FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...

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16 CHRISTOPHER HEIL The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space. Theorem 2.18. If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banach space. Proof. Assume that {An}n∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy in operator norm. For any given f ∈ X, we have Amf − Anf ≤ Am − An f, so we conclude that {Anf}n∈N is a Cauchy sequence of vectors in Y . Since Y is complete, this sequence must converge, say Afn → g ∈ Y . Define Af = g. This gives us a candidate limit operator A. Exercise: Show that A defined in this way is a linear operator. To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we must have C = sup An < ∞. If f ∈ X, then since Anf → Af we have Af = lim Anf ≤ sup n→∞ n∈N Anf ≤ sup An f = C f. n∈N Hence A is bounded, and A ≤ C. Finally, we must show that An → A in operator norm. Fix any ε > 0. Then there exists an N such that m, n > N =⇒ Am − An < ε 2 . Choose any f ∈ X with f = 1. Then since Amf → Af, there exists an m > N such that Af − Amf < ε 2 . Hence for any n > N we have Af −Anf ≤ Af −Amf+Amf −Anf ≤ Af −Amf+Am−An f < ε ε + = ε. Taking the supremum over all unit vectors, we conclude that A − An ≤ ε for all n > N. Thus An → A. Corollary 2.19. If X is a normed space, then its dual space X ′ = B(X, F) is a Banach space. The next exercise deals with the problem of extending an operator defined only a dense subspace to the entire space. Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed space X, and let Z be a Banach space. (a) Suppose that L: Y → Z is a bounded linear operator. Show that there exists a unique bounded linear operator ˜ L: X → Z whose restriction to Y is L. Prove that ˜ L = L. (b) Prove that if L: Y → range(L) is a topological isomorphism, then ˜ L: X → range(L) is also. 2 2
CHAPTER 3. BANACH SPACES 17 (c) Prove that if L: Y → range(L) is an isometry, ˜ L: X → range(L) is also. Solution (a) Fix any f ∈ X. Since Y is dense in X, there exist gn ∈ Y such that gn → f. Since L is bounded, we have Lgm − Lgn ≤ L gm − gn. But {gn}n∈N is Cauchy in X, so this implies that {Lgn}n∈N is Cauchy in Z. Since Z is a Banach space, we conclude that there exists an h ∈ Z such that Lgn → h. Define ˜ Lf = h. To see that ˜ L is well-defined, suppose that we also had g ′ n → f for some g ′ n ∈ Y . Then Lg ′ n−Lgn ≤ L g ′ n−gn → 0. Since Lgn → h, it follows that Lg ′ n = Lgn+(Lg ′ n−Lgn) → h + 0 = h. Thus ˜ L is well-defined. To see that ˜ L is linear, suppose that f, g ∈ X are given and c ∈ F. Then there exist fn, gn ∈ Y such that fn → f and gn → g. Since cfn + gn → cf + g and L(cfn + gn) = cLfn + Lgn → c ˜ Lf + ˜ Lg, by definition we have that ˜ L(cf + g) = c ˜ Lf + ˜ Lg. To see that ˜ L is an extension of L, suppose that g ∈ Y is fixed. If we set gn = g, then gn → g and Lgn → Lg, so by definition we have ˜ Lg = Lg. Hence the restriction of ˜ L to Y is L. Consequently, ˜ L = sup f∈X, f=1 ˜ Lf ≥ sup f∈Y, f=1 ˜ Lf = sup Lf = L. f∈Y, f=1 Finally, suppose that f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf, so ˜ Lf = lim n→∞ Lgn ≤ lim n→∞ L gn = L f. Hence ˜ L ≤ L. Combining this with the opposite inequality derived above, we conclude that ˜ L = L. (b) Suppose that L: Y → range(Y ) is a topological isomorphism. We already know that ˜L: X → range( ˜ L) is bounded. We need to show that ˜ L is injective, that ˜ L −1 : range( ˜ L) → X is bounded, and that range( ˜ L) = range(L). Fix any f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf. Since L is a topological isomorphism, we have by Exercise 2.16 that gn ≤ L −1 Lgn. Hence ˜ Lf = lim n→∞ Lgn ≥ lim n→∞ gn L −1 Consequently, ˜ L is injective and for any h ∈ range( ˜ L) we have = f L −1 . ˜ L −1 h ≤ L −1 ˜ L( ˜ L −1 h) = L −1 h. Therefore ˜ L −1 : range( ˜ L) → X is bounded. It remains only to show that the range of ˜ L is the closure of the range of L. If f ∈ X, then by definition there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf. Hence ˜ Lf ∈ range(L), so range( ˜ L) ⊂ range(L).
Page 1 and 2: FUNCTIONAL ANALYSIS LECTURE NOTES C
Page 3 and 4: CHAPTER 3. BANACH SPACES 3 Exercise
Page 5 and 6: CHAPTER 3. BANACH SPACES 5 Thus fn
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Page 21 and 22: CHAPTER 3. BANACH SPACES 21 Lemma 3
Page 23 and 24: CHAPTER 3. BANACH SPACES 23 Since t
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