Upper bound for the norm of inverse Fourier tansform

Question

Recall Hausdorff-Young inequality:

For any $f\in L^p(\mathbb{R}^n)$, we have $||\hat{f}||_q\le ||f||_p$, where $p$ and $q$ are conjugate exponents and $p\in[1,2]$.

It seems to me that it follows that $||f||_q\le ||\check{f}||_p$ by Fourier inversion. My question is:

Is there an inequality that we can use to bound the norm of $\check{f}$ by the norm of $f$?

Answer 1

Be more careful with the exponents of the $L^p$ spaces, in which $f$ and $\hat f$ belong.

If $f\in L^1$, then $\hat f\in L^\infty$, and $\|\hat f\|_\infty\le\|f\|_1$. For a dense subset of $L^2$, the $\hat f$ is definable as an element of $L^2$, and $\|f\|_2=\|\hat f\|_2$, and the Fourier transformation is hence definable using a density argument in the whole $L^2$. Hausdorff-Young is an interpolation of the two inequalities.

The inverse Fourier transform has almost identical definition, and thus satisfies the same inequalities, and it can only belong in $L^p$, $p\ge 2$. Note that the inversion, $\mathcal F^{-1}{\mathcal F}f=f$, holds in $L^2$.