Distribution $f$ such that (a) $\widehat{f}$ has compact support, (b) $\mathbb{E}(|X|)$ is minimal?

Question

(What follows is motivated by an answer to Fourier optimization problem related to the Prime Number Theorem)

Let $f:\mathbb{R}\to [0,\infty)$ be such that
(a) $\int_{\mathbb{R}} f(x) dx = 1$,
(b) $\widehat{f}(t)=0$ for all real $t$ with $|t|>1$.

What is the choice of $f$ such that $$\int_{\mathbb{R}} |x| f(x) dx$$is minimal? What is that minimum?

Remarks:

1. It is easy to see that we can assume $f$ to be an even function.
2. Yes, this seems to be yet another incarnation of the uncertainty principle.

Answer 1

One can prove that under your assumptions $$A:=\int_{-\infty}^\infty|x|f(x)dx\geq 9/(4\pi),$$ but estimate is not exact.

The proof is based on the formula $$A=-\lim_{y\to 0+}\frac{1}{\pi}\frac{d}{dy}\left(y\int_{-1}^1\frac{\hat{f}(t)}{t^2+y^2}dt\right),$$ which is easy to obtain. Now for $\hat{f}$ supported on $[-1,1]$, Cramer obtained the estimate $\hat{f}(t)\leq 1-t^2/8$, and using this estimate and $\hat{f}(0)=1$, we obtain the result.

Ref. H. Cramer, Random variables and probability distributions, Cambridge UP, 1970.

One can use a better, exact estimate for $\hat{f}$ in the work of

A. Fryntov, An extremal problem in the theory of Hermitian positive functions, Func Anal., Appl, 10, 1 (1976) 91-92;

this will improve the estimate, but still will not give the exact minimum of $A$.

Comment. A somewhat simpler but similar problem would be to miminize $$|\hat{f}''(0)|=\int_{-\infty}^\infty x^2f(x)dx.$$ For this quantity, Fryntov implies the estimate $A\geq \pi^2$, which is probably also not exact.

Answer 2

Warning: the following answer is (a) maybe a bit careless in a nineteenth-century sort of way, (b) missing its final step (which may be obvious to others, and/or amount to looking things up in the right table)

We are looking for an even function $g:\mathbb{R}\to \mathbb{R}$ with support on $[-1,1]$; we will define $f$ to be its Fourier transform. The condition $\int_{\mathbb{R}} f(x) dx=1$ becomes $g(0)=1$. What we need to minimize is the quantity $$I = \int_{\mathbb{R}} |x| \widehat{g}(x) dx.$$ Here are two attempts to express $I$ more directly in terms of $g$.

1. The Fourier transform of $\frac{g'(t)}{2\pi i}$ equals $x \widehat{g}(x)$. We can write $|x| \widehat{g}(x) = \mathrm{sgn}(x) x \widehat{g}(x)$. The Fourier transform of $\mathrm{sgn}(x)$ is $\frac{1}{i \pi t}$ (in some sense). Hence, the Fourier transform of $|x| \widehat{g}(x)$ should be the convolution of $\frac{1}{i \pi t}$ and $\frac{g'(-t)}{2\pi i} = \frac{g'(t)}{2\pi i}$. In particular, $I$ should equal the value of the Fourier transform of $|x| \widehat{g}(x)$ at $0$, i.e., $$I = \int_{\mathbb{R}} \frac{1}{i \pi t} \frac{g'(t)}{2\pi i} dt = - \frac{1}{2 \pi^2} \int_{\mathbb{R}} \frac{g'(t)}{t} dt.$$ We can assume $g'(0)=0$, so the integral above should make sense.

2. Since $I = - 2 \int_{-\infty}^0 x \widehat{g}(x) dx$, we can write $I = - 2\int_{-\infty}^0 \int_{-\infty}^y \widehat{g}(x) dx dy$. Now, the second antiderivative of $\widehat{g}(x)$ should have Fourier transform $\frac{g(t)}{(2\pi i)^2 t^2}$. Its value at $0$ equals both $-I/2$ (by definition) and $$\int_\mathbb{R} \frac{g(t)}{(2\pi i)^2 t^2} dt = -\frac{1}{4\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt;$$ hence, $$I = \frac{1}{2\pi^2} \int_{\mathbb{R}} \frac{g(t)}{t^2} dt.$$ Of course this diverges.

At the same time, by integration by parts, $-\int_\mathbb{R} \frac{g'(t)}{t} dt$ equals $\int_\mathbb{R} \frac{g(t)-1}{t^2} dt$, which converges. So, it looks like $$I =\frac{1}{2\pi^2} \int_\mathbb{R} \frac{g(t)-1}{t^2} dt.$$

We recall we must also fulfill the constraint that $f$ take non-negative values. This will certainly be true if $g$ is defined as a convolution $h\ast h$, with $h$ symmetric and real-valued, as then $\widehat{h}$ will be real valued, and $\widehat{h\ast h} = \widehat{h}^2$. We can require that $h$ also be an even function, and that the support of $h$ be contained in $[-1/2,1/2]$. I think these are all necessary conditions, so I am not making the search space for my optimum any smaller, but I'd be delighted if others can double-check and confirm.

So, we've reduced our problem to: find a symmetric function $h:[-1/2,1/2]\to \mathbb{R}$ with $(h\ast h)(0) = |h|_2^2 = 1$ such that $$\int_{\mathbb{R}} \frac{(h\ast h)(t)-1}{t^2} dt$$ is minimal.

A bit of calculus variations seems to show that the optimal $h(t)$ has to have $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h(t+t_0)+h(t-t_0)) - h(t_0)}{t^2} dt$$ equal to a constant independent of $t_0$, for $t_0\in (-1/2,1/2)$. Again by integration by parts, this is just $$\frac{1}{h(t_0)} \int \frac{\frac{1}{2} (h'(t+t_0)+h'(t-t_0))}{t} dt,$$ which equals $\frac{1}{h(t_0)} H(h')(t_0)$, where $H$ is our old friend the Hilbert transform.

In other words: we are to find a (continuous) function $h:\mathbb{R}\to \mathbb{R}$, supported on $[-1/2,1/2]$, with $|h|_2=1$, such that $H(h')(t) = \lambda h(t)$ for all $t\in (-1/2,1/2)$ and some $\lambda$.

Surely such a function must be known (if it exists)?