# Proof Fourier Series Converge

1. Aug 19, 2013

### Swimmingly!

Hey. I'm looking for a proof of:
Theorem: If $f \in C^1(\mathbb{T})$, then the Fourier series converges to f uniformly (and hence also pointwise.)

I have looked around for it, googled, etc, but I only found proofs which used theorem they did not prove. (Or I misunderstood what they said.)
I'd really like to truly understand why they converge, be it uniformly or pointwise. If anyone could either link me to a proof or do it, it'd be great. Thanks.

2. Aug 19, 2013

### micromass

Staff Emeritus
Check Folland's "Fourier analysis and applications", Theorem 2.5

3. Aug 28, 2013

### brmath

Sorry, I don't know what C^1(T) is. Are these complex functions? And what domain is T?

4. Sep 8, 2013

### Swimmingly!

Completely answered my question. Thanks a lot!

C^k is the set of functions such that: There exist continuous derivatives of 0th, 1st, 2nd... and kth order.
C^1(T) probably means that f is periodic of period 2π or something of the sorts. The number of senseful meanings is not that big.

5. Sep 8, 2013

### brmath

Thanks Swimmingly. Then$$C^1$$ would be real functions that have just one continuous derivative and you guess T means they are periodic. I would guess if they are not periodic one could construct examples where the Fourier series wouldn't converge at all. There is always the famous $${x^2sin1/x}$$ which has exactly one continuous derivative at 0.

Last edited: Sep 8, 2013
6. Sep 8, 2013

### Office_Shredder

Staff Emeritus
Bold faced T usually refers to a torus - in this case I assume the one dimensional torus (which is the circle, equivalently we are discussing periodic functions on the real line)

7. Sep 8, 2013

### brmath

Office_Shredder- Thanks for the clarification. He most likely meant periodic functions on the real line. I personally am more than willing to consider functions on the unit circle -- quite often it helps.

Also, for all, please substitute $$x^3sin1/x$$ for $$x^2sin1/x$$ as per my previous post. The first does have one continuous derivative at x = 0. The second has only a discontinuous derivative at x = 0.