On maximal rearrangement inequalities for the Fourier transform

Abstract:

Suppose that $w$ is a measurable function on ${{\mathbf {R}}^n}$ and denote by $W = {w^ \ast }$ the decreasing rearrangement of $\left | w \right |$ (provided that it exists). We show that the $n$-dimensional Fourier transform $\hat f$ satisfies (1) \[ {\left \| {w\hat f} \right \|_q} \leqslant {\left \| {W{{(\hat f)}^ \ast }} \right \|_q} \leqslant C\left \| {W(t)\int _0^{1/t} {{f^ \ast }} } \right \|\quad (C\ {\text {absolute constant}}),\] if $1 < q < \infty$ and ${t^{2/q - 1}}W(t) \searrow$ for $t > 0$. We also show that (2) \[ {\left \| {w\hat f} \right \|_q} \geqslant {c_{n,q}}{\left \| {w(x)\int _{\left | v \right | \leqslant 1/\left | x \right |} {f(y)} dy} \right \|_q}\quad (f\ {\text {nonnegative),}}\] if $1 < q < \infty$ and $w$ is nonnegative and symmetrically decreasing. Inequality (2) implies that (1) is maximal in the sense that the left side reaches the right side if $f$ is nonnegative and symmetrically decreasing. Hence, (1) implies all other possible estimates in terms of $W$ and ${f^ \ast }$. The cases $q \ne 2$ of (1) can be derived from the case $q = 2$ (and same $f$) by a convexity principle which does not involve interpolation. The analogue of (1) for Fourier series is due to H. L. Montgomery if $q \geqslant 2$ (then the extra condition on $W$ is automatically satisfied).