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Let $f$ be a function on $\mathbb R^n$ whose Fourier transform $\hat f$ exists. Is there any relation between integrability of $\hat f$ and summability of the series $\sum_{n \in \mathbb Z^n} \hat f(n)$? Under what circumstances does one of these conditions (integrability or summability) imply the other? If the answer is not "always", what is an example of a function for which one condition holds, but the other does not?

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Somewhat related question:… – Jonas Meyer Jan 9 '12 at 6:47
up vote 2 down vote accepted

For example (with $n=1$) $$f(x) = \cases{{{\rm e}^{-\pi }}\cosh \left( x \right) & if $\left| x \right| <\pi$ \cr -{{\rm e}^{- \left| x \right| }}\sinh \left( \pi \right) &\text{otherwise}\cr}$$ which has $$\hat{f}(s) = \frac{2 s \sin(\pi s)}{1+s^2}$$ Thus $\hat{f}$ is not Lebesgue integrable, but $\hat{f}(n) = 0$ for all integers $n$.

EDIT: In the other direction, $$f(x) = \sum_{j=1}^\infty \frac{\exp(ijx - x^2/(4 j^4))}{2 \sqrt{\pi} j^2}$$ has Fourier transform $$\hat{f}(s) = \sum_{j=1}^\infty e^{-j^4 (s-j)^2}$$ which is integrable but not summable.

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