FIVE MAJOR RESULTS IN ANALYSIS AND TOPOLOGY Aaron ...

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4. THE HAHN-BANACH THEOREM 26 The proof will be completed in two parts. First, we prove a lemma which allows us to extend any linear functional defined on a linear subspace M of the normed linear space N in an additional single dimension, while preserving the norm of the functional. We then conclude with a brief reducio ad absurdum argument that is equivalent to the conclusion of the theorem. Lemma (Extension of Linear Functionals). Let N be a normed linear space, M a linear subspace of N, and f ∈ M ∗ . If y ∈ N − M, let M0 = M + [y] be the linear subspace of N spanned by M and y. Then f can be extended to a functional f0 ∈ M ∗ 0 so that f0 = f. Proof. First we prove the theorem in the case where N is a real normed linear space. We will use this result to establish the theorem for complex normed linear spaces. Suppose, without loss of generality, that f = 1. Since y /∈ M, every v ∈ M0 can be written uniquely as v = x + αy, where x ∈ M and α ∈ R. If f0 is a linear extension of f to M0, then f0(v) = f(x) + αf0(y). Hence, it we are done when we find a value for f0(y) that ensures that f0 = 1. This equivalent to the condition that (0.1) |f0(x + αy)| ≤ x + αy for every x ∈ M,α = 0. Since f0(x + αy) = f(x) + αf0(y), we use (0.1) to write (0.2) −f Now, we note that if x1,x2 ∈ M, then so that x − α x α + y ≤ f0(y) ≤ −f x + α x + y. α f(x2) − f(x1) ≤ fx2 − x1 = x2 − x1 ≤ x2 + y + x1 + y, (0.3) −f(x1) − x1 + y ≤ −f(x2) + x2 + y. Define a = sup {−f(x) − x + y} , and b = inf {−f(x) + x + y}. By (0.3), a ≤ b, so we x x need merely choose f0(y) ∈ [a,b] to complete the proof for real normed linear spaces. Now, suppose that N is a complex normed linear space. Let f be a complex functional on M with f = 1. Then f(x) = g(x) + ih(x), where g and h are real linear functionals on M. Since f = 1, g ≤ 1.
4. THE HAHN-BANACH THEOREM 27 Note that since f(ix) = if(x), f(ix) = g(ix) + ih(ix) and if(x) = ig(x) − h(x), h(x) = −g(ix) so that f(x) = g(x)−ig(ix). Since g is a real linear functional on M, we can extend g to a real functional g0 on M0 in such a way that g0 = g. Define f0(x) = g0(x)−ig0(ix) for all x ∈ M0. Certainly f0 is an extension of f from M to M0, and f0(αx+βy) = αf0(x)+βf0(y) for all α,β ∈ R. This last equality holds for complex α,β as well, since f0(ix) = g0(ix) − ig0(i 2 x) = i (g0(x) − ig0(ix)) = if0(x), so that f0 is a complex linear functional on the complex space M0. It remains to be shown that f0 = 1. Clearly, f0 ≥ f = 1. Let x ∈ M0 such that x = 1. If f0(x) ∈ R, then f0(x) = g0(x) and g0 ≤ 1, so that |f0(x)| ≤ 1. If f0(x) ∈ C, then f0(x) = reiθ , r > 0, so |f0(x)| = r = e−iθ −iθ −iθ f0(x) = f0 e x . Since e x = x = 1, the proof is complete. Definition. Let {fα} α∈J be a set of functions defined on subsets {Aα} α∈J of a set X, such that fα = fβ in Aα ∩ Aβ for each α,β ∈ J. The the union of the fα is a function f defined on Aα such that f agrees with fα on Aα for each α ∈ J. α∈J Theorem (Hahn-Banach). Let L be a normed linear space, and M a linear subspace of L. Then every bounded linear functional f on M has an extension F on L such that F = f. Proof. Let f ∈ M ∗ be bounded, and let G be the class of all complex functionals which extend f to a linear subspace of L containing M, where g ∈ G if, and only if, g = f. Define on G the relation ≤, where g1 ≤ g2 if g2 is an extension of g1. This relation is a partial ordering on the set of all linear extensions of f with the same norm as f. Since the union of every chain of functionals in G is a functional on L which agrees with each of the functionals in the chain on their domains, and the domain of each functional is bounded by L, then every chain in G is bounded above. By Zorn’s lemma, there exists a maximal extension F ∈ G of f. If the domain of F is not L, then we could extend F via our previous theorem, contradicting the fact that F was maximal.
Page 1: FIVE MAJOR RESULTS IN ANALYSIS AND
Page 4 and 5: Preface During the analysis of scie
Page 6 and 7: N = {1, 2, 3,...} R = {x| x is a re
Page 8 and 9: (1) x + y ∈ G, (2) (x + y) + z =
Page 10 and 11: TOPOLOGY 6 implies that f is contin
Page 12 and 13: A ⊂ β∈J ∗ TOPOLOGY 8 Uβ, t
Page 14 and 15: TOPOLOGY 10 ensures that f(x0) ∈
Page 16 and 17: 2. THE ASCOLI-ARZELÀ THEOREM 12 Th
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Page 20 and 21: that 1 −1 cn(1 − x 2 ) n dx =
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Page 34 and 35: CHAPTER 5 The Baire Category Theore
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