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Let $\phi:\mathbb{R}\to\mathbb{R}$ be an even function with support on $[-1,1]$. Assume that it is in $L^1\cap L^2$ and that its Fourier transform is also in $L^1\cap L^2$. Assume as well that $|\phi|_\infty = \phi(0)=1$. Define $F(x) = \int_x^\infty \widehat{\phi}(t) dt$.

Given these constraints, what is the choice of $\phi$ that minimizes $$\int_0^\infty |F(x)| dx?$$

(a) What is the value of that minimum?

(b) A possibly easier problem is to minimize $$\int_0^\infty F(x) dx,$$ which of course equals $$\int_0^\infty t \widehat{\phi}(t) dt.$$

(c) A possibly easier problem (in some sense a special case of (b)) is the following: let $g:\mathbb{R}\to \mathbb{R}$ be an even function with $L^2$-norm $|g|_2=1$ and support on $[-1/2,1/2]$. Choose $g$ so as to minimize $$\int_{\mathbb{R}} |x| |\widehat{g}(x)|^2 dx.$$

These are clearly questions of uncertainty-principle type.

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    $\begingroup$ @KevinCasto I think you are forgetting "support on $[-1,1]$"? $\endgroup$
    – H A Helfgott
    Commented Nov 25 at 0:30
  • $\begingroup$ It may not be hard to solve (b). I have a not terribly rigorous argument that can probably be made rigorous - if you restrict yourself to $\phi(x)$ decreasing for $x>0$, the integral in part (b) can be made as close as you want to $1/2\pi^2$ - a function $\phi(x)$ that is $1$ for $|x|<1-\epsilon$ and then plunges to $0$ for $|x|>=1$ will do the trick. If you don't restrict yourself to $\phi(x)$ decreasing for $x>0$, the integral can be as close to $-\infty$ as you want, I think. $\endgroup$
    – H A Helfgott
    Commented Nov 25 at 23:21
  • $\begingroup$ Right, that works for (b). Now, for (c), the problem becomes: given those constraints on $g$, choose $g$ such that $-\int_{\mathbb{R}} (g\ast g)'(x)/x dx$ is minimal. Or, equivalently, so that $\int_{\mathbb{R}} (1-(g\ast g)(x)) dx/x^2$ is minimal. Or, equivalently, so that $\int_{\mathbb{R}} g(t) (Ag)(t) dt$ is minimal, where $Ag(t) = \int_{\mathbb{R}} \frac{g(t) - \frac{1}{2} (g(t+x) + g(t-x))}{x^2} dx$. I feel a solution is one step away... $\endgroup$
    – H A Helfgott
    Commented Nov 26 at 10:58
  • $\begingroup$ For (a), it would be reasonable to consider candidates such that $F$ is non-negative but $\widehat{\varphi}$ isn't (as then one is taking advantage of cancellation). $\endgroup$
    – H A Helfgott
    Commented Dec 8 at 9:58

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