Let $w$ be a rapidly decaying function on $\mathbb{R}$ such that $$ \sum_{n \in \mathbb{Z}} w(x+n) = 0$$ for all $x \in \mathbb{R}$. Does that imply that $w$ is identically zero? What if we assume that $w$ is continuous?
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asked Oct 15, 2018 at 6:34
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$\begingroup$ The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $\lim_{m\to\infty}\sum_{-m}^m$? $\endgroup$– Gerry MyersonCommented Oct 15, 2018 at 6:42
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$\begingroup$ Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value. $\endgroup$– Matthias LudewigCommented Oct 15, 2018 at 8:53
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3$\begingroup$ Indeed if you assume say that $w$ is also smooth this periodization equals $\sum \hat w(n) e(n x)$, so the condition is equivalent to $\hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them). $\endgroup$– RodrigoCommented Oct 15, 2018 at 10:24
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1 Answer
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No, the Haar wavelet is a counter example (i.e. $w(x) = \chi_{[0,1[}(x) - \chi_{[1,2[}(x)$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.
