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1 vote

Conditions for continuity of an integral functional

Since $\nu$ does not appear anywhere else in the question, I suppose that $L^1(X)=L^1(\nu)$. In order that the functional be defined, one should then assume (probably without loss of generality) that …
Martin Väth's user avatar
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1 vote

Structure of the inverse of a Fredholm integral operator of the second kind

Not really an answer, but some remarks: Even if the spectral radius of $K$ is less than $1$, there are counterexamples that the resolvent need not have the required form: This is related to the fact …
Martin Väth's user avatar
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