Convergence of Uniformly Convergent Functions: Proof for Infinity Case

[thebetapirate]

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

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

 

[HallsofIvy]

What do you mean "I can prove the first-differentiable continuous case"? If f and f_n are C_\infty they are certainly C₁ so any proof for "first differentiable" works here.

 

[thebetapirate]

The problem statement is much simpler with the single-differentiable continuous case. It should read like this:

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

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

~Thanks!

 

[Vid]

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.

 

[thebetapirate]

Isn't it the other way around?...

 

[Vid]

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).

 

[thebetapirate]

Your illustration f\left(x\right)=\left|x\right|\in C_{0}\notin C_{1} implies C_{0}\subset C_{1}, which is exactly what I'm trying to say...

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

In general, if any function f\left(x\right)\in C_{n}\left(\left[a,b\right]\right), then 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}. Therefore if a a function f\left(x\right)\in C_{\infty}\left(\left[a,b\right]\right), then it is also a member of the set C_{0}\left(\left[a,b\right]\right).

So how does proving it for C_{1} prove it for the C_{\infty} case if C_{1}\subset C_{\infty}?

 

[thebetapirate]

Ok, yeah, please ignore what I wrote...I'm not thinking so clearly. Heh. Thanks Vid...