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Suppose we are given an arbitrary two-sided sequence of complex numbers, say ${a_n}$. After forming the formal Fourier series $\sum_{n=-\infty} ^{\infty} a_n e^{inx}=a_0 + \sum_{n=1} ^{\infty} (a_n e^{inx} + a_{-n} e^{-inx} )$, suppose the series converges to zero for all x in $\mathbb{R}$. Then can we say that every ${a_n}$ is actually zero? Here, the convergence is pointwise.

If there is a counterexample for the above question, can we impose some conditions on the prescribed sequence ${a_n}$ (e.g. it is in $l^2$) so that the abovementioned property holds?

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This a uniqueness result due to Cantor. You can find a proof in this article by J. Marshal Ash.

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Nice question. If $a_n\in \ell^2$, then letting $S_n(x)$ denote the partial Fourier sum we have that $S_n(x)$ converges in $L^2(\mathbb T)$ to some function $f$. Now, because of the $L^2$ convergence, there is a subsequence of $S_n$ that converges pointwise almost everywhere to $f$, and by the assumption, $f=0$. The Bessel inequality now allows you to conclude that $a_n=0$ for all $n$.

This argument works even if the convergence to zero at all points is relaxed to the convergence to zero at almost all points.

I don't know what happens if $a_n$ is not in $\ell^2$.

A trivial example: Taking $a_n\equiv 1$ produces a Fourier series that converges to $0$ at all $x\in (0, 2\pi)$, and that diverges to $+\infty$ at $x=0, x=2\pi$. This shows that, if you require only almost everywhere convergence, then the statement may fail.

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  • $\begingroup$ Thanks for your nice answer on the $l^2$ case! I was totally ignoring the relation between L^p convergence and a.e. convergence. $\endgroup$ – John Doe Aug 24 '18 at 15:29

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