Applied Analysis (APPM 5440): Final exam 1:30pm 4:00pm, Dec. 14, Closed books.

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1 Applied Analysis APPM 44: Final exam 1:3pm 4:pm, Dec. 14, 29. Closed books. Problem 1: 2p Set I = [, 1]. Prove that there is a continuous function u on I such that 1 ux 1 x sin ut 2 dt = cosx, x I. Define the operator T via [T u]x = 1 x sin ut 2 dt + cosx. Clearly T u is continuous whenever u is continuous so T maps CI to CI. Now 1 takes the form 2 u = T u. Define Ω = {u CI : u M} where the norm is the uniform norm and M is a real number to be determined. Ω is a closed subset of the complete space CI, so if we can determine M so that T maps Ω to Ω, and T is a contraction on Ω, then the contraction mapping theorem will apply. First we observe that sin α α and sin α sin β α β for any real numbers α and β. Ensure that T maps Ω to Ω: Suppose that u Ω. Then T u = sup 1 x sinut 2 dt + cosx 1 x x I sup sinut 2 dt + cosx 1 We see that T u Ω if M 6/. x I 1 dt+1 = 6 Ensure that T is a contraction: If u, v Ω, then T u T v = sup 1 x sin ut 2 sin vt 2 dt 1 sin ut 2 sin vt 2 dt x I 1 1 ut 2 vt 2 dt = 1 ut vt ut + vt dt 1 sup x I We see that T is a contraction if M < /2. ut vt ut + vt dt u v M + Mdt = 2M u v. Conclusion: Picking, say, M = 2, we see that T becomes a contraction on the complete metric space Ω. Consequently, 2 has a unique solution.

2 Problem 2: 2p Let X be a set. a 6p State the definitions of a metric on X and a topology on X. b 4p Given a metric d on X, define the topology T induced by d. c 8p Prove that the T that you defined in b satisfies the axioms of a topology. d 7p Set Y = R 2, and define on Y Y the function bx, y = b[x 1, x 2 ], [y 1, y 2 ] = x 1 y 1 1/2 + x 2 y 2 1/2 2. Is Y, b a metric space? Motivate. a See text book. b A subset G of X belongs to T if and only if the following condition holds: For any x G, there is an ε > such that B ε x G, where B ε x = {y X : dx, y < ε}. c and X obviously satisfy the criterion in b. Let {G α } α A be an arbitrary collection of sets in T. Set G = α A G α. We need to show that G T. Suppose x G. Then x G β for some β A. Since G β T, there is a ε > such that B ε x G β. But then B ε x G β α A G α = G. n Let {G j } n j=1 be a finite collection of sets in T. Set G = G j. We need to show that G T. Pick x G. Then x G j for every j, so for each j we can pick an ε j > such that B εj x G j. Set ε = min ε j. Since the min is over a finite number of elements, ε >. Moreover, B ε x G j for 1 j n every j, so B ε x G. d No, the given b does not satisfy the triangle inequality. Consider the points We have and j=1 x = [1, ], y = [, 1], z = [1, 1]. bx, y = 1 1/ /2 2 = 4, bx, z + bz, y = 1/ / / /2 2 = 2.

3 Problem 3: 1p a p State the Hahn-Banach theorem. b p Define what it means for a sequence x n n=1 in a Banach space X to converge weakly. a Let M be a linear subspace of a NLS X, and let ψ be a bounded linear functional on M. Then there exists a bounded linear functional φ on X such that and φx = ψx, x M, φ X = ψ M. Alternative formulation: Every bounded linear functional of a subspace of a NLS can be extended to the entire space without enlarging the norm. b A sequence x n n=1 converges weakly if there exists an element x X such that φx n φx for every φ X.

4 Problem 4: 2p Let H be a Hilbert space with a closed convex subset M. a 13p Suppose that x H and that x / M. Prove that there exists a unique z M such that x z = inf x y. y M b 12p Now consider the particular case of H = L 2 I where I = [, 1]. The set H is equipped with the usual inner product f, g = ft gt dt. Let M denote the linear space of polynomials of degree two or less, and set ft = t 3. Evaluate d = distm, f = inf f g. g M a See lecture notes for Chapter 6. Observe that M is only a subset of H, not a subspace. b The unique minimizer z assured in part a takes the form zt = a + b t + c t 2 where a, b, and c are some numbers to determined. We know that f z M. Since the functions u 1 t = 1, u 2 t = t, and u 3 t = t 2 form a basis for M, the condition that f z M is satisfied iff Solving the linear system we find that = f z, u 1 = = f z, u 2 = = f z, u 3 = t 3 a b t c t 2 dt = 1 4 a 1 2 b 1 3 c, t 4 a b t 2 c t 3 dt = a 1 3 b 1 4 c, t a b t 3 c t 4 dt = a 1 4 b 1 c. 1 1/2 1/3 1/2 1/3 1/4 1/3 1/4 1/ a b c = 1/4 1/ 1/6 a = 1 2, b = 3, c = 3 2. Now the laborious part is to evaluate the norm of the residual. After some work, we find distm, f = z = t t 3 1/2 2 t2 2 dt = = 1 = Alternate solution: It is possible to use Gram-Schmidt to orthonormalize {u 1, u 2, u 3 } to form an orthonormal set {v 1, v 2, v 3 }; this is a bit of work, but results in the functions v 1 t = 1, v 2 t = 2 3t 1/2, v 3 t = 6 t 2 t + 1/6.

5 Then z = v 1, f v 1 + v 2, f v 2 + v 3 f v 3, and since f z M, Pythagoras theorem yields f z 2 = f 2 z 2 = 1 7 v 1, f 2 v 2, f 2 v 2, f 2 = = Problem : 2p Let X, d be a compact metric space. Let C b X denote the set of all bounded real-valued continuous functions on X, equipped with the uniform norm, Prove that C b X is complete. f u = sup fx. The assumption that X is compact is a red herring this property is not required for the statement to be true. Let f n n=1 be a Cauchy sequence in X, d. We will construct a limit function, and then prove that it is bounded, that it is indeed the limit of the sequence in the uniform norm, and finally that it is continuous. Step 1 construct the limit point f: Fix x X. Since f n x f m x f n f m and f n n=1 is Cauchy, the sequence f n x n=1 is Cauchy in R. Since R is complete, the sequence is convergent, we can therefore define a function f via fx = lim n f nx. Step 2 prove that f is bounded: We have sup fx = sup lim f nx lim inf n n sup f n x = lim inf f n <, n where in the last step we used that f n is Cauchy, and therefore bounded. Step 3 prove that f n f uniformly: Fix ε >. Pick N such that f m f n < ε/2 when m, n N. Then for n N, we have f n f = sup f n x fx = sup lim f nx f m x m lim inf sup f n x f m x = lim inf f n f m ε/2 < ε. m m Step 4 prove that f is continuous: This follows directly from the fact that each f n is continuous and f n f uniformly since uniform convergence preserves continuity. Steps 2 and 4 prove that f C b X, and step 3 proves that f is the limit point of f n. The proof is therefore complete.