Banach Spaces, HW Problems & Solutions from Hunter

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1 Problem 1: Banach Spaces, HW Problems & Solutions from Hunter Ex 5.3: Let δ: C ([,1]) -> R be the linear functional that evaluates a function at the origin: δ(f ) = f (). If C ([,1]) is equipped with the sup-norm, f = sup x 1 f (x). Show that δ is bounded and compute its norm. If C ([,1]) is equipped with the one-norm f 1 = 1 f (x) dx, show that δ is unbounded. δ(f ) = f () f => d(f) f 1 => f 1 Let f 1; then δ(f ) = f = 1. Also δ d(f) f = 1. Therefore, δ =1. Let f n(x) = { 2 2n ( 1 2 n x); x < 1 2 n ; 1 2 n x Clearly, f 1 = 1 2 ; but δ(f ) = f n() = 2 n -> => δ is not bounded, with. 1 norm. 1

2 Problem 2: Ex 5.5: Define K :C ([,1])-> C ([,1]) by Kf(x) = ˆ 1 k(x, y)f(y)dy, where k:[,1]x[,1]->r is continuous. Prove that K is bounded and K = max {ˆ 1 } k(x, y) dy x 1 Let A = max 1 x 1 k(x, y) dy = 1 k(x, y) dy Kf(x) = 1 k(x, y)f(y)dy 1 k(x, y) f(y) dy 1 k(x, y) f Cdy = f C 1 k(x, y) dy f C A => Kf C = sup x Kf(x) A f C => K A Let { 1; k(x, y) > f (x) = 1; f (x) = ; otherwise k(x, y) < Then, 1 k(x, y)f (y)dy = 1 k(x, y) dy = A By density theorem, f n ɛ C([, 1]); s.t. f n = 1 and f n (x) > f (x) a.e. => K Kfn(x) f n(x) Kf n (x ) = 1 k(x, y)f n (y)dy > 1 k(x, y)f (y)dy = A (LDCT ) => K A Thus, K = A. 2

3 Problem 3: Ex 5.6: Let X be a normed linear space. Use the Hahn-Banach theorem to prove the following statements. (a) For any non zero x ɛ X, there is a bounded linear functional ϕ ɛ X such that ϕ = 1 and ϕ(x) = x. (b) If x, y ɛ Xand ϕ(x) = ϕ(y) ϕ ɛ X, then x = y. (a) Let Y = {λx λ ɛ R} (assuming x ) We define: ϕ Y : Y > R by ϕ Y (λx) = λ x ; ϕ Y is linear and ϕ Y (λx) λx = 1 => ϕ Y = 1 By the Hahn-Banach theorem; we can extend ϕ Y to ϕ : X > R s.t. ϕ(λx) = ϕ Y (λx) and ϕ = 1, Now, by definition, ϕ = sup ϕ(x) x x = 1 => ϕ(x) = x. (b) If x y => x y From part (a) ϕ ɛ X s.t. ϕ (x y) = x y => ϕ (x) ϕ (y) which contradicts the assumption stated in (b). Therefore, indeed x = y. 3

4 Problem 4: Ex 5.7: Find the kernel and range of the linear operator K : C ([,1]) -> C ([,1]) defined by Kf(x) = ˆ 1 sin π(x y)f(y)dy. sin π(x y) = sin (πx πy) = cos πy sin πx cos πx sin πy Therefore, Kf(x) = 1 sin π(x y)f(y)dy = a sin πx + b cos πx And hence, Range(K) = {a sin πx + b cos πx a, b ɛ R} Kernel(K) = {f ɛ C([, 1]) 1 cos πy f(y) dy = 1 sin πy f(y) dy = } 4

5 Problem 5: Ex 5.8: 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. Let. 1 and. 2 be the 2 equivalent norms on a normed linear space, X. Here all C s are constants. i.e. c x 1 x 2 C x 1 => C 1 x 2 x 1 => 1 x 1 1 C 1 x 2 Let X ɛ B(X); X (x) Therefore, X 1 = sup X (x) x x 1 sup x C 1 x 2 => C 1 X 1 X 2 Similarly, c 1 X 1 X 2 C 1 X 1 or equivalently, c X 2 X 1 C X 2. 5

