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Math Help - Show uniform convergence implies equicontinuity and uniform boundedness

  1. #1
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    Show uniform convergence implies equicontinuity and uniform boundedness

    Hi guys! I was wondering if you could look at my proof and tell me if you think its correct/rigorous enough. I'm having a little difficulty with this.

    The question is: Let (f_n) be a sequence of functions in C([0,1]), with f_n uniformly converging to f in [0,1]. Show without using the Heine-Borel theorem that the sequence is also uniformly bounded and equicontinuous in [0,1].

    Equicontinuity:
    By uniform convergence, the sequence of functions meets the Caunchy criterion, so there exists a natural number N such that when n,m>=N:

    |f_n(x)-f_m(x)|<E/3 (for every x in [0,1])

    By continuity of each f_k, there exists a d such that:

    |x-y|<d implies |f_k(x)-f_k(y)|<e/3 (for every x,y in [0,1])

    Combining these two facts, we have that for any e>0, when n,k>=N and |x-y|<d:

    |f_n(x)-f_n(y)| =< |f_n(x)-f_k(x)| + |f_k(x)-f_k(y)| + |f_k(y)-f_n(y)|
    |f_n(x)-f_n(y)| =< e/3 + e/3 + e/3
    |f_n(x)-f_n(y)| =< e (for every x,y in [0,1])

    Which is exactly the definition of equicontinuity. So, f_n is equicontinuous on [0,1].


    Uniform Boundedness:
    Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

    I don't think this last part works, but I'm not sure what else to do. Maybe something like:

    If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

    Any comments/helpful tips are much appreciated. =)
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by steph236 View Post
    H
    Uniform Boundedness:
    Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

    I don't think this last part works, but I'm not sure what else to do. Maybe something like:

    If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

    Any comments/helpful tips are much appreciated. =)
    You can prove this using the following theorem:

    Theorem: Let \{f_n\}_{n\in\mathbb{N}} be a sequence of bounded on [a,b] functions and f_n{\xrightarrow{\text{unif.}}}f with f bouned. Then, \{f_n\}_{n\in\mathbb{N}} is uniformly bounded.

    Proof: We know that f_n{\xrightarrow{\text{unif.}}f if and only if f_n\to f with the \|\cdot\|_{\infty} norm. So, choose N\in\mathbb{N} such that N\leqslant n\implies \|f_n-f\|_{\infty}<1. Then, let M=\max\{\|f_1\|_{\infty},\cdots,\|f_{N-1}\|_{\infty},\|f\|_{\infty}+1\}. Then, let f_k be arbitrary. If k<N then evidently \|f_k\|_{\infty}<M if k\geqslant N then \|f_k\|_{\infty}-\|f\|_{\infty}\leqslant \|f_k-f\|_{\infty}<1\implies \|f_k\|<\|f\|_{\infty}+1\leqslant M. Regardless, for every n\in\mathbb{N} we have that \|f_k\|_{\infty}\leqslant M. Or, that for every x\in[a,b] and every n\in\mathbb{N} we have that |f_n(x)|\leqslant M. \blacksquare

    Remark: The above proof is not necessarily the most general but will work for our purposes.

    Thus, noticing that the continuity of f_n and compactness of [0,1] guarantees that each f_n is bounded, and since the uniform limit of continuous functions is continuous we see that f is a continuous function on a compact space and thus also bounded. Thus, the theorem applies
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