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If $$ \sum_{k=-\infty}^{\infty} |\hat f (k) | < \infty $$ does it implies $$ S_n(t)=\sum_{k=-n}^{n} \hat f (k) e^{ikt} \to f(t) \; ? $$

I know $S_n$ converges for each $t$ to some function $S$.
Can we say that $S$ is equal to $f$?
From what I read in our book (Katznelson), it's not clear for me.

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Well, it should be clear (by direct computation) that the Fourier series of $S$ is the same as that of $f$. So all Fourier coefficients of $S-f$ are zero. This in turns implies that $S-f=0$ almost everywhere. So, yes.

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