 ) M 2. Prove that M 1 T (M 2 ). 4. Let a, b be two positive real numbers. Let x be a vector in a normed space X and assume that f(x) a holds for all f X with f b. Prove that x a/b. Students taking Functional Analysis as a 6 point course should also solve the following problems: 5. Let Y 1, Y 2, Y 3, be closed linear subspaces of the Hilbert space H, such that Y j Y k for all 1 j < k, and Y j = {}. Prove that for every vector v H there is a unique choice of vectors y 1 Y 1, y 2 Y 2, y 3 Y 3, such that y j = v in H. 6. Let Y be a subspace of a Banach space X. The annihilator Y a is defined as the subspace Y a := {f X : f(y) =, y Y } of X (cf. 2.1, problem 13). Hence Y aa = (Y a ) a is a subspace of X. Let C : X X be the canonical map. Prove that C(Y ) Y aa. Also prove that if X is reflexive and Y is closed then C(Y ) = Y aa. Solutions to problems 1-4 should be handed in by Friday, February 24. Solutions to problems 5-6 should be handed in by Monday, March 13. (Either give the solutions to me directly or put them in my mailbox, third floor, House 3, Polacksbacken.) 1

2 2 Functional Analysis Solutions to homework assignment 2 1. For all x 1, x 2 C[, 1] and all α, β R and all t [, 1] we have hence (T (αx 1 + βx 2 ))(t) = t = αt (αx 1 + βx 2 )(s) ds x 1 (s) ds + βt x 2 (s) ds = αt x 1 (t) + βt x 2 (t) = (αt x 1 + βt x 2 )(t); T (αx 1 + βx 2 ) = αt x 1 + βt x 2, and this shows that T is linear. Furthermore, for each x C[, 1] we have: T x = max t x(s) ds max t [,1] max t [,1] t t t [,1] x(s) ds x ds = max t [,1] t2 x = x. Hence T is bounded with T 1. In fact if we take x as the constant function x(t) = 1 then x = 1 and (T x)(t) = t x(s) ds = t2, hence T x = max t [,1] t 2 = 1. But T x T x, i.e. 1 T. Hence we have proved both T 1 and T 1. It follows that T = 1. Now assume that x C[, 1] satisfies T x =, i.e. t x(s) ds = for all t [, 1]. It then follows that x(s) ds = for all t (, 1], and thus by differentiation with respect to t we get x(t) = for all t (, 1]. Since x(t) is continuous we then also have x() =. Hence x =. We have thus proved ( ) x C[, 1] : T x = = x =. Hence by Theorem 2.6-1(a), T 1 exists. Given any n Z + we let x n (t) = t n. Then x n C[, 1] and x n = max t [,1] t n = 1. We let y n (t) = T x n (t) = t s n ds = (n + 1) 1 t n+2.

3 Then y n = max t [,1] (n + 1) 1 t n+2 = (n + 1) 1. Also, by construction, y n R(T ) and T 1 y n = x n ; thus T 1 y n = x n = 1. This shows that T 1 cannot be bounded. (For if T 1 were bounded then we would have T 1 y n T 1 y n, i.e. 1 T 1 (n + 1) 1, for all n Z +. This is impossible.) 3 2. Let f 1, f 2, f 3 L 2 [, 1] be given by f 1 (t) = 1, f 2 (t) = t, f 3 (t) = t 2. The definition of M says that x L 2 [, 1] belongs to M if and only if x is orthogonal to f 1, f 2, f 3. That is: M = {f 1, f 2, f 3 } = (Span{f 1, f 2, f 3 }). (The last identity holds since x, f 1 = x, f 2 = x, f 3 = implies x, c 1 f 1 + c 2 f 2 + c 3 f 3 =, for all c 1, c 2, c 3 C.) Let Y = Span{f 1, f 2, f 3 } (so that M = Y ). This is a closed subspace of L 2 [, 1] since it is finite dimensional (Theorem 2.4-3), and hence by Theorem 3.3-4, L 2 [, 1] decomposes as a direct sum L 2 [, 1] = Y Y = Y M. This means that given any x L 2 [, 1] there exist unique vectors y Y and z M such that x = y+z. It is easy to see that in this situation z is the vector in M which lies closest to x, 1 i.e. v M : v x z x. [Proof: If v is an arbitrary vector in M then also v z M = Y, and since y Y we then have v z, y = ; hence we may use Pythagoras theorem: v x 2 = v z+z x 2 = v z y 2 = v z 2 + y 2 y 2 = z x 2, and the proof is complete.] 1 This is also, in principle, seen in the book in the proof of Theorem Note that y is the orthogonal projection of x on Y, and z is the orthogonal projection of x on M; this concept is discussed in the book on p. 147.

