(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.

(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.

(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.

(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.