I've been musing about the following question for a while now. Given an integral operator $G$ defined by $$ (Gf)(x) = \int_0^1 G(x,u) f(u)\,du $$ Is it possible to decompose this into two separate "onesided" integral operators $L$ and $R$ such that $$ (Gf)(x) = \int_x^1 L(x,u) \int_0^x R(u,v) f(v)\,dv\,du $$ If so, under what assumptions on $G$? And are there expressions for the left and right operators $L$ and $R$ in terms of $G$?

imho, if you can rewrite the equation in a suitable basis, you obtain an infinite matrix and may try the QR decomposition. Try wavelet basis or a parent of this family of functions like curvelets, ridglets... regarding properties of the operators $G$. If $G$ is a pseudo or Fourier integral operator, these bases are known to give almost diagonal representations. 


To mimic multiplication, the middle integral you wrote down should be from $0$ to $u$, not from $0$ to $x$. The way you have written it, $f(v)$ does not affect $Gf(x)$ unless $v \leq x$. Assuming $G, L,$ and $R$ are sufficiently nice smooth functions, we should have some kind of uniqueness result where an equality of integral operators is an equality of integrals, and so we can write $G(x,v) = \int_{\max (x,v)}^1 L(x,u) R(u,v) du$ If $L$ and $R$ are honest bounded functions, this gives $G(x,v)=0$ if $x=1$ or $v=1$, which is problematic. So you must put some vanishing condition on $G$ or use unbounded $L$, $R$. 

