Functional Analysis Exercise Class

Transcription

1 Functional Analysis Exercise Class Week 9 November 13 November Deadline to hand in the homeworks: your exercise class on week 16 November 20 November Exercises (1) Show that if T B(X, Y ) and S B(Y, Z) are bounded operators then and x X : T x, ST S T. Solution: Recall the definition of the operator norm, T := sup x X\{0} x. Then obviously for all x X \ {0}, we have x sup x X\{0} x = T, and hence T x, as claimed. trivial, as both sides are equal to 0. For the second part, clearly, The inequality T 0 T 0 is ST (x) = S(T x) S S T x. (2) Show that for a bounded linear map T : X Y the following expressions for the norm are equivalent: a) T = inf{c 0 : C x for all x X}, b) T = sup x 0 x, c) T = sup x 1, 1

2 d) T = sup x =1. Solution: Let I = inf{c 0 : C x for all x X}, I 1 = sup x 0 x, I 2 = sup, x 1 I 3 = sup. x =1 Straight from definition we have that I 3 I 2, and since have I 1 I 3. Now if x 1 we have x x. Then I 2 I 1 and therefore I 1 = I 2 = I 3. = T ( x ) we x Note that I 1 x for all x X. Then I I 1. Moreover by definition of sup there exists a sequence {x n } n such that T x n / x n I 1 1/n for all n. By definition of I we have that for all n, I T x n / x n. Therefore I = I 1. (3) a) Let T B(X, Y ) be a bounded linear operator. Show that if λ is an eigenvalue of T then λ T. (0.1) b) A matrix T R n n is called stochastic if all its entries are non-negative, and the sum of each row is 1. Show that all eigenvalues of a stochastic matrix have absolute value at most 1. Solution: a) Let λ be an eigenvalue with eigenvector x X \ {0}. Then λ x = λx = T x, from which (0.1) follows. 2

3 b) Equip R n with the -norm x := max 1 i n x(i). We have seen in the lecture that in this case the induced operator norm is T = max 1 i n n j=1 T ij. Hence, for a stochastic matrix T = 1, and by the previous point, every eigenvalue λ of T satisfies λ T = 1. (4) Let δ : C([0, 1]) R be the linear functional that evaluates a function at the origin: δ(f) = f(0). a) Show that if C([0, 1]) is equipped with the sup-norm f = sup f(x) 0 x 1 then δ is bounded, and compute its norm. b) Show that if C([0, 1]) is equipped with the one-norm then δ is unbounded. f 1 = 1 0 f(x) dx Solution: a) By one of the equivalent definitions of the norm we have δ = sup δ(f) = sup f(0). f =1 f =1 Since f = 1, it means that 0 x 1: f(x) 1. In particular, f(0) 1. Therefore δ 1. Taking the constant 1 function f(x) = 1, x [0, 1], we get that f = 1, and hence δ δ(f) = 1. Combining the two inequalities we get δ = 1. b) We are going to construct a sequence of functions f n, n 2, such that f n 1 = 1 for all n, and δ(f n ), proving the unboundedness of δ. For every n N, n 2, define { n 2 f n (x) = x + n 0 x 2/n 2 0 2/n x 1. Clearly, f n 1 = 1 f(x) dx = 1 for all n, and f(0) = n. Therefore 0 δ(f n ) = n as n. So δ is unbounded. 3

4 (5) Define the linear functional ϕ(x) := x(1) 4x(2) + 5x(2015), x l p := l p (N, K). Show that ϕ is a bounded linear functional on l p for all 1 p +, and compute its norm. Solution: Linearity is obvious, and ϕ is of the form ϕ(x) = + 1, i = 1, 4, i = 2, y(i)x(i), with y(i) = 5, i = 2015, 0, otherwise. Clearly, y l q (N, K) for any 1 q +. Assume first that 1 < p < +. As we have seen in the lecture, ϕ defines a bounded linear functional on l p, and its norm is ϕ = y q = (1 + 4 q + 5 q ) 1/q For p = 1, we have for ϕ(x) x(1) + 4 x(2) + 5 x(2015) 5 1 q = 1 1 p. + x(i) = 5 x 1, x l 1, and hence ϕ 5 = y. On the other hand, choosing x l 1 such that x(2015) = 1 and x(i) = 0 for i 2015, we get x 1 = 1 and ϕ(x) = 5, and hence ϕ 5. Combining the two inequalities, we get ϕ = 5. For p = +, we have for every x l ϕ(x) x(1) + 4 x(2) + 5 x(2015) x ( ) = x y 1, and hence ϕ y 1 = 10. On the other hand, choosing 1, i = 1 or i = 2015, x(i) = 1, i = 2, 0, otherwise, we see that x = 1, and ϕ(x) = 10, and thus ϕ 10. Combining the two inequalities, we get ϕ = 10. 4

