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This is a question of uncertainty-principle type stemming from Eigenvalue of a convolution and a restriction?

Let $f:\mathbb{R}\to \mathbb{R}$ be even, absolutely continuous and supported in $[-\frac{1}{2},\frac{1}{2}]$, with $f'$ of bounded total variation. Let $|f|_2=1$. What is the minimal possible value of $$I(f) = \int_{-\infty}^\infty |t| |\hat{f}(t)|^2 dt?$$ Which choice or choices of $f$ minimize $I(f)$?

Here "absolutely continuous" and "$f'$ of bounded total variation" are technical conditions that guarantee that $f(t)=O(1/t^2)$ as $t\to \pm \infty$.

PS. MathOverflow just noticed something I had forgotten, viz., I asked a question that turns out to be equivalent two years ago in Distribution $f$ such that (a) $\widehat{f}$ has compact support, (b) $\mathbb{E}(|X|)$ is minimal? . Clever robot! That question got 9 upvotes but no full answers, so it seems legitimate to ask a not-so-very-obviously equivalent question.

On another note: $f(x) = \cos(\pi x)|_{[-1/2,1/2]}$ is a natural choice - can one do better?

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  • $\begingroup$ Here is a possible approach. Let $B$ the linear operator such that $\widehat{Bf}(t) = |t| \widehat{f}(t)$. (Namely: $B$ is given by $Bf = f'\ast \frac{1}{2\pi^2 x}$.) We are being asked to minimize $\langle Bf, f\rangle$ given $|f|_2=1$, i.e., find the lowest eigenvalue. Since the eigenvalues are clearly all positive, this is the same as minimizing $|B f|_2$. However, $$|B f|_2 = \int_{-\infty}^\infty ||t| \widehat{f}(t)|^2 dt = \int_{-\infty}^\infty |t \widehat{f}(t)|^2 dt,$$ which equals both $\frac{1}{(2\pi)^2} |f'|_2^2$ and $- \frac{1}{(2\pi)^2} \langle f,f''\rangle$. (continued below) $\endgroup$
    – H A Helfgott
    Commented yesterday
  • $\begingroup$ Here I want to conclude by saying 'the eigenfunctions of $f\mapsto f'$ are the exponentials' but of course they are not of compact support. The issue is that $e^{i a x}\cdot 1_{[-b,b]}$ is almost an eigenfunction of $f\mapsto f'$, except it's not continuous at $b$ or $-b$, and $\cos(a x)\cdot 1_{[-1/2a,1/2a]}$ is continuous and almost an eigenfunction of $f\mapsto f''$, except it is not differentiable at $-1/a$ or $1/a$. $\endgroup$
    – H A Helfgott
    Commented yesterday
  • $\begingroup$ However, for $\cos(a x)\cdot 1_{[-1/2a,1/2a]}$, $|f'|_2^2$ is still meaningful, and in the sense of distributions or measures, so is $\langle f,f''\rangle$. $\endgroup$
    – H A Helfgott
    Commented yesterday
  • $\begingroup$ ... though $f'\ast \frac{1}{x}$ blows up at both $-1/2a$ and $1/2a$ for $f(x) = \cos(ax) 1_{[-1/2a,1/2a]}$, meaning we have to tread carefully. $\endgroup$
    – H A Helfgott
    Commented yesterday
  • $\begingroup$ I am confused you speak of support of $f$ contained in $[-1/2,1/2]$ and $f=O(1/t^2)$. $\endgroup$
    – juan
    Commented yesterday

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