this post was submitted on 22 Jan 2024
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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)
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.
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.