m425b-hw2soln-s08

m425b-hw2soln-s08 - MATH 425b ASSIGNMENT 2 SOLUTIONS SPRING...

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MATH 425b ASSIGNMENT 2 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (8) Since | c n I ( x - x n ) | ≤ | c n | and | c n | < , the series defining f converges uniformly by the Weierstrass M -test (Theorem 7.10.) Letting f n ( t ) = n k =1 c n I ( x - x n ), this means that f n f uniformly. Let x be a point that is not one of the x n ’s; then each f n is continuous at x , that is, lim t x f n ( t ) = f n ( x ). Since f n ( x ) f ( x ) as n → ∞ , Theorem 7.11 (applied with A n = f n ( x )) says lim t x f ( t ) = lim n →∞ f n ( x ) = f ( x ) , that is, f is continuous at x . (16) Suppose { f n } is equicontinuous on K , and f n f pointwise. Then { f n ( x ) } , being convergent, is bounded for each x , so { f n } is pointwise bounded. Suppose (to get a contradiction) that the convergence of f n to f is not uniform: || f n - f || 6→ 0. This means there exists ± > 0 and a subsequence { f n k } with || f n k - f || > ± for all k . But as a set, { f n k } is pointwise bounded and equicontinuous (since { f n } is), so { f n k } has a uniformly convergent subsequence by Theorem 7.25, say f m j g uniformly for some g , where { f m j } ⊂ { f
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m425b-hw2soln-s08 - MATH 425b ASSIGNMENT 2 SOLUTIONS SPRING...

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