Consider a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$, supported on $[-1,1]$, of positive type. Assume $f(0) = 1$; what is the "largest" area $\int f\,dx$ that can be achieved?
To be more precise, let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying:
- $f$ is supported on $[-1,1]$,
- for all $x$, $0 \leq f(x) \leq 1 = f(0)$, and
- $f$ has positive type: for any finite family of points $x_1 < \cdots < x_n$ in $\mathbb{R}$, the matrix $(f(x_i - x_j))_{ij}$ is positive semi-definite.
How large can $\int f(x)\,dx$ be?
Remarks
- One example of such a function is the "triangle" $t(x) = \max(1 - |x|,0)$. This achieves $\int t(x)\,dx = 1$. Is that the best one can do?
- There are functions $g$ of positive type (and satisfying the other requirements above) for which $g(x) > t(x)$ for some $x$. (However, I do not know of any for which $\int g(x)\,dx > 1$.)
I asked a (near-) duplicate of this question on mathoverflow, where it was answered; indeed, it turns out that the triangle is optimal. The proof of this is a simple and very pretty application of the Poisson summation formula.