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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 $\max_{\alpha \in \mathbb{R}} |\widehat{f}(\alpha)|$?

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.

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(If $f$ is a finite linear combination of delta functions (point masses), then one can just take derivatives and solve a polynomial equation - but of course this doesn't quite solve the general problem.) – H A Helfgott May 29 '12 at 10:25
I think that if f(x) = d mu(x) where mu(x) is a non-lattice distribution, then you can get some bounds < 1. I've seen results of this type in Esseen's (of Berry-Esseen fame) thesis. – kolik May 29 '12 at 10:28
Just to give an idea - the f I am working with is a nice smooth function plus three point masses. – H A Helfgott May 29 '12 at 11:22
What do you mean by something "better"? For positive real valued function $f$, you necessarily have $\hat{f}(0) = |f|_1$, so the inequality is sharp. (The same argument also shows that in the class of distributions which can be decomposed into a sum of a smooth function and three point masses, the inequality is sharp. So you probably need to say more.) Do you mean you want a bunch of statements of the form "If $f$ satisfies conditions blah and blah then we can improve the inequality to (some inequality)"? – Willie Wong May 29 '12 at 12:01
Obviously I am talking about a function that is not positive everywhere. Say somebody gives you a function that is piecewise defined as segments of the form $x\mapsto C/x$ (say) and then you get a distribution f by summing that function to three point masses, one or two of them negative. How do you go about finding $\max_{\alpha\in \mathbb{R}} |\widehat{f}(\alpha)|$ (rigorously and to a certain accuracy)? – H A Helfgott May 29 '12 at 13:34

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, $$ \hat u(\xi)(1+\vert\xi\vert)^{n/2}\ln(2+\vert\xi\vert)\in L^2\Longrightarrow u\in L^\infty. $$

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