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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 '12 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 '12 at 13:53
@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 '12 at 19:18
@MarcPalm: of course, I'not saying that. The OP ask under what assumptions on G it is possible; I'm saying that the infinite dimensional problem includes the analogous problem for matrices, etc. – Pietro Majer Sep 27 '13 at 18:17
The upper limit $x$ screws the problem up because then $Gf(x)$ depends only on $f(t)$ with $t\le x$, which is possible only if the operator is triangular from the beginning, i.e., when there is nothing to do really. You'd better put $u$ there as Uday suggested. – fedja Oct 28 '13 at 1:52

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.

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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 $L$, $R$.

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