let $f \in L^2(\mathbb{T})$ whose Fourier coefficients $c_k=0$ except for multiples of $3$. The Fourier series is given by $f(x)=\sum_{k\in \mathbb{Z}}c_ke^{ikx}$

I know that for every $f \in L^2(\mathbb{T})$, it's associated Fourier series converges almost everywhere

Question: Does this mean that the Fourier series is zero almost everywhere? If so, does that imply $f(x)=0$ almost everywhere? I'm struggling a bit with the intuition. Can someone provide a proof of this, or an appropriate counter-example?

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    $\begingroup$ Why would it be zero? Consider the function $f(x)=\exp(3ix)$. $\endgroup$ – Mariano Suárez-Álvarez Mar 5 '17 at 5:21
  • $\begingroup$ the Fourier series would necessarily have to be then $exp(3ix)=3exp(3ix)+6exp(6ix)+...+3kexp(kix)$. I am not sure how this helps. Plus, if the Fourier coefficients are all zero, then $f(x)=0$, so I am trying to see if the same holds for the almost everywhere case. $\endgroup$ – Kernel_Dirichlet Mar 5 '17 at 5:36
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    $\begingroup$ The Fourier series of $\exp(3ix)$ is just $\exp(3ix)$: all the Fourier coefficients vanish except for $c_3$. (This is an even more extreme example than you asked about, so it illustrates an even stronger point.) What does this fact tell you about whether the vanishing of lots of Fourier coefficients implies that the Fourier series converges to a function that is zero a.e.? $\endgroup$ – symplectomorphic Mar 5 '17 at 5:56
  • $\begingroup$ now I see it, the coefficients $exp(3ix)$ vanishes everywhere except for $c_3$. So even if only one coefficient is non-zero, the function $f(x)$ can fail to be zero everywhere. Thanks! $\endgroup$ – Kernel_Dirichlet Mar 5 '17 at 13:54
  • $\begingroup$ Here's another phrasing of your question. "Let $v$ be a vector whose entries are zero except for every third coordinate. Does this mean that the vector is equal to zero?" :) $\endgroup$ – Neal Mar 5 '17 at 17:35

Here is another argument : the Fourier transform is an isometry : $L^2(T) \to L^2(T)$. In particular, $f = 0$ a.e iff $\hat f = 0$ a.e. So if your function is non-zero, its Fourier transform will be non-zero.

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  • $\begingroup$ But the q assumes that the fourier coefficients are zero, only. $\endgroup$ – Behnam Esmayli Mar 5 '17 at 17:29
  • $\begingroup$ The OP assumes that some coefficients are non-zero. If at least one coefficient is non-zero, then the function is non-zero. $\endgroup$ – user171326 Mar 5 '17 at 17:36

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