5 (6) Let c 00 := { x K N : #{i : x(i) 0} < + } be the space of K-valued sequences that are non-zero only at finitely many places. a) Show that c 00 is dense in l p (N, K) for any 1 p < +. b) Equip c 00 with the p-norm x p := ( + x(i) p) 1/p for some 1 < p < +, and define the linear functional ϕ : c 00 K as ϕ(x) := + 1 i x(i). Solution: For which values of p is ϕ bounded? a) Let 1 p < +. We have to show that for every x l p := l p (N, K) and every ε > 0, there exists an x ε c 00 such that x x ε p < ε. If x l p then, by definition, + x(i) p < +, and hence there exists an N ε N such that + i=n ε+1 x(i) p < ε p. Define x ε (i) := x(i), i N ε, and x ε (i) := 0, i > N ε. Then x ε c 00, and ( + ) 1/p ( + x x ε p = x(i) x ε (i) p = i=n ε+1 x(i) p ) 1/p < ε. b) Let y(i) := 1 i such that ϕ(x) = + y(i)x(i), and let q be the Hölder conjugate of p, i.e., = 1. Then q p + y(i) q = + 1 i q/2, and hence y l q (N, K) q > 2. If q > 2 then x + y(i)x(i) defines a bounded linear functional on l p, and ϕ is its restriction to c 00, and hence ϕ is bounded, too. On the other hand, if q 2 then y q = +, and hence for any K > 0, there exists an N K N such that N K y(i) q K. Let y K (i) := y(i), 1 i N K, and y K (i) := 0, i > N K. Then y K can be considered as an element of l q ([N K ], K), and we have seen (week 2) that there exists an 5

6 x K l p ([N K ], K) such that x K p = 1 and N K y K(i)x K (i) = y K q. This x K can be naturally considered as an element in c 00 with x K p = 1 and ϕ(x K ) = N K y(i)x K (i) = y K q K 1/q. This shows that ϕ sup K>0 K 1/q = +. (7) a) Show that a linear operator from a normed space X to a normed space Y is bounded if and only if the image of the unit ball around 0 is a bounded set in Y. b) Show that if X is finite-dimensional then every linear operator to any normed space Y is bounded. c) Is it true that if Y is finite-dimensional then any linear operator from any normed space X is bounded? d) The rank of a linear operator T : X Y is the dimension of its range, i.e., rk T := dim{t x : x X}. Is it true that every finite-rank operator is bounded? Solution: a) Let T : X Y be a linear operator. Note that T is bounded if and only if its norm T = sup x X\{0} x = sup T x x X\{0} x = sup x B 1 (0)\{0} is finite, from which the assertion follows. b) Let X denote the norm on X. Let e 1,..., e d be a basis in X, and for every X x = d x ie i, define ( d x 2 := x i 2 ) 1/2. As we have seen in the lecture, all norms on a finite-dimensional space are equivalent, and hence there exists a constant C > 0 such that x 2 C x X, x X. Then ( d d d ) 1/2 ( d ) 1/2 = T x i e i x i T e i x i 2 T e i 2 ( d ) 1/2 C x X T e i 2, 6

7 where the first inequality is due to the triangle inequality, and the second one follows from the Cauchy-Schwarz inequality. Hence, T is bounded, and T C ( d T e i 2) 1/2. c) No. Consider the examples in (4)b) and (6)b), where Y is one-dimensional, and the given linear functional is unbounded. d) No, for the same reason as in the previous point. (8) In the lecture it was proved that for p, q (1, ) with 1/p + 1/q = 1 the space l q := l q (N, K) is isometrically isomorphic to l p (N, K), i.e. l q = l p. Prove the same for p = 1, i.e. prove that l 1 = l. Solution: We follow the same argument as in the lecture. Define T : l l 1, by (T x)(y) := k=1 x ky k for x l and y l 1. Then (T x)(y) x k y k x y 1. k=1 This means that for all x l we have T x l 1 and = y l 1 \{0} (T x)(y) y 1 x. (0.2) On the other hand, for every k such that x k 0, let y (k) (k) := x k /x k, and let y (k) (i) := 0, i k. Then y (k) 1 = 1, and (T x)(y (k) ) = x k implies that x k. Since this holds for every k N, we obtain sup k N x k = x. Combining the two inequalities, we get = x, i.e., T is an isometry. This immediately implies that T is injective, and hence the only thing we are left to show is the surjectivity of T. To this end, let ϕ l 1 and define x = (x 1, x 2,... ) by x k := ϕ(e (k) ), where e (k) = (0,..., 0, 1, 0,... ) has 1 at the k-th place and 0 everywhere else. We need to show that T x = ϕ, i.e. for all y l 1 : (T x)(y) = ϕ(y). Since c c := {y K N : #{i : y(i) 0} < + } is dense in l 1, it suffices to consider y c c. By construction, for all y c c we have (T x)(y) = ϕ(y). (9) Prove that equivalent norms on a normed linear space X lead to equivalent norms on the space B(X) of bounded linear operators on X. Solution: Let 1 and 2 be the two equivalent norms on a normed linear space X, i.e. there exist constants c and C such that for all x X c x 1 x 2 C x 1. 7

