TITLE: Solution $y(x)$ for $\sin(x) = \int_0^{2\pi} \max(y(t), y(x+t)) dt$
QUESTION [6 upvotes]: I've been banging my head against a wall for a few weeks to find a feasible solution for $y(x)$.
$$\sin(x) = \int_0^{2\pi} \max(y(t), y(x+t)) dt$$
I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.
Does anybody have any ideas?

REPLY [3 votes]: If $x=0$ then
\begin{equation*}
0= \sin(0)=\int_{0}^{2\pi}\max(y(t),y(t))\, \mathrm{d}t = \int_{0}^{2\pi}y(t)\, \mathrm{d}t .
\end{equation*}
If $x = \frac{3\pi}{2}$ then
\begin{equation*}
-1 = \sin\left(\frac{3\pi}{2}\right) = \int_{0}^{2\pi}\max\left(y(t),y\left(\frac{3\pi}{2}+t\right)\right)\, \mathrm{d}t \ge \int_{0}^{2\pi}y(t)\, \mathrm{d}t = 0.
\end{equation*}
Consequently, it does not exist any solution $y(x)$.