Fall TMA4145 Linear Methods. Solutions to exercise set 9. 1 Let X be a Hilbert space and T a bounded linear operator on X.
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1 TMA445 Linear Methods Fall 26 Norwegian University of Science and Technology Department of Mathematical Sciences Solutions to exercise set 9 Let X be a Hilbert space and T a bounded linear operator on X. a Suppose x and x are two elements in X. If we have that x, y x, y for all y X, then show that x x. b Show that the operator norm of T can be expressed in terms of the innnerproduct: T sup{ T x, y : x, y X with x y }. Solution. a We have x, y x, y, y X x, y x, y, y X x x, y, y X (by linearity of the innerproduct x x X (by the definition of orthogonal complement x x {} x x x x b We will first show that sup{ x, y : y X with y } x for all x X. ( By the Cauchy-Schwarz inequality we have It follows that x, y x, y x y x, for all x, y X with y. sup{ x, y : y X with y } x for all x X. It remains to show the inequality sup{ x, y : y X with y } x for all x X. This clearly holds when x, since, y for all y X. Now suppose x. Let y and notice that y. We have x x The inequality follows. x, y x, x x x, x x x x 2 x. November 2, 26 Page of 7
2 We will now show that the norm of T can be expressed in terms of the innerproduct. We have T sup{ T x, x X with x } (Lemma 2., part c sup{ sup{ T x, y : y X, y } : x X, x } (equation ( sup{ T x, y : x, y X with x y } 2 We define on C[, ] the following innerproduct: f, g f(tg(tdt. Show that (C[, ],.,. is an innerproduct space, but that it is not complete with respect to the norm f 2 ( f(t 2 dt /2. Solution. We need to show that.,. defines an innerproduct on C[, ]. Linearity Let f, f 2, g C[, ] and let λ, λ 2 R. We have λ f + λ 2 f 2, g (λ f (x + λ 2 f 2 (x g(xdx ( λ f (xg(x + λ 2 f 2 (xg(x dx λ f (xg(xdx + λ 2 f 2 (xg(xdx λ f (xg(xdx + λ 2 f 2 (xg(xdx λ f, g + λ 2 f 2, g Symmetry Let f, g C[, ]. We have f, g f(xg(xdx f(xg(xdx g(xf(xdx g, f. Positive definiteness Let f C[, ]. We have f, f f(xf(x dx f(x 2 dx. November 2, 26 Page 2 of 7
3 The zero vector in C[, ] is the constant function x. We have, dx. Conversely, we have that f(x whenever f(x 2 dx. We will now show why (C[, ],.,. is not complete with respect to the induced norm. We will do this by constructing a Cauchy sequence {f n } n N in C[, ] that doesn t converge in C[, ]. We define 2, x < 2 n f n (x n(x 2, 2, 2 n x < 2 + n 2 + n x We show that this is a Cauchy sequence. Pick an ɛ >. Let N N be an integer greater than 2ɛ and let N n m N. We have f n f m f n f m, f n f m 2 2n 2N ɛ (f n (x f m (x 2 dx f n (x f m (x dx ( n m and conclude that {f n } is a Cauchy sequence. Now assume that f n f C[, ]. We see that f n converges point-wise to the function { f(x 2, x < 2 2, 2 x, but f is not continuous, contradicting the assumption that f C[, ]. Hence, C[, ] is not complete. 3 Let T be a bounded operator from a Hilbert space X to a Hilbert space Y. Define the adjoint T of T and show that it exists and is unique. Describe the domain and the codomain of T. Hint: In class we treated the case X Y in detail. Solution. The adjoint T is defined as the unique operator from Y to X such that T x, y x, T y for all x X, y Y. We show that the adjoint exists and is unique. The proofs follows closely the ones given in the lectures for the case X Y. November 2, 26 Page 3 of 7
