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Math Help - Functional Analysis Proof: complete metric space

  1. #1
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    Functional Analysis Proof: complete metric space

    Define the space
    β1([a,b])
    as the space of functions
    f:[a,b]R
    which are everywhere differentiable and whose derivative
    f
    is a bounded function. One equips this space with the metric
    d(f,g)=sup|f(x)g(x)|+sup|f(x)g(x)|
    Prove that this turns
    β1([a,b])
    into a complete metric space.
    Please help. I know what a complete metric space is, but I'm having trouble getting started with this proof. I would prefer a hint that would point me in the right direction as I'd like the opportunity to work through the proof myself, but I'll be thankful for any help I receive.
    Thanks
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  2. #2
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    Re: Functional Analysis Proof: complete metric space

    I can't make things look pretty yet and I've spent way to long trying to put this question up
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  3. #3
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    Re: Functional Analysis Proof: complete metric space

    To show completeness in this metric, all Cauchy sequences converge to a point in your metric space. Take an arbitrary cauchy sequence and show that this is the case.

    EDIT: That is, show that the limit is a differentiable function from [a,b] to R with bounded derivative.
    Last edited by Gusbob; April 10th 2013 at 08:05 PM.
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    Re: Functional Analysis Proof: complete metric space

    So, let {fn} be a cauchy sequence in b. To show that b is complete in [a,b] we must show that given a Cauchy sequence fn of everywhere differentiable functions there exists h contained in b such that lim fn equals h with respect to d(f,g). We must prove this limit is a differentiable function from [a,b] and has a bounded derivative.

    How do I prove this limit is differentiable, except that it's contained in b, so it must be everywhere differentiable (same being true for bounded derivative)? Do I need to prove this limit is contained in b, or can I just assume that?
    Thanks for the help
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  5. #5
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    Re: Functional Analysis Proof: complete metric space

    This proof may be greatly simplified depending on how much analysis you know. The proof I gave is long, but should be able to be understood by most people with even a passing knowledge of analysis.

    I'll change notation a bit. Let C_b^1([a,b]}) be the space differentiable functions on [a,b] with bounded derivative. Also, denote C_b([a,b]) to be the space of continuous function on [a,b]. We make use of the fact that C_b([a,b]) is complete. I will also refer to the given norm as \|\cdot\|_{C^1_b}

    The space in your question is acutally C_b^1([a,b]}) since [a,b] is compact and every continuous function defined on a real valued function on a compact set is bounded.

    Now take a Cauchy sequence \{f_n\} in C_b^1([a,b]}). Then certainly \{f_n\} is Cauchy in C_b([a,b]}), so there is a continuous function f on [a,b] such that f_n \ to f uniformly. Now

    \|f'_m -f'_n\|_\infty \leq \|f_m - f_n\|_\infty + \|f'_m -f'_n\|_\infty = \|f_m -f_n\|_{C^1_b}

    which shows that \{f'_n\} is Cauchy with respect to the uniform norm. In particular, \{f'_n\} is Cauchy in C_b([a,b]). Using the completeness of C_b([a,b]), there is a g\in C_b([a,b]) such that f'_n \to g uniformly. It remains to show that g=f'.



    Now here comes the fun part. Fix any x\in [a,b] and \epsilon >0. THere is some N>0 such that \|f'_m-f'_n\|_\infty < \epsilon for m,n > N. Given any y\in [a,b] with y\not=x and suppose m,n > N. Applying the Mean value theorem to f_m-f_n shows that there is some point c between x and y such that

    (f_m-f_n)(y)-(f_m-f_n)(x)=(y-x)(f'_m-f'_n)(c)

    which is the same thing as saying

    \left| \frac{f_m(y)-f_m(x)}{y-x}-\frac{f_n(y)-f_n(x)}{y-x}    \right| = |f'_m(c)-f'_n(c)| \leq \|f'_m-f'_n\|_\infty < \epsilon

    Taking one of the variable to infinity, say m\to \infty, we obtain for n>N
    \left| \frac{f(y)-f(x)}{y-x}-\frac{f_n(y)-f_n(x)}{y-x}    \right|\leq \epsilon

    Using the definition of differentiability (of f_n), there is some \delta >0 such that
    |x-y|<\delta \implies \left| \frac{f(y)-f(x)}{y-x}-f'_n(x)} \right|
    Furthermore, the fact that f'_n \ to g uniformly implies that there is some \tilde{N} such that \|f'_n-g\|_\infty < \epsilon for n>\tilde{N}.



    Here is where I ask you to do some work. If you truly understood what I just said, and the implications thereof, it should be doable. Choosing n>N,\tilde{N} and |x-y|<\delta, show that
    \left| g(x)-\frac{f(y)-f(x)}{y-x}    \right|\leq 3\epsilon

    This would imply that
    g(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=f'(x)
    Last edited by Gusbob; April 12th 2013 at 03:19 AM.
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