Let $g : \mathbb{R}_{\geq 0}^2 \to \mathbb{R}$ satisfy $g(a, b) = g(b, a)$. Is there necessarily some $f : \mathbb{R}_{\geq 0}^2 \to \mathbb{C}$ such that $$g(a, b) = \int_0^\infty f(a, t) \, f(b, t)\,\mathrm{d}t?$$

Some examples:

  • For $g(a, b) = h(a)\,h(b)$, we may take $f(a, t) = h(a)\,\mathbf{1}_{\leq 1}(t)$.
  • For $g(a, b) = \min(a, b)$, we may take $f(a, t) = \mathbf{1}_{\leq a}(t)$.
  • For $g(a, b) = a+b$, we may take $f(a, t) = (a+1)\,\mathbf{1}_{[0, 1]}(t) + ia\,\mathbf{1}_{[1, 2]}(t) + i\,\mathbf{1}_{[2, 3]}$.
  • For $g(a, b) = \max(a, b)$, we may take $f(a, t)$ as for the $a+b$ case, and add $i\,\mathbf{1}_{[3, a+3]}(t)$.

If we replace the domain of $f$ and $g$ by $X^2$, where $X$ is a finite (and replace the integral by a sum), I already have a constructive proof:

View $g$ as a symmetric matrix. We diagonalize $g = Q \Lambda Q^T$ for some orthogonal matrix $Q$ and diagonal matrix $\Lambda$. We may pick $\Lambda'$ so that $\Lambda'^2 = \Lambda$, and then $f = Q \Lambda'$ suffices. (The RHS is simply $ff^T$).

However, I do not know how to approach the continuous case.

Any help is appreciated! If this has already been proven, I would be grateful for a link to the relevant literature.

  • $\begingroup$ You'd want to show some $Q$ satisfies $\int Q(a,\,t)Q(b,\,t)dt=\delta(a-b),\,g(a,\,b)=\int Q(a,\,t)\Lambda(t,\,u)Q(b,\,u)dtdu,\,\Lambda(t,\,u):=\lambda(t)\delta(t-u)$ for some function $\lambda$, with all integrals on $[0,\,\infty]$. $\endgroup$ – J.G. Nov 8 '18 at 14:43

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