By Lokenath Debnath-
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Additional resources for Hilbert Spaces with Applications
31. Compact sets are closed and bounded. Proof: Let S be a compact subset of a normed space E. Suppose x1 , x2 , . . ∈ S and xn → x. Then (xn ) contains a subsequence (xpn ), which converges to some y ∈ S. On the other hand, we have xpn → x. Thus, x = y and x ∈ S. This shows that S is closed. 18 Chapter 1 Normed Vector Spaces Suppose now that S is not bounded. Then there exists a sequence x1 , x2 , . . ∈ S such that xn ≥ n for all n ∈ N. Clearly, (xn ) does not contain a convergent subsequence, and hence S is not compact.
The following conditions are equivalent: (a) (xn ) is a Cauchy sequence; (b) xpn − xqn → 0 as n → ∞, for every pair of increasing sequences of positive integers (pn ) and (qn ); 20 Chapter 1 (c) Normed Vector Spaces xpn+1 − xpn → 0 as n → ∞, for every increasing sequence of positive integers (pn ). The proof is left as an exercise. Observe that every convergent sequence is a Cauchy sequence. In fact, if xn − x → 0, then xpn − xqn ≤ xpn − x + xqn − x → 0, for every pair of increasing sequences of indices (pn ) and (qn ).
Prove that f attains the minimum and the maximum values on S. 38. Is the sequence of functions fn (x) = C ([0, 1]) ? nx 1+nx2 a Cauchy sequence in 39. 2. 40. Show that the sum of two Cauchy sequences is a Cauchy sequence. 41. Prove completeness of lp for arbitrary p ≥ 1. 42. Give an example of an incomplete normed space. 43. Show that, in a Banach space, every sequence (xn ) convergent to zero contains a subsequence (xpn ) such that the series ∞ n=1 xpn converges. 44. Consider the space C ([a, b]) with the norm defined as f = Is this a Banach space ?
Hilbert Spaces with Applications by Lokenath Debnath-