# Proof Involving Convergence

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

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!

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.

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

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).

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 $$C_{1}\subset C_{\infty}$$?

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