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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 (?)

hassan
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    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. – PhoemueX Oct 26 '15 at 22:09

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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 $\xi=\pm1$.

Julián Aguirre
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