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Members of an equicontinuous set are trivially

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(members of an equicontinuous set are trivially continuous) must be continuous on [ a,b ]. Because x 0 is in the interior of [ a,b ], this implies that the original f : R R is continuous at x 0 . Quickie Proof. Let ² > 0 be given. There is δ > 0 such that | f n ( x ) - f n ( y ) | < ²/ 2 whenever | x - y | < δ . Let x,y R and assume | x - y | < δ . Then | f ( x ) - f ( y ) | = lim n →∞ | f n ( x ) - f n ( y ) | ≤ ²/ 2 < ². 2
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Note on the quickie proof, and its variants. A popular variant of the quickie proof was to use an ²/ 3 argument, writing | f ( x ) - f ( y ) | ≤ | f ( x ) - f n ( x ) | + | f n ( x ) - f n ( y ) | + | f n ( y ) - f n ( x ) | . If one uses it, to use it correctly, given ² > 0 first you select δ > 0 such that if x,y R , then | f n ( x ) - f n ( y ) | < ²/ 3 for all n N . Next you fix x,y such that | x - y | < δ . That is, x,y are arbitrary with | x - y | < δ , but will not be moved anymore. Now you can find N x ,N y such that n N x implies | f ( x ) - f n ( x ) | < ²/ 3 and n N y implies | f ( y ) - f n ( y ) | < ²/ 3. Then let n max( N x ,N y ) and conclude | f ( x ) - f ( y ) | < ² , with n having disappeared in the end result. Some additional notes: 1. I would also have accepted a much shorter proof without so much symbolism: By Arzela-Ascoli, on every closed and bounded interval the sequence { f n } has a subsequence converging uniformly, since the limit has to be f , this implies that f is continuous when restricted to bounded intervals, hence continuous. QED 2. Equicontinuity on a bounded interval plus uniform boundedness at one single point implies uniform bounded- ness. Suppose { f n } is a sequence of real valued functions on [ a,b ], -∞ < a b < . Assume { f n } is equicontinuous and there is c [ a,b ], M R , M 0 such that | f n ( c ) | ≤ M for all n . This happens, for example, if the sequence of real numbers { f n ( c ) } converges. We apply equicontinuity with any ² > 0, for example, with ² = 1. By equicontinuity there exists δ > 0 such that if x,y [ a,b ] and | x - y | < δ , then | f n ( x ) - f n ( y ) | < 1. Now every point x [ a,b ] is at distance at most b - a from c . So if x [ a,b ] we can find a sequence of at most m = [ b - a ]+1 points ([ b - a ] = integer part of b - a ) leading from x to c with any two consecutive points at distance at most 1 from each other. That is, we can find points x 1 ,...,x k where k m , x 1 = x , x k = c , and | x i - x i - 1 | < 1 for i = 2 ,...,k . Then, for all n N , | f n ( x ) | ≤ | f n ( x ) - f n ( x 2 ) | + | f n ( x 2 ) - f n ( x 3 ) | + ··· + | f n ( x k - 1 ) - f n ( c ) | + | f n ( c ) | ≤ ( k - 1) + M ( b - a ) + M proving the sequence is uniformly bounded. (b)
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members of an equicontinuous set are trivially continuous...

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