• @[email protected]
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    310 months ago

    What is meant here, I believe, is (1 - Int)^-1. Writing 1/(1 - Int) is an abuse of notation, especially when the numerator isn’t just 1 but another operator, which loses the distinction between a left and a right inverse. But for a bounded linear operator on a normed vector space, and I think Int over an appropriately chosen space of functions qualifies, (1 - Int)^-1 equals the Neumann series \sum_k=0^∞ Int^k, exactly as in the derivation.

    Int is injective: Take Int f = Int g, apply the derivative, and the fundamental theorem gives you f = g. I think you can make it bijective by working with equivalence classes of functions that differ only by a constant.

    • @[email protected]
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      310 months ago

      Int is definitely not injective when you consider noncontinuous functions (such as f(X)={1 iff X=0, else 0}). If you consider only continuous functions, then unfortunately 1-Int is also not injective. Consider for example e^x and 2e^x. Unfortunately your idea with equivalence classes also fails, as for L = 1 - Int, L(f) = L(g) implies only that L(f-g) = 0, so for f(X)=X and g(X)=X + e^x L(f) = L(g)

      • @[email protected]
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        210 months ago

        Sets of measure zero are unfair. But you’re right, the second line in the image is basically an eigenvector equation for Int and eigenvalue 1, where the whole point is that there is a subspace that is mapped to zero by the operator.

        I’m still curious if one could make this work. This looks similar to problems encountered in perturbation theory, when you look for eigenvectors of an operator related to one where you have the spectrum.

        • @[email protected]
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          210 months ago

          Well, sets of measure 0 are one of the fundamentals of the whole integration theory, so it is always wise to pay particular attention to their behaviour under certain transformations. The whole 1 + int + int^2 + … series intuitively really seems to work as an inverse of 1 - int over a special subspace of R^R functions, I think a good choice would be a space of polynomials over e^x and X (to leave no ambiguity: R[X, e^X]). It is all we need to prove this theorem, and these operators behave much more predictably in it. It would be nice to find a formal definition for the convergence of the series, but I can’t think of any metric that would scratch that itch.