## Decomposition of an integral operator into a composition

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 "one-sided" 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 $G$?

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So you want a continuous analog of the decomposition of matrices into lower and upper triangular matrices. I'm not a specialist, but it seems to me that for reasonably nice kernels (e.g. Hilbert-Schmidt), the natural approach should be: pass to the limit in the decomposition for finite dimensional approximations, and show that the sequence of factors admits a converging subsequence. – Pietro Majer Feb 16 2012 at 9:58
@Pietro Majer: Matrixes can not be decomposed in a product of upper triangular and lower triangular matrices (look up Bruhat decomposition). You have to allow permutation matrices. But nice interpretation of the question. – Marc Palm Feb 16 2012 at 13:53
Is the upper limit of the inner integral $x$ or $u.$ ? – Uday Feb 16 2012 at 13:58
@Uday: The upper limit is $x$. @Pietro: This was indeed inspired by the LU decomposition. I'm wondering how to construct the analogous functional form of the decomposition. – Victor Liu Feb 16 2012 at 19:18