As it is known that if 1 ≤ p < 2d/d+1, d ≥2, the Fourier transform of L^p($\mathbb{R^d}$) radial function is a continuous function away from origin. How do we prove that this range for p cannot be extended?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
For $p>\frac{2d}{d+1}$, the function $f(x)=\|x\|^{-d/2}J_{d/2}(\|x\|)$ is in $L^p(\mathbf R^d)$ by standard Bessel asymptotics, and its Fourier transform is the (obviously discontinuous) characteristic function of the unit ball.
answered Apr 13, 2017 at 5:36
-
-
$\begingroup$ @OwenKING Then this $f$ is barely not in $L^p$ ($\int_{\|x\|\leqslant r}|f|^p\simeq\log r$), so it's no longer a counterexample. $\endgroup$ Commented Apr 14, 2017 at 1:21