Cyclic convolution of a periodic function with itself is a constant? Let $f(x)$ be a periodic function of period $T$. Now let us define $c(x)$ the cyclic convolution (on the same period T) of $f$ with itself:
$$c(x)=\int_t^{t+T}f(\tau)f(x-\tau)d\tau$$
I have the impression, and kind of confirmed numerically, that $c(x)$ is a constant function, independent of $x$.
Is this true and if yes how can we prove it? And can we express the value of that constant in a simpler way in terms of $f$?
 A: You need to be careful with your notation - you have $t$ as part of the bounds of integration (so it's a free variable) but it's also the variable of integration. We can fix it by writing something like:
$c(x) = \int_t^{t+T} f(\tau) f(x - \tau) d\tau$
With that fix, let's check what happens with an example function, so let $f(t) = \sin t$:
$\begin{eqnarray} c(x) & = & \int_t^{t+2\pi} f(\tau) f(x - \tau) d\tau \\
& = & \int_t^{t+2\pi} \sin (\tau) \sin (x - \tau) d\tau \\
& = & \int_t^{t+2\pi} \frac{1}{2}\left(\cos(\tau - (x - \tau)) - \cos(\tau + (x - \tau)) \right) d\tau & \text{using a trig identity} \\
& = & \frac{1}{2} \int_t^{t+2\pi} \left( \cos(2\tau - x) - \cos x \right) d\tau \\
& = & \frac{1}{2} \left[\frac{1}{2} \sin(2\tau - x) - \tau \cos x \right]_t^{t+2\pi} \\
& = & \frac{1}{4} \left(\sin(2t + 4\pi - x) - \sin(2t - x) \right) - \frac{1}{2}((t+2\pi)\cos x - t \cos x) \\
& = & - \pi \cos x \end{eqnarray}$
This is clearly not a constant, so it's likely that you got caught up by something in your notation.
