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How to prove that if Fourier series of function $f$ converge uniformly, then function is continuous?

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If you mean that the Fourier series converges uniformly to $f$, then this simply follows from the fact that a uniform limit of continuous functions is continuous. – M Turgeon May 23 '12 at 19:24
I'm sorry, Fourier series, not coefficient. If for $f(t)$ exists Fourier series that converge uniformly $\sum_{k\in \mathbb{Z} }{c_k e^{(-2i\pi kt)}}$ on $[0,1]$ then $f(t)$ is continious. – Aram Malkhasyan May 23 '12 at 19:35
@M Turgeon yes i know this theorem, but is this theorem true under $\mathbb{C}$ - functions? – Aram Malkhasyan May 23 '12 at 19:39
@user31919 see my answer below – M Turgeon May 23 '12 at 19:41

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).

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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.

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