FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...

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6 CHRISTOPHER HEIL Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and only if every absolutely convergent series in X converges in X. Proof. ⇒. This is Exercise 1.21. ⇐. Suppose that every absolutely convergent series is convergent. Let {fn}n∈N be a Cauchy sequence in X. By Exercise 1.18, there exists a subsequence {fnk }k∈N such that fnk+1 − fnk < 2−k for every k. The series (fnk+1 k − fnk ) is absolutely convergent, because ∞ − fnk ≤ ∞ 2 −k = 1 < ∞. fnk+1 k=1 k=1 By hypothesis, we conclude that (fnk+1 k − fnk ) converges in X, say to f. In terms of the partial sums, this says that fnk+1 − fn1 = k j=1 (fnj+1 − fnj ) → f as k → ∞, or fnk → g = f +fn1 as k → ∞. Thus {fn}n∈N is a Cauchy sequence which has a subsequence that converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X is complete. Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and absolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Let xn = 1 . Then n n xn = 1 n n is the harmonic series, and the partial sums sN = N n=1 this series are unbounded. Thus the harmonic series does not converge. Since sN → ∞, we usually say that 1 1 n diverges to infinity, and write n n = ∞. n On the other hand, consider the alternating series 1 n (−1)n . Since the terms alternate n signs and since 1 → 0, it follows that this series does converge to a finite scalar (in fact, it n converges to − ln 2). However, it does not converge absolutely. Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let {fn}n∈N be a sequence of elements of X. The series f = ∞ n=1 fn is said to converge unconditionally if every rearrangement of the series converges. That is, f = ∞ n=1 fn converges unconditionally if for each bijection σ : N → N the series ∞ converges. n=1 fσ(n) Remark 1.25. It is not obvious, but it can be shown that if ∞ n=1 fσ(n) is unconditionally convergent, then every rearrangement of the series must converge to the same sum, i.e., there is a single f such that f = ∞ n=1 fσ(n) for every permutation σ. 1 n of
CHAPTER 3. BANACH SPACES 7 Exercise 1.26. Let X be a Banach space. Prove that if a series f = ∞ n=1 fn converges absolutely, then it converges unconditionally. Remark 1.27. In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finitedimensional and a series f = ∞ n=1 fn converges unconditionally, then it converges absolutely. However, this fails in infinite dimensions. Example 1.28 (The Harmonic Series Revisited). To illustrate the importance of unconditional convergence, again consider X = F and the alternating series 1 n (−1)n . We know n that this series converges, but does not converge absolutely. Now consider what happens if we change the order of summation. Let pn = 1 2n and qn = 1 2n+1 , i.e., the pn are the positive terms from the alternative series and the qn are the absolute values of the negative terms. Each series n qn diverges. Hence there must exist an m1 > 0 such that p1 + · · · + pm1 > 1. Then, there must exist an m2 > m1 such that Continuing in this way, we see that n pn and p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 > 2. p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 − q2 + · · · is a rearrangement of 1 n (−1)n which diverges to +∞. n In the same way, we can construct a rearrangement which diverges to −∞, which converges to any given real number r, or which simply oscillates without ever converging. Moreover, the same can be done for any series of real scalars which converges conditionally. The following is an equivalent formulation of unconditional convergence. Proposition 1.29. Let X be a Banach space and let {fn}n∈N be a sequence of elements of X. Then the series f = ∞ n=1 fn converges unconditionally if and only if it converges with respect to the net of finite subsets of N, i.e., if ∀ ε > 0, ∃ finite F0 ⊆ N such that ∀ finite F ⊇ F0, f − < ε. Definition 1.30 (Topology). Let X be a normed linear space. (a) The open ball in X centered at x ∈ X with radius r > 0 is (b) A subset U ⊆ X is open if Br(x) = B(x, r) = {y ∈ X : x − y < r}. ∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U. (c) A subset F ⊆ X is closed if X \ F is open. n∈F fn
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