# Does every commutative operation have an integral representation?

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.

• 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 $[0,\,\infty]$. – J.G. Nov 8 '18 at 14:43