Q:

show $ \sum^{\infty}_{-\infty} | c_n | < \infty \implies F_N f \to f \text{ uniformly}$

I need to show that :

$$F_n f = \sum_N^N c_n e^{inx}$$ with $$c_n = \frac 1 {2 \pi }
 \int^{\pi}_{- \pi} f(x) e^{-inx} dx $$
$f \in C^0 ( \mathbb R )$, $2 \pi$ periodic
$$ \sum^{\infty}_{-\infty} | c_n | < \infty  \implies  F_N f \to f
 \text{ uniformly}$$

how to do that ?

 I don't understand why in the correction, they are computing the
 difference between $F_N$ and $F_M$. Are we trying to show that the
 serie is Cauchy ? If it is, then what ? In the correction they are just saying that then $F_N f$ converges uniformly to a continuous function g define as the sum.
 Moreover, at the end, they are calling $$g(x) = \sum^{\infty}_{-\infty} | c_n |$$ in order to conclude that $f=g$ a.e.. Why do you need to do
 that ? They then say that since g and f has the same Fourier coefficients, they are equal.

Thanks a bunch. 

A:

How you prove this depends on what you already know. It  should be trivial  to verify that there is some function $g$ with $F_Nf\to g$ uniformly, and that it follows that $\hat g(k) = \hat f(k)$ for every $k$. Now if you've already shown that $\hat g = \hat f$ implies $g=f$ almost everywhere you're done - if you don't already know that then that's the not-entirely-trivial part.

