Constancy of an integral function Fix some $\ell\in\mathbb{R}^+$.  Say that $f:\mathbb{R}^2\to\mathbb{R}_{\geq0}$ and $\mu:\mathbb{R}\to\mathbb{R}^+$ are functions satisfying the following:


*

*$f$ and $\mu$ are continuous.

*$f$ is symmetric, that is $f(s,t)=f(t,s)$.

*$\mu$ is $\ell$-periodic, that is $\mu(t+\ell)=\mu(t)$ for all $t$.

*$f$ is $\ell$-periodic in both variables, that is $f(s+\ell,t)=f(s,t+\ell)=f(s,t)$ for all $s,t$.

*$f$ is diagonally constant, that is $f(s+a,t+a)=f(s,t)$ for all $s,t,a$.


If I then define $F:\mathbb{R}\to\mathbb{R}$ by $$F(s)=\int_0^\ell f(s,t)\mu(t)\,dt,$$
will $F$ be constant?
Experimental evidence (from MATLAB) has led me to believe that the answer is yes, but I can't figure out how to prove it.  Any advice or suggestions (especially counterexamples) are most appreciated.
 A: To construct counterexamples we need a better understanding of "diagonally constant."  Let $g(x)$ be given, then $f(x,y)=g(x-y)$ is "diagonally constant" since $g((x+a)-(y+a))=g(x-y)$.  Conversely if $g(x)=f(x,0)$ with $f$ diagonally constant, then we have $f(x,y)=f(x-y,0)=g(x-y)$.  If $g(x)$ is $\ell$ periodic then $g(x-y)$ is $\ell$ periodic separately in $x$ and $y$.  Thus our integral is
$$\int_0^\ell g(s-t)\mu(t)\,dt$$
a convolution of two periodic functions.  Symmetry of $f(x,y)$ is the same as $g$ being even.
Now for the counterexample.  Make $\mu(x)=1$ for $0\le x\le \ell/2$, $0$ for $\ell/2 < x< \ell$, and periodic.  Pick some $0<a<\ell/4$.  Make $g(x)=1$ for $a\le x\le \ell-a$, otherwise $0$ in $[0,\ell]$, and periodic.  Then the convolution starts out at $\ell/2-a$ for $s=0$, becomes $\ell/2$ when $s$ increases so that the positive part of $\mu$ overlaps the positive part of $g$ to the max, and then descends back to $\ell/2-a$.
A: Does this count as a counterexample? Here $\ell=2\pi, f(s,t)=\cos(s-t)+1, \mu(t)=\sin(t)+1$. Then
$$
F(s)=\int_0^{2\pi} f(s, t)\mu(t)\, dt=\pi\left(2+\sin(s)\right).$$(Evaluated with Maple). 
