Strong limit of operator to spectral projection 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.
 A: 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.
