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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?

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I think in this case the Fourier transform is not even defined. –  littleO Nov 1 '12 at 19:14
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The Fourier transform is defined, at least in the sense of tempered distributions, as long as $|a_k|$ is bounded by a polynomial. –  Robert Israel Nov 1 '12 at 19:46
    
Ah, I see, thanks Robert. –  littleO Nov 2 '12 at 0:57
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up vote 1 down vote accepted

What it says is there is no square-integrable function on $[0,2\pi]$ whose inverse Fourier transform is $a_k$.

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