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. (Equivalently--by Bochner's theorem--the Fourier transform is pointwise nonnegative.)

How large can $\int f(x)\,dx$ be?


  1. 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?
  2. 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 $\int g(x)\,dx > 1$.)
  3. I asked this question the "Mathematics" stack exchange site, but didn't have any takers.
  • $\begingroup$ The Poisson summation formula would give $\sum_{n\in {\Bbb Z}} f(n) = \sum_{k\in {\Bbb Z}} {\hat f}(k)$. The LHS is $1$, and since ${\hat f}(k)\ge 0$ by assumption, it follows that ${\hat f}(0) \le 1$. $\endgroup$
    – Lucia
    Commented Nov 6, 2014 at 14:19
  • $\begingroup$ Awesome! That's just right. If you turn it in to an "answer," I can accept it. $\endgroup$
    – acr
    Commented Nov 6, 2014 at 14:38

1 Answer 1


As noted in my comment above, the Poisson summation formula would give $$ 1= f(0) = \sum_{n\in {\Bbb Z}} f(n) = \sum_{k\in {\Bbb Z}} {\hat f}(k) \ge {\hat f}(0), $$ since ${\hat f}(k) \ge 0$ by assumption. Since ${\hat f}(0) = \int_{-1}^{1} f(x) dx$, this proves the desired bound.

If you are worried about using the Poisson summation formula on just a continuous function, use the above argument with $f$ convolved with $\phi_{\epsilon}$ where $\phi_{\epsilon}$ is smooth (or just in $C^2$, say) and non-negative, supported on $(-\epsilon,\epsilon)$, has non-negative Fourier transform, and ${\hat \phi_{\epsilon}}(0)=1$. Now use the above argument with $f*\phi_{\epsilon}$, to obtain ${\hat f}(0)\le 1+O(\epsilon)$, and then let $\epsilon \to 0$.


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