most recent 30 from http://mathoverflow.net 2013-06-20T01:22:02Z http://mathoverflow.net/feeds/question/98263 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98263/max-of-fourier-transform H A Helfgott 2012-05-29T10:22:48Z 2012-05-29T21:38:42Z

<p>Let f be a real-valued function (or distribution) on $\mathbb{R}$. (You can assume it is nice in one way or another.) What would be some practical ways to bound $\max_{\alpha \in \mathbb{R}} |\widehat{f}(\alpha)|$?</p> <p>Obviously $\max_{\alpha \in \mathbb{R}} |\widehat{f}(\alpha)| \leq |f|_1$, but I am looking for something a bit better. Either a numerical method or a bag of tricks is fine, as long as the answer is rigorous.</p>

http://mathoverflow.net/questions/98263/max-of-fourier-transform/98312#98312 Bazin 2012-05-29T21:38:42Z 2012-05-29T21:38:42Z

<p>Let me reformulate your question. How can we control the $L^\infty$ norm of $u$ by some behavior of the Fourier transform? The most classical thing that could be said is $$ H^s(\mathbb R^n)\subset L^\infty(\mathbb R^n)\quad\text{when $s>n/2$ and then $\Vert u\Vert_{L^\infty}\lesssim \Vert u\Vert_{H^s}=\Vert (1+\vert \xi\vert)^s\hat u(\xi)\Vert_{L^2}$}, $$ where $H^s$ is the standard Sobolev space based on $L^2$. It is known and easy to check that $H^{n/2}(\mathbb R^n)\not\subset L^\infty(\mathbb R^n)$. There are some refinement of the injection above: writing $$ u(x)=\int e^{2i\pi x\cdot \xi}\hat u(\xi)d\xi= \int e^{2i\pi x\cdot \xi}\omega(\xi)\hat u(\xi)\frac{1}{\omega(\xi)}d\xi, $$ we get $ \Vert u\Vert_{L^\infty}\le \Vert \omega (D) u\Vert_{L^2}\left(\int\frac{d\xi}{\omega(\xi)^2}\right)^{1/2}, $ which is useful if $1/\omega$ belongs to $L^2$. In particular, $$ \hat u(\xi)(1+\vert\xi\vert)^{n/2}\ln(2+\vert\xi\vert)\in L^2\Longrightarrow u\in L^\infty. $$</p>