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Let $\phi:\mathbb{R}\to \lbrack 0,\infty)$ be piecewise continuous, symmetric ($\phi(x)=\phi(-x)$) and with support on $(-1,1)$. Let $\Phi(x)=\int_{-\infty}^x \phi(u) du$; assume $\Phi(1)=1$.

What is $\phi$ such that $$\frac{|\Phi|_2^2}{1-\max_{|t|\geq 1} |\widehat{\phi}(t)|}$$ is minimal?

Note that $\Phi(1)=1$ implies $|\widehat{\phi}(t)|\leq 1$ for all $t$.

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  • $\begingroup$ I've asked a related question in mathoverflow.net/questions/285670/uncertainty-principle-redux $\endgroup$
    – H A Helfgott
    Commented Mar 2, 2019 at 3:07
  • $\begingroup$ Isn't $\Phi$ a constant for $x>1$? $\endgroup$
    – Piero D'Ancona
    Commented Mar 2, 2019 at 5:38
  • $\begingroup$ Ah, sure - I am writing $|\Phi|_2$ for its norm on $L^2(\lbrack -1,1\rbrack)$. Thanks for the remark. $\endgroup$
    – H A Helfgott
    Commented Mar 2, 2019 at 5:40
  • $\begingroup$ Is there a finite version of this problem? $\endgroup$
    – Seva
    Commented Mar 2, 2019 at 7:13
  • $\begingroup$ I suppose one can give a finite version in the obvious way. Perhaps it is better to discuss a discrete version: find a function $\phi:\mathbb{R}/\mathbb{Z}\to \mathbb{Z}$ with support on $\lbrack -1/4,1/4\rbrack$ and $L^2$ norm equal to $1$ such that $\max_{n\ne 0} |\widehat{\phi}(n)|$ is minimal. It's all the uncertainty principle in one way or the other. $\endgroup$
    – H A Helfgott
    Commented Mar 2, 2019 at 18:37

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