Can a function $f:G\to\mathbb{C}$ in $L^p,\ p>1, p\neq 2$ have a Fourier transform $F:\hat{G}\to\mathbb{C}$, where $\hat{G}$ is the Pontryagin dual space of $G$? I believe it can be shown that such a transform exists such that $F$ is in $L^q$, with $1/p+1/q=1$. However, does this not violate Parseval's identity, since $p\neq q\neq 2$?
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Parseval's theorem, or more generally the Plancherel theorem, only guarantees an isometry $L^2(G) \rightarrow L^2(\hat{G})$. In short, you are absolutely right. Take your favorite function and Fourier transform it, and you can easily see that the $L^p$ norm of one doesn't equal the $L^q$ norm of the other.
Christopher A. Wong
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Thanks. My confusion was that what if the $L^2$ norm doesn't exist for either $f$ or $F$? Would that be considered a violation? – f16 Oct 31 '12 at 19:45
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No, it would not. The Fourier transform is, in general, defined for any $L^1(G)$ function. – Christopher A. Wong Nov 01 '12 at 00:52