The relationship between Fourier coefficients of function $f$ and its continuity How to prove that if Fourier series of function $f$ converge uniformly, then function is continuous?
 A: In general, a uniform limit of continuous functions is continuous (this is proven in any introductory course in analysis, and it holds in any metric space). Now, if the Fourier series of $f$ converges uniformly to $f$, we can thus conclude that $f$ is continuous. However, in general, your Fourier series could converge uniformly without converging to $f$, and so your result, as stated, is false (for example, you can modify your function on a set of measure zero without changing your Fourier series).
A: Perhaps what you really look for is the following
Lemma Given a sequence $f_n$ of continuous complex valued functions defined on a compact set $E$, then $f_n$ converges uniformly if and only if $\operatorname{Re}f_n$ and $\operatorname{Im}f_n$ converges uniformly.
Proof To see this, just take the supremum in the following 
$$|\operatorname{Re}f_n(x) - \operatorname{Re}f(x)|^2\leq|f_n(x)-f(x)|^2=|\operatorname{Re}f_n(x) - \operatorname{Re}f(x)|^2+|\operatorname{Im}f_n(x) - \operatorname{Im}f(x)|^2$$
The imaginary part in the same.
Thus the complex valued version of the theorem mentioned by M Turgeon, follows from the real version and vice versa. 
