here's a claim made in Reed & Simon which I do not know how to prove :
Given $B$ a (positive) operator of some $L^2(M, d\nu)$ such that $\Vert B \Vert = 1$ and $1$ is an (non-degenerate) eigenvalue of $B$ (with stricly positive eigenvector). Then : $$ \Vert B^n - P_{\{1\}} \Vert \longrightarrow 0 $$ where $P_{\{1\}}$ is the orthogonal projection onto $Ker(B - Id)$. This claim appears in some proof where some hypothesis are made on B (those indicated in parenthesis). Maybe those are useless, but I've put them anyway. According to authors, the claim follows from functional calculus.
Any help is greatly appreciated.
One property of the functional calculus (see e.g. this book, theorem VII.2) is that if $B$ is a self-adjoint bounded linear operator, $f_n(x)\to f(x)$ for each $x\in \sigma(B)$ and $\|f_n\|_{\infty}=\sup_{x\in \sigma(B)}|f_n(x)|$ is bounded (uniformly in $n$) then $f_n(B)\to f(B)$ strongly.
Now let $f_n(x)=x^n$. since $\|B\|=1$ and $B$ is positive we have $\sigma(B)\subset [0,1]$ and therefore $$\|f_n\|_{\infty}\leq \sup_{x\in [0,1]}x^n=1 $$ and $$f_n(x)\to \chi_{\left\{1\right\}}(x) $$ for all $x\in \sigma(B)\subset [0,1]$.
(notice that here it makes perfect sense to say that the limit is $\chi_{\left\{1\right\}}$ and not $0$ because $x=1$ is actually an element of the spectrum, since by assumption it is an eigenvalue. But even if that were not the case, i.e. $1\notin \sigma(B)$, the thesis would still hold, and we would have $P_{\left\{1\right\}}=0$).
Hence the assumptions on the result are satisfied and we conclude that $f_n(B)=B^n\to \chi_{\left\{1\right\}}B=P_{\left\{1\right\}}$ strongly.