Weird integral equation with non convolution kernel Let $f$ and $g$ two rugular functions. My question is the following:
Under what condition can we say that for given $g$, there exists $f$ such that we have:
$$\int\limits_0^1 {f(x - s,s)ds = g(x)} $$
Thank you.
 A: This type of functional equation resembles Wigner transforms in physics, except there one has lots of constraints and an infinite integration domain which result in unique solutions. 
Here, without further information, the problem is hopelessly open-ended. For starters, it is not clear why you are writing f in this convoluted way, if you do not know something about the dependence of its two arguments which you are not telling. 
Why don't you simply redefine it as $f(x-s,s)\equiv h(x,s)$, in which case the enormity of the space of solutions is apparent?
$$
\int\limits_0^1  \!ds~~ h(x,s) = g(x) ~ .
$$
Typically, for regular functions, compare the respective Taylor expansions around 0,
$$
g(x)=  \sum_{j=0}^{\infty} g_j x^j ~,
$$
with 
$$
h(x,s)= h_0(x) +  \sum_{j=0}^{\infty}   x^j    \sum_{k=1}^{\infty} h_{j,k} ~k ~s^k ,    
$$
where the coefficient k is pulled out of the respective $h_{j,k}$ s for convenience in integration. That is to say,
$$ g(x)=  h_0(x) +  \sum_{j=0}^{\infty}   x^j  \left ( \sum_{k=1}^{\infty} h_{j,k}  \right) .$$
At this point, you may go haywire... Slosh any degrees of freedom of the problem between $h_0 (0)$ and the series on the r.h.s.  Unless you had further constraints.
