If $f(x)$ is a function on the real line, and $\widehat {f}(\xi )$ is its Fourier transform, then
$$\displaystyle \int _{-\infty }^{\infty }|f(x)|^{2}\,dx=\int _{-\infty }^{\infty }|{\widehat {f}}(\xi )|^{2}\,d\xi $$
is the Plancherel theorem. As stated by Wikipedia:
"A more precise formulation is that if a function is in both $L_1( \mathbb R)$ and $L_2( \mathbb R)$, then its Fourier transform is in $L_2( \mathbb R)$, and the Fourier transform map is an isometry with respect to the L2 norm. This implies that the Fourier transform map restricted to $L_2( \mathbb R)\cap L_1( \mathbb R)$ has a unique extension to a linear isometric map $L_2( \mathbb R)$ $\rightarrow$ $L_2( \mathbb R)$, sometimes called the Plancherel transform."
I wonder whether there is a Plancherel identity also on $L_p$. I do not think so because the Fourier transform of functions in $L_p$ for $ p>2$ requires the study of distributions. Is there perhaps an inequality that generalizes that identity in $L_p$?
Best Answer
It is easy to see that the Fourier transform is not bounded in $L^1$. Just take something close to $\delta_0$ but in $L^1$, its Fourier transform should be something approaching a constant.
You can exploit this trick to see that the Fourier transform is not isometric in $L^p$. Try to write down how the $p$-norm changes with the dilation $f(x) \mapsto f_\lambda(x) = f(\lambda x)$ and then write down how the $L^p$-norm of the Fourier transform of $f_\lambda$ scales with lambda.