4 Existence Fix y Y and let ϕ : X C be defined by ϕ(x T x, y. Then ϕ is linear and by Cauchy-Schwarz bounded: ϕ(x T x, y T x y T x y. Hence ϕ is a bounded linear functional on X and so by the Riesz representation theorem there exists a unique ξ X such that ϕ(x x, ξ for all x X. The vector ξ depends on the vector y Y. In order to keep track of this fact we set T y : ξ. Hence we have defined an operator T from Y to X based on the structure of bounded linear functionals on X. In summary, we have demonstrated the existence of an operator T with domain Y and codomain X such that T x, y x, T y for all x X, y Y. Uniqueness Suppose there exists another S B(Y, X such that T x, y x, Sy for all x X, y Y. Then we have x, Sy x, T y for all y Y and by a well-known fact about innerproducts we deduce that T y Sy for all y Y. Hence T is unique. 4 Consider on (L 2 [, ],.,. the integral operator T k T k f(x for k C([, ] [, ]. a What is T? k(x, yf(ydy b Which kernels k give rise to unitary operators? When is T k a selfadjoint operator? Solution. a Let k C([, ] [, ]. We will show that T k defined as Tk f(x k(y, xf(y dy is the adjoint to T k. We need to show that T k f, g f, Tk g for all f, g L2 ([, ]. November 2, 26 Page 4 of 7
5 Let f, g L 2 [, ]. We have b When is T k unitary? T k f, g ( k(x, yf(y dy g(x dx k(x, yf(yg(x dy dx k(x, yf(yg(x dx dy ( f(y ( f(y f, T k g. k(x, yg(x dx dy k(x, yg(x dx dy By definition, T k is unitary if and only if T kt k T k T k I, or equivalently that T k T k f(x T k T kf(x f(x for all f L 2 [, ] and x [, ]. We have that ( T k Tk f(x k(x, z and T k T kf(x k(y, zf(y dy dz k(x, zk(y, zf(y dy dz ( k(z, x k(z, yf(y dy dz k(z, xk(z, yf(y dy dz. By inserting these two integrals into the equation above, we get the conclusion that Tk is unitary if and only if k(x, zk(y, zf(y dy dz for all f L 2 [, ] and x [, ]. When is T k selfadjoint? k(z, xk(z, yf(y dy dz f(x By definition, T k is selfadjoint if and only if T k T k, or equivalently that T k f(x T kf(x for all f L 2 [, ] and x [, ]. By inserting the definitions of Tk and T k in the equation above we get the condition that k(x, yf(y dy k(y, xf(y dy November 2, 26 Page 5 of 7
6 for all f L 2 [, ] and x [, ]. This is equivalent to saying that k(x, y k(y, x for all x, y [, ]. 5 Let M be a closed subspace of a Hilbert space (X,, and consider X M M. Show that the projection onto M is selfadjoint. Solution. Let P be the projection of X onto M. We need to show that P is selfadjoint, meaning that P x, y x, P y for all x, y X. Let x, y X. By the Projection Theorem, we can write x x + x 2 and y y + y 2, where x, y M and x 2, y 2 M. We have and P x, y P (x + x 2, y + y 2 P x, y + y 2 + P x 2, y + y 2 P x, y + y 2 +, y + y 2 P x, y + y 2 x, y + y 2 x, y + x, y 2 x, y x, P y x + x 2, P (y + y 2 x + x 2, P y + x + x 2, P y 2 x + x 2, P y + x + x 2, x + x 2, P y x + x 2, y x, y + x 2, y x, y and hence P x, y x, P y, which was what we needed to show. 6 Let M be the subspace of l 2 that consists of all sequences with finitely many non-zero terms. Show that best approximation fails for M. Why does this not contradict the best approximation theorem from class? Hint: For x (, /2, /3,... show that inf{ x m : m M} and that there exists no m M that achieves this infimum. November 2, 26 Page 6 of 7
7 Solution. Using the hint, we let x (, /2, /3,... We start by showing that inf{ x m : m M}. Since the norm is always non-negative, we have that inf{ x m : m M}. We will show equality by constructing a sequence {m n } n N in M such that x m n. Let m n (, /2, /3,..., /n,,,... We have lim x m n lim (,,..., /(n +, /(n + 2, /(n + 3,... n n lim n k 2 lim n k kn+ ( k k 2 lim n x x n k 2 k n k k 2 k 2 which was what we needed to prove. Now, suppose there exist a sequence m M such that x m. Then we have m x, but x is not in M since it has no non-zero terms. This does not contradict the best approximation theorem because the conditions for the theorem are not satisfied. Indeed, the subspace M is not closed. To see this, observe that that the sequence {m n } n N converges to x, but x is not in M. November 2, 26 Page 7 of 7
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