Problems and Solutions in - Mathematics - University of Hawaii

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1.5 2001 November 26 1 REAL ANALYSIS (i) If 1 and 1 1 p + q = 1, then fg1 = fpgq. Thus, if f ∈ Lp and g ∈ Lq , then fg ∈ L1 (ii) If p = ∞ and if f ∈ L ∞ and g ∈ L 1 , then |fg| ≤ f∞|g|, so fg1 ≤ f∞g1. (d) (Riesz representation theorem for L p ) 25 Suppose 1 and 1 p + 1 q = 1. If Λ is a linear functional on Lp , then there is a unique g ∈ L q such that Λf = fg dµ (∀f ∈ L p ). (e) (Hahn-Banach theorem) Suppose X is a normed linear space, Y ⊆ X is a subspace, and T : Y → R is a bounded linear functional. Then there exists a bounded linear functional ¯ T : X → R such that ¯ T (y) = T (y) for all y ∈ Y , and such that ¯ T X = T Y , where ¯ T X and T Y are the usual operator norms, ¯ T X = sup{| ¯ T x| : x ∈ X, x ≤ 1} and T Y = sup{|T x| : x ∈ Y, x ≤ 1}. 4. (a) State the Baire category theorem. (b) Prove the following special case of the uniform boundedness theorem: Let X be a (nonempty) complete metric space and let F ⊆ C(X). Suppose that for each x ∈ X there is a nonnegative constant Mx such that |f(x)| ≤ Mx for all f ∈ F. Prove that there is a nonempty open set G ⊆ X and a constant M > 0 such that |f(x)| ≤ M holds for all x ∈ G and for all f ∈ F . Solution: 26 (a) (Baire category theorem) If X is a complete metric space and {An} is a collection of open dense subsets, then ∞ n=1 An is dense in X. Corollary 1. If X is a complete metric space and G ⊆ X is a non-empty open subset and G = ∞ o = ∅ for at least one n ∈ N. ¯ Gn Corollary 2. A nonempty complete metric space is not a countable union of nowhere dense sets. n=1 Gn then (b) Define Am = {x ∈ X : |f(x)| ≤ m, ∀f ∈ F }. Then X = ∞ m=1 Am, since for every x there is a finite number Mx such that |f(x)| ≤ Mx for all f ∈ F . Now note that Am = f∈F {x ∈ X : |f(x)| ≤ m}, and, since f and a ↦→ |a| are continuous functions, each {x ∈ X : |f(x)| ≤ m} is closed, so Am is closed. Therefore, corollary 2 of the Baire category theorem implies that there must be some m ∈ N such that A◦ m = ∅, so the set G = A◦ m and the number M = m satisfy the given criteria. ⊓⊔ 25 Note to self: add case p = ∞ 26 See Royden [6], § 7.8, for an excellent treatment of this topic. Part (b) of this problem appears there as theorem 32, and another popular exam question is part c of problem 37. 22
1.5 2001 November 26 1 REAL ANALYSIS 5. Prove or disprove: (a) L 2 convergence implies pointwise convergence. (b) ∞ sin(x lim n→∞ 0 n ) xn dx = 0. (c) Let {fn} be a sequence of measurable functions defined on [0, ∞). If fn → 0 uniformly on [0, ∞), as n → ∞, then lim fn(x) dx = lim fn(x) dx. Solution: [0,∞) [0,∞) (a) This is false, as the following example demonstrates: For each k ∈ N, define fk,j = χ [ j−1 k and let {gn} be the sequence defined by g1 = f1,1, g2 = f2,1, g3 = f2,2, g4 = f3,1, g5 = f3,2, g6 = f3,3, g7 = f4,1, . . . j , k ) for j = 1, . . . , k, Then |fk,j| 2 dµ = 1/k for each j = 1, . . . , k, so fk,j2 = 1/ √ k → 0, as k → ∞. Therefore gn2 → 0 as n → ∞. However, {gn} does not converge pointwise since, for every x ∈ [0, 1] and every N ∈ N, we can always find some k ∈ N and j ∈ {1, . . . , k} such that gn(x) = fk,j(x) = 1 with n ≥ N, and we can also find a k ′ ∈ N and j ′ ∈ {1, . . . , k ′ } such that gn ′(x) = fk ′ ,j ′(x) = 0 with n′ ≥ N. ⊓⊔ sin t t (b) For any fixed 0 < x < 1, limn→∞ xn = 0. Also, Together, these two facts yield sin x lim n→∞ n xn = 1. → 1, as t → 0, which can be proved by L’Hopital’s rule. Now, recall that | sin θ| ≤ |θ| for all real θ. Indeed, since sin θ = θ cos x dx, we have, for θ ≥ 0, 0 and, for θ < 0, In particular, for any 0 < x < 1, | sin xn | |xn | function sin xn x n , to obtain | sin θ| ≤ θ 0 | cos x| dx ≤ θ 0 1 dx = θ, | sin θ| = | sin(−θ)| ≤ | − θ| = |θ|. ≤ 1. Therefore, we can apply the dominated convergence theorem to the 1 lim n→∞ 0 sin xn dx = xn 1 0 1 dx = 1. (9) Next consider the part of the integral over 1 ≤ x < N, for any real N > 1. Fix n ≥ 2. The change of variables u = xn results in du = nxn−1dx, and, since u1/n = x, we have xn−1 1 1− = u n . Therefore, N 1 sin xn n N sin u dx = xn 1 u 23 du 1 nu1− n = 1 n N n 1 sin u 1 u2− n du.
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A.1 Real Analysis A MISCELLANEOUS D
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A.1 Real Analysis A MISCELLANEOUS D
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A.2 Complex Analysis A MISCELLANEOU
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B List of Symbols F an arbitrary fi
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INDEX INDEX Riemann mapping theorem
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Problems and Solutions in - Mathematics - University of Hawaii

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