Question on Parseval's theorem

Question

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?

I think in this case the Fourier transform is not even defined. — littleO, Nov 01 '12 at 19:14
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 01 '12 at 19:46

score 1 · Accepted Answer · answered Nov 01 '12 at 19:47

1

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

answered Nov 01 '12 at 19:47

Robert Israel

1 Answers1