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Proof Involving Convergence

  1. May 2, 2008 #1
    I ran into this proof in one of my textbooks and was wondering if anybody could lead me in the right logical direction. I can prove the first-differentiable continuous case but the infinity case throws me off. Please help if you can!

    Thanks!

    Suppose [tex]\left\{f}\right\}\subset C_{\infty}\left(\left[a,b\right]\right)[/tex] such that [tex]\left{f\right}_{n}[/tex] converges uniformly to some [tex]\left{f\right}\in C_{\infty}\left(\left[a,b\right]\right)[/tex]. Prove that:

    [tex]\int^a_b\left{f\right}_{n}\left(x\right)dx \rightarrow \int^a_b\left{f\right}\left(x\right)dx[/tex]
     
  2. jcsd
  3. May 3, 2008 #2

    HallsofIvy

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    What do you mean "I can prove the first-differentiable continuous case"? If f and fn are [itex]C_\infty[/itex] they are certainly C1 so any proof for "first differentiable" works here.
     
  4. May 3, 2008 #3
    The problem statement is much simpler with the single-differentiable continuous case. It should read like this:

    Suppose [tex]\left\{f_{n}}\right\}\subset C\left(\left[a,b\right]\right)[/tex] such that [tex]\left{f\right}_{n}[/tex] converges uniformly to some [tex]\left{f\right}\in C\left(\left[a,b\right]\right)[/tex]. Prove that:

    [tex]\int^a_b\left{f\right}_{n}\left(x\right)dx\rightarrow\int^a_b\left{f\right}\left(x\right)dx[/tex]

    ~Thanks!
     
  5. May 3, 2008 #4

    Vid

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    C^(inf) is a subset of C^(1). So if you've proven it for C^(1), you've proven it for C^(n) for n>=1.
     
  6. May 3, 2008 #5
    Isn't it the other way around?...
     
  7. May 3, 2008 #6

    Vid

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    No, think of f(x) = |x|. It is obviously continuous so it is an element of C^(0), but it isn't differentiable everywhere so it's not an element of C^(1).
     
  8. May 3, 2008 #7
    Your illustration [tex]f\left(x\right)=\left|x\right|\in C_{0}\notin C_{1}[/tex] implies [tex]C_{0}\subset C_{1}[/tex], which is exactly what I'm trying to say...

    If [tex]C_{0}[/tex] were in [tex]C_{1}[/tex], then [tex]\left|x\right|\in C_{1}[/tex], but that can't be true. In the previous post before the last one you state [tex]C_{\infty}\subset C_{1}[/tex], but this contradicts what you last posted.

    In general, if any function [tex]f\left(x\right)\in C_{n}\left(\left[a,b\right]\right)[/tex], then [tex]f\left(x\right)\in C_{k}\left(\left[a,b\right]\right), \forall k\in\left[0,n\right]\in\mathbb{Z} \rightarrow C_{0}\subset C_{1}\subset C_{2}\subset... \subset C_{\infty}[/tex]. Therefore if a a function [tex]f\left(x\right)\in C_{\infty}\left(\left[a,b\right]\right)[/tex], then it is also a member of the set [tex]C_{0}\left(\left[a,b\right]\right)[/tex].

    So how does proving it for [tex]C_{1}[/tex] prove it for the [tex]C_{\infty}[/tex] case if [tex]C_{1}\subset C_{\infty}[/tex]?
     
  9. May 3, 2008 #8
    Ok, yeah, please ignore what I wrote...I'm not thinking so clearly. Heh. Thanks Vid...
     
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