I started to read about Fourier analysis. I was just reading about uniqueness of the Fourier representation. Apparently if the Fourier series $\sum\limits_{n\in \mathbb{Z}}c_n e^{2\pi i n x}$ converges uniformly to $f(x)$ then the values of $c_n$ are forced to be the value $\hat{f}(n)=\int_0^1 f(x)e^{-2 \pi n x}dx$. This is stated like if it was obvious but I can't understand why. I don't see how the uniform convergence comes into play.
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2Without uniform convergence, there may be a set of measure zero, where the series does not converge to the function. – herb steinberg Aug 16 '20 at 18:15
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If it converges uniformly, then $\int_0^1 \left(\sum_{n\in\mathbb{Z}}c_n e^{2\pi inx} \right) e^{-2\pi imx}dx = c_m$ – Disintegrating By Parts Sep 15 '20 at 03:24
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The uniform convergence implies that $$\int_0^1 e^{-2i\pi kx}\sum\limits_{n\in \mathbb{Z}}c_n e^{2\pi i n x}dx=\sum\limits_{n\in \mathbb{Z}} \int_0^1 e^{-2i\pi kx}c_n e^{2\pi i n x}dx = c_k$$
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1Oh, uniform convergence is actually for swapping series with Riemann integral? – roi_saumon Aug 16 '20 at 23:03