I am trying to solve this integral equation for $f$ (can be assumed to be positive)
$$ f(y) \int_0^1 \frac{e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx =1, \quad y \in [0,1]$$
Any insight is welcome, I have no idea how this kind of equations is called and how to solve them. I thought about differentiating both side with regard to $y$, we get
$$f(y)\int_0^1 \frac{(x-y)e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx + f'(y)\int_0^1 \frac{e^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx =0$$
Plugging in the first equation, we get
$$ f(y)\int_0^1 \frac{xe^{-\frac{(x-y)^2}{2}}}{\int_0^1 e^{-\frac{(x-z)^2}{2}} f(z) dz} dx -y + \frac{f'(y)}{f(y)} = 0$$
But it only seems to make the problem harder.
EDIT: As this question seems to interest some people, here is some more info. We can rewrite the problem as a system of two integral equations:
$$ f(y) \int_0^1 e^{-\frac{(x-y)^2}{2}} \hat f(x) dx =1, \quad y \in [0,1]$$
$$ \hat f(x) \int_0^1 e^{-\frac{(x-y)^2}{2}} f(y) dy =1, \quad x \in [0,1]$$
The existence of solutions $f,\hat f$ is a deep result in stochastic processes and probability theory. The trained eye recognized the heat kernel (Brownian transition density). The product $f(y)\hat f(x)$ is actually the density (not w.r.t Lebesgue, it's complicated) of a certain coupling of 2 given random variables. So $f,\hat f$ contain all the information about the dependence between those 2 random variables. They are unique up to multiplicative constant (when you multiply one by $c$, you divide the other one by $c$)