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If $X$ is a Hilbert space with scalar product/norm of $Y$ (that is: complete, and thus closed in $Y$) the argument is fine: In this case $A$ (and equivalently $\beta I-A$) is closed, that is, its grap …

The spectrum (as a multivalued map) is upper semicontinuous with respect to the operator norm. With respect to strong convergence no analogous assertion is true, as can be seen in $\ell_2$ by the oper …

As Michael already observed: It is simple to construct counterexamples if $B$ has a nontrivial kernel $N(B)$ and $A$ maps $N(B)$ into itself, since on $N(B)$ the size of $c$ plays no role. But even if …

I do not have a counterexample, but a strong feeling that the conjecture is false, based on the following positive proof. If you require slightly more, namely Lebesgue integrability of $t\mapsto f(t,x …

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 …

Maybe the simplest classical example is a weakly singular kernel $$K(x,y) = |x-y|^{-\lambda}$$ with some fixed $\lambda\in(0,1)$. In this example $\int_{\mathbb R^2}K(x,y)^qdx=\infty$ for every $q>0$ …

In Biberdorf, E. A. and Väth, M., On the spectrum of orthomorphisms and Barbashin operators, Z. Anal. Anwendungen 18, 1999(4), 12-31 it is shown that even in the more general case of an orthomorphism, …

It should not be hard to prove - e.g. by some minmax-characterization - that $\lambda_1$ is measurable. (As already remarked in the comments by Christian Remling, $\lambda_1$ is actually continuous, b …