8 Let T B(X). Then T (x) 1 T 1 = sup x 0 x 1 c 1 2 sup = C x 0 C 1 x 2 c T 2. Similarly, In other words, T (x) 2 T 2 = sup x 0 x 2 C 1 sup = C x 0 c x 1 c T 1. c C T 1 T 2 C c T 1. 8

9 Homework with solutions (1) Define the right shift operator S B(l p ) on l p := l p (N, K) by Sx := (0, x 1, x 2,... ). Define the left shift operator T B(l p ) by T x := (x 2, x 3,... ). a) Show that S is an injection but not a surjection, and T is a surjection, but not an injection. b) Show that S = T = 1. c) Show that T S = I (the identity operator), but ST I. Solution: a) If Sx = 0 then for all n we have x n = 0, therefore x = 0. So S is injection. Since the sequence (1, 0,... ) has no pre-image, S is not a surjection. Since for the sequence x = (1, 0,... ), T x = 0, T is not injective. For every y l p define x = (0, y 1, y 2,... ), then T x = y, and so T is surjective. b) For every x l p we have Sx p = S = sup x lp\{0} Sx p / x p = 1. ( ( j=1 x j p ) 1/p = x p. Therefore ) 1/p For every x l p, p = j=2 x j p x p. Therefore T 1. But for every x = (0, x 1, x 2,... ) we have p = x p. Thus T = 1. c) For x = (x 1, x 2,... ), we have T S(x) = T ((0, x 1, x 2,... )) = (x 1, x 2,... ) = x. But ST (x) = S((x 2, x 3,... )) = (0, x 2, x 3,... ) x, so ST I. (2) Let X = (C([0, 1]), ), and define a linear operator K on X by (Kf)(x) := 1 0 k(x, y)f(y)dy, where k : [0, 1] [0, 1] R is continuous. Prove that K is a bounded operator. Solution: Note that k is continuous on the compact set [0, 1] [0, 1], and hence it is bounded, i.e., k := max x,y [0,1] k(x, y) < +. Consider 1 Kf = sup (Kf)(x) = sup k(x, y)f(y)dy 0 x 1 1 sup 0 x x 1 k(x, y) f(y) dy f k. 9 0

10 Therefore K is bounded, and K k. Score: 3 points. (3) Let c 0 := {x K N : lim n + x(n) = 0}. Show that c 0 = l 1. Solution: We follow the same argument as in the lecture. Define T : l 1 c 0, by (T x)(y) := x k y k (0.3) k=1 for x l 1 and y c 0. The above sum is well-defined, as it is absolute convergent due to k=1 x ky k x 1 y. Then (T x)(y) x k y k x 1 y. k=1 This means that for all x l 1 we have T x c 0, and = (T x)(y) sup x 1. (0.4) y c 0 \{0} y On the other hand, for every k such that x k 0, let y k := x k /x k, and let y i := 0 otherwise. For every N N, let y (N) i := y i, 1 i N, and y (N) i := 0, otherwise, i.e., y (N) = (y 1,..., y N, 0, 0,...). For all N N, y (N) c 0 and y (N) 1, and hence (T x)y (N) = N x i. (0.5) Taking the limit N +, we get x 1, and combining it with (0.4) yields = x 1. That is, T is an isometry. This immediately implies that T is injective. Hence, we are left to prove that T is also surjective. To this end, let ϕ c 0 and define x = (x 1, x 2,... ) by x k := ϕ(e (k) ), where e (k) = (0,..., 0, 1, 0,... ) has 1 at the k-th place and 0 everywhere else. Note that (T x)(y) := + k=1 x ky k defines a linear functional on c c := {y K N : #{i : y(i) 0} < + } such that T x = ϕ cc, and the same argument as in (0.5) shows that x 1 = ϕ cc ϕ < +, i.e., x l 1. Hence, T x can be extended to a bounded linear functional on l 1 by the formula given in (0.3). Since T x and ϕ are both bounded linear functionals on l 1, and they cocincide on the dense subspace c c, we finally obtain that T x = ϕ. 10

11 Score: Continuous embedding of l 1 into c 0: 3 points. The embedding is isometry+injectivity: 4 points. Surjectivity: 3 points. Total score: 10 points. Total score for HW: 13 points. 11