How do I solve $ f(y) \int_0^1 \tfrac{\exp(-\frac{(x-y)^2}{2})}{\int_0^1 \exp(-\frac{(x-z)^2}{2}) f(z) dz}\,dx =1$? 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 $c$)
 A: Things begin to clear up, a bit: you have a system of equations for $f$ and $\hat f,$ and while $f=\hat f$ would give a solution, that's probably nonsense from the point of view of your model. I investigated your problem numerically, starting, however, from the form
$$g(y)\int_0^1 \frac{e^{xy}}{\int_0^1 e^{xz}\,g(z)\:dz}\:dx = 1$$ proposed by Ninad Munshi. Here, $g(y) = f(y)e^{-\frac{y^2}{2}},$ so $g(1) = e^{-\frac{1}{2}}g(0)$ corresponds to $f(0)=f(1),$ and that's a symmetry present in the original problem, obviously (your equations are invariant under $x\to1-x, y\to1-y,$ so you must have $f(x)=f(1-x)$ and $\hat f(1-x)=\hat f(x)$ due to uniqueness).
So we have a system of equations $$h(x)=\frac1{\int_0^1 e^{xz}\,g(z)\:dz},$$
$$g(x)=\frac1{\int_0^1 e^{xz}\,h(z)\:dz}.$$
The most obvious try to solve that numerically would be a simple iteration,
$$h_{n+1}(x)=\frac1{\int_0^1 e^{xz}\,g_n(z)\:dz},$$
$$g_{n+1}(x)=\frac1{\int_0^1 e^{xz}\,h_{n+1}(z)\:dz},$$ approximating the integrals by some sort of quadrature formula. Interestingly, this process (with the initial guess $g_0(x)=e^{-x^2/2}$) converges with impressive speed, the difference to the previous value is $\le10^{-15}$ after about five iterations. And even though that numerical process knows nothing about $f(0)=f(1),$ the relation $g(1) = e^{-\frac{1}{2}}g(0)$ is satisfied with said accuracy.
So we can calculate those functions rather precisely, but... I doubt there is a closed solution in elementary functions. And while there are algorithms identifying constants known with sufficient accuracy (quite beyond 15 or 16 digits, though), I don't know of anything similar for functions.
EDIT: as it turns out, the functions $g$ and $h$ differ only by a multiplicative constant. So there is a solution with $f=\hat f$ for your equations! Here is a graph:

We have $f(0)\approx 1.1234005998770296,$ but I can't identify the constant.
EDIT 2: Factorizations aren't always obvious, if $q(x,y)=e^{-(x-y)^2/2},$  we have $$\frac{q(x,y)}{e^{xy}}=e^{-\frac{x^2}2}\cdot e^{-\frac{y^2}2}.$$ If the integration interval were $(-\infty, \infty)$ instead of $[0,1],$ the solution would be just constant, and I'm pretty sure that fact has a probabilistic interpretation. And since that constant is sufficiently close to the solution you seek, that may explain the rapid convergence. However, that's all mere speculation without knowing more details from the probabilistic background.
