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I am working with transfer operators and I reach a point where would be nice if I could use a result from Tosio Kato's book about perturbation theory of linear operators. I think I am able to use Kato's result if the question below has an affirmative answer. I asked this question to some of my colleagues in the math department, but no one was able to help me, so I decided to post it here in hope that some specialist point to me the way to answer it or a reference.

Let $(\Omega,d)$ be a compact metric space, $C(\Omega)$ the space of all real continuous functions and $T:C(\Omega)\to C(\Omega)$ a bounded linear positive operator ($T(f)\geq 0$ if $f\geq 0$) having an isolated positive eigenvalue $\lambda_M$ such that $\lambda_M>\sup\{|\beta|: \beta\in \sigma(T)\setminus\{\lambda_M\} \}$, where $\sigma(T)$ denotes the spectrum of $T$ and suppose additionally that the eigenspace associated to $\lambda_M$ is one-dimensional.

Assume that there is a Borel probability measure $\nu$ over $\Omega$ so that for all $u\in C(\Omega)$ we have $\|T(u)\|_{L^1(\nu)}\leq \lambda_M\|u\|_{L^1(\nu)}$. Since $C(\Omega)$ is dense in $L^1(\nu)$ we can naturally extend $T$ to a bounded linear positive operator on $L^1(\nu)$, which will be denoted by $\overline{T}$.

Question. Is it true that $\lambda_M>\sup\{|\beta|:\beta\in\sigma(\overline{T})\setminus\{\lambda_M\}\}$ ?

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  • $\begingroup$ $\overline{T}$ can be viewed as the operator closure of $T$, so if $\|\overline{T}x\| \ge (\lambda-\epsilon)\|x\|$, then there are $x_n\in C(\Omega)$ with $x_n\to x$, $Tx_n\to\overline{T}x$, and everything is clear now. $\endgroup$ Dec 19, 2015 at 0:17
  • $\begingroup$ Dear @ChristianRemling first of all thanks for your comment. From the hypothesis it follows that $T-(\lambda-\varepsilon)$ has bounded inverse and therefore $\exists \ c>0$ such that $c\|x\|_{\infty}\leq \| [T-(\lambda-\varepsilon)]x\|_{\infty}$. From your hint should I be able to conclude that there is a positive constant $d$ such that $d\|x\|_{L^1}\leq \| [T-(\lambda-\varepsilon)]x\|_{L^1}$ ? Could you elaborate a little bit more on your comment ? $\endgroup$
    – Leandro
    Dec 21, 2015 at 23:41
  • $\begingroup$ No, what I'm saying is that $x$ is a multiple of the (original) eigenvector. $\endgroup$ Dec 22, 2015 at 0:36
  • $\begingroup$ Thanks again @ChristianRemling but I don't understand why. I will think about it more. $\endgroup$
    – Leandro
    Dec 22, 2015 at 0:44

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