6 Problem 6: Ex 5.9: Prove proposition 5.43: let X,Y,Z be Banach spaces. (a) If S,T ɛ B(X,Y) are compact, then any linear combination of S and T is compact. (b) If (T n ) is a sequence of compact operators in B(X,Y) converging uniformly to T, then T is compact. (c) If T ɛ B(X,Y) has finite dimensional range, then T is compact. (d) Let S ɛ B(X,Y), T ɛ B(Y,Z ). If S is bounded and T is compact, or S is compact and T is bounded, then TS ɛ B(X,Z) is compact. (a) Assume S,T are compact and a,b are real numbers for any bounded sequence x n ɛ X. S(x n ) has a convergent subsequence say, S(x nk ), and T is compact => T (x nk ) has a convergent subsequence say T (x nkl ), then (at + bs)(x nkl ) converges => (at + bs) is compact. (b) Assume T n compact and T n - T -> : Then B bounded subset of X, for any ε >, T n, s.t. T n T ε 3M, (B B M () = {x ɛ X x M}) T n B pre-compact => ɛ 3, net on T nb, say {T n x 1,..., T n x k }. Then {Tx 1,..., Tx k } is an ɛ-net of TB. Thus TB is precompact and T is compact. in fact, x ɛ B, x i, T n x T n x i ɛ 3 => T x T x i T x T n x + T n x T n x i + T n x i T x i T n T x + ε 3 T n T x i ε 3M M + ε 3 + ε 3M M = ε (c) T has finite dimensional range => T (B) is bounded and finite dimensional => T (B) is pre-compact, when B is bounded. (d) B bounded set B X, => SB is bounded in X, T compact => T (SB) is pre-compact or (TS)(B) is pre-compact. 6

7 Problem 7: Ex 5.1: Suppose k:[,1]x[,1]->r is a continuous function. Prove that the integral operator K :C ([,1])->C ([,1]) defined by is compact. Kf(x) = ˆ 1 k(x, y)f(y)dy This is an application of Arzella Ascolli Theorem. We already have shown in problem 2 that KB = {Kf bounded, for B, a bounded subset of C ([,1]). f ɛ B} is We now show KB is also equi-continuous. Since k is continuous => ɛ > δ > and x 1 x 2 < δ s.t. k(x 1, y) k(x 2, y) ε M Therefore, Kf(x 1 ) Kf(x 2 ) 1 ɛ M Mdy = ε Hence, KB is pre-compact. By definition K is compact. 7

8 Problem 8: Ex 5.11: Prove tha if T n > T uniformly, then T n > T. T n T n T + T T T n T + T n => T T n T n T T T n => T n T T T n > => lim( T n T ) = or lim T n = T. 8

9 Problem 9: Ex 5.12: Prove tha if T n > T uniformly, then T n > T strongly. strongly. x ɛ X, T n x T x T n T x >. Hence, T n > T 9

10 Problem 1: Ex 5.13: Suppose that is the diagonal matrix (nxn) and N is the Nilpotent matrix (nxn). N = = λ 1 λ 2. λ n (a) Compute the 2 norms and spectral radii of and N. (b) compute e λt and e Nt (a) has eigen values: λ 1, λ 2,..., λ n => spectral radii, R = sup 1 i n λ i. N has zero as the only eigen values, therefore R =. (b) e λt = ) e λ1t. k= and. ent = k= N k t k k! = k=n k= N k t k k! (since N n+1 = e λnt 1

11 Problem 11: Ex 5.14 Suppose that A is nxn matrix. For t ɛ R we define f(t) = det e ta. (a) Show that lim f(t) 1 = tr A t > t (b) Deduce thatf : R > R is differentiable, and is a solution of the ODE df = (tr A)f dt (c) Show that det e A = e tr A (a) proof by induction. (1) d dt eta11 t= = a 11 (2) Assume lim det e ta n 1 1 t > t = tr A n 1 let A n = A n 1 a nn Then, lim det e tan 1 t > t = lim det(i + ta n) 1 t > t =... = tr A n 1 + a nn = tr A n. (b) lim det e (t+h)a det e ta h > h = lim h > f(t) = f() e (tr A)t = e (tr A)t det e ta (det e ha 1) h = f(t) tr A or df dt = tr A f => (c) Let t = 1, f(1) = det e A = e tr A. 11