Let $f \in L^1 (\mathbb R^n)$. Suppose that $f$ is continuous at zero and that the fourier transform $\hat f$ of $f$ is non-negative. Does this imply that $\hat f \in L^1$ (and hence, by the inversion theorem, that $f$ is continuous)? If so, how could I go about proving it?
A condition for $\hat f$ to be integrable
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fourier-analysis
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6http://math.stackexchange.com/questions/193114/a-positive-fourier-transform-is-integrable?rq=1 – 2015-04-30
2 Answers
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In A positive "Fourier transform" is integrable Zarrax gave a great answer to this question using mollifiers. He/she also provides a counterexample in the case $f$ is not continuous at $x=0$.
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I don't think that $\hat f \in L^1$, but I think that you can get easily that $f$ is continuous.
If I remember right, $\hat{f}$ non-negative implies that $f$ is positive definite, and then Krein's inequality takes care of continuity...