Proof Involving Convergence

  • #1

Main Question or Discussion Point

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]
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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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.
 
  • #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!
 
  • #4
Vid
401
0
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.
 
  • #5
Isn't it the other way around?...
 
  • #6
Vid
401
0
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).
 
  • #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]?
 
  • #8
Ok, yeah, please ignore what I wrote...I'm not thinking so clearly. Heh. Thanks Vid...
 

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