1
$\begingroup$

As is well known the Fourier transform of a $L^1(\mathbb R)$ function is continuous, bounded and vanishes at infinity. It is also well-known that not every such function is a Fourier transform of some $L^1(\mathbb R)$. Can You provide me with an example (as explicit and as simple as possible) of a $L^1(\mathbb R)$ function, whose Fourier transform is continuous, bounded, vanishes at infinity but is not Lipschitz. Or this is impossible (?)

$\endgroup$
  • 1
    $\begingroup$ It is easy to see that if $x \cdot f(x)\in L^1$, then the Fourier transform is Lipschitz. Thus,a possible counterexample example has to violate this condition. But at the moment I seem to be too tired to calculate the Fourier transform of $x^{-1-\epsilon} \cdot 1_{(1,\infty)}$ for $0<\epsilon \leq 1$, which would be the first example violating the condition from above. $\endgroup$ – PhoemueX Oct 26 '15 at 22:09
1
$\begingroup$

Let $$ f(x)=\frac{J_1(x)}{x}\text{ if }x\ne0,\quad f(0)=\lim_{x\to0}\frac{J_1(x)}{x}=\frac12, $$ where $J_1$ is a Bessel function. Then $f\sim x^{-3/2}$ as $x\to\infty$, so that $f\in L^1(\mathbb{R})$. $$ \hat f(\xi)=\sqrt{\frac2\pi}\,\sqrt{1-\xi^2}\text{ if }|\xi|\le1,\quad \hat f(\xi)=0\text{ if }|\xi|>1. $$ $\hat f$ fails to be Lipschitz at $\xi=\pm1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.