If $\sum_{k=-\infty}^{\infty}|a_k|^2$ is not finite, does Parseval's theorem say that the Fourier transform of $a_k$ is also not finite?
Question on Parseval's theorem
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fourier-analysis
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0I think in this case the Fourier transform is not even defined. – 2012-11-01
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1The Fourier transform is defined, at least in the sense of tempered distributions, as long as $|a_k|$ is bounded by a polynomial. – 2012-11-01
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0Ah, I see, thanks Robert. – 2012-11-02
1 Answers
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What it says is there is no square-integrable function on $[0,2\pi]$ whose inverse Fourier transform is $a_k$.