4 4 To determine a formula for z as a function of x we first use Gram- Schmidt to find an orthonormal basis in Y = Span{f 1, f 2, f 3 }: ẽ 1 = f 1 = 1; e 1 = ẽ1 ẽ 1 = 1; ẽ 2 = f 2 f 2, e 1 e 1 = t = t 1 2 ; e 2 = ẽ2 ẽ 2 = 3(2t 1); ẽ 3 = f 3 f 3, e 1 e 1 f 3, e 2 e 2 = t (2t 1) = t 2 t e 3 = ẽ3 ẽ 3 = 5(6t 2 6t + 1). Now since y Y we must have y = c 1 e 1 +c 2 e 2 +c 3 e 3 for some constants c 1, c 2, c 3 C. We also have x y = z M = Y and hence for each j = 1, 2, 3, since e j Y, we have: = x y, e j = x, e j c 1 e 1 + c 2 e 2 + c 3 e 3, e j = x, e j c j. Hence c j = x, e j. It follows that i.e. z = x y = x (c 1 e 1 + c 2 e 2 + c 3 e 3 ) = x x, e 1 e 1 x, e 2 e 2 x, e 3 e 3. Answer: The vector z M which lies closest to x is z(t) = x(t) z = x x, e 1 e 1 x, e 2 e 2 x, e 3 e 3, 5(6t 2 6t + 1) x(s) ds 3(2t 1) x(s)(2s 1) ds x(s)(6s 2 6s + 1) ds. 3. Let v be an arbitrary vector in T (M2 ). Then there is some w M 2 such that v = T (w). Since w M2 we know that w, x = for every vector x M 2. Now let y be an arbitrary vector in M 1. Then v, y = T (w), y = w, T (y). But we have T (y) M 2, since y M 1 and T (M 1 ) M 2. Also recall w M2. From these two facts T (y) M 2 and w M2 it follows that

5 w, T (y) =. Hence from the above computation we see: v, y =. This is true for every y M 1. Hence v M 1. We have proved that for every v T (M 2 ) we have v M 1. Hence T (M 2 ) M 1, Q.E.D By Theorem there exists an f X such that f = 1 and f (x) = x. Let f = bf ; then f = b f = b. In particular f b and hence by the assumption in the problem we have f(x) a. On the other hand f(x) = bf (x) = b f (x) = b x. Hence b x a, i.e. x a/b, Q.E.D. Alternative solution: 2 By Corollary we have f(x) ( ) x = sup f X {} f. Now let f be an arbitrary element in X {}, as in the above supremum. Set c = f ; then c > since f. Set f = (b/c)f X ; then f = (b/c) f = (b/c) c = b. Hence by the assumption in the problem text we have f (x) a. But f = (c/b)f, hence f(x) = (c/b) f (x) (c/b)a = ca/b, and f(x) f = f(x) c ca/b c = a b. We have proved that this is true for every f X {}. Hence the supremum in ( ) is a/b, i.e. we have proved Q.E.D. x a b, 2 In some sense this is actually exactly the same solution as the first one, but in a different language.

6 6 5. We first prove uniqueness. Let v H be given. Assume that the vectors y 1 Y 1, y 2 Y 2, y 3 Y 3, are such that y j = v, i.e. lim N N y j = v. Take any k {1, 2, 3,. } and any vector w Y k ; we then have (using Lemma 3.2-2) v, w = lim N N y j, w = lim N y j, w = lim N N N y j, w. But for each j k we have y j, w = since y j Y j Y k. Hence we can continue the computation: Y j, w Y k and = lim N y k, w = y k, w. Hence we have proved v, w = y k, w, i.e. v y k, w =. This is true for every w Y k. Hence v y k Yk. But Theorem says that we have a direct sum H = Y k Yk, and now from y k Y k, v y k Yk we see that v = y k + (v y k ) is the unique decomposition of v in this direct sum. Hence y k is the orthogonal projection of v on Y k (cf. p. 147). This proves that y k is uniquely determined from v. This is true for every k {1, 2, 3, }. We next prove that every vector can actually be expressed as a sum in the stated way. Let v H be given. For each k {1, 2, 3, } we let y k be the orthogonal projection of v on Y k (this construction is of course suggested by the uniqueness proof above). We now wish to prove y j = v. For each j with y j we let e j = y j 1 y j ; then these vectors e j (where we throw away those indices j for which y j = ) form an orthonormal sequence, and hence by part (c) of the main theorem about Hilbert bases as I formulated it in my lecture, we have v, e j 2 v 2 (this is Bessel s inequality, Theorem in the book), and (hence) v, e j e j is a convergent sum (cf. Theorem 3.5-2(a) in the book). But by definition of orthogonal projection we have v y j Yj for each j, and in particular v y j y j, thus v y j, y j =. This gives v, y j = y j, y j = y j 2 and thus if y j : v, e j e j = y j 1 v, y j e j = y j 1 y j 2 e j = y j. Hence what we have proved is that the sum y j is convergent! Let us write v = y j H.

7 We now have for every k 1 and every w Y k, by arguing as in the first part of this solution: v, w = y k, w. Hence v v, w = v y k, w =, since v y k Y k because y k is the orthogonal projection of v on Y k. This is true for every w Y k, hence This is true for every k 1, hence Hence Q.E.D. v v Y k. v v k=1y k = {}. v = v = y j, 7 6. Take an arbitrary vector y Y. Then for every f Y a we have (C(y))(f) = f(y) =. Hence C(y) Y aa. This proves that C(Y ) Y aa. Next assume that X is reflexive and Y is closed. Take an arbitrary vector y Y aa. (Thus y X.) Since X is reflexive C is surjective, hence there is some vector x X such that y = C(x). Since y Y aa we have, for all f Y a : = y (f) = (C(x))(f) = f(x). Now assume x / Y. Then (since Y is closed!) by Lemma there exists a g X such that g = 1, g(y) = for all y Y (i.e. g Y a ), and g(x) = δ = inf y Y y x >. Hence we have both g Y a and g(x) > ; this contradicts the fact from above that f(x) =, f Y a! Hence the assumption x / Y must be discarded. Thus x Y. Hence from y = C(x) we see y C(Y ). This is true for every y Y aa. This proves Y aa C(Y ). Together with C(Y ) Y aa this proves C(Y ) = Y aa.