Let $A$ be a positive self-adjoint linear operator (not necessarily bounded) in the Hilbert space $\mathcal{H}$. Then $a=(1+A)^{-1}\in\mathcal{B}(\mathcal{H})$. For a given $n$, let us consider the projection $e_n:={\chi}_n(a)$ in $\mathcal{R}(\{a\})$ ($\mathcal{R}(\{a\})$ stands for the von Neumann algebra generated by the element $a$). Then both of $Ae_n$ and $ae_n$ are bounded self-adjoint operators in $\mathcal{R}(\{a\})$.
Problem: Prove that for $\lambda\neq0$, $\frac{1-\lambda}{\lambda}\in\sigma(Ae_n)\implies\lambda\in\sigma(ae_n)$.
P.S. I got stuck in this while reading Section $9.10$ of the book 'Lectures on von Neumann algebras' by Strătilă and Zsidó. More specifically, where they consider the map $\mathcal{B}(\sigma(Ae_n))\ni f\mapsto F_f(ae_n)\in\mathcal{B}(\mathcal{H})$. To make sense of this, we need to have $F_f$ defined on $\sigma(ae_n)$ for $f$ defined on $\sigma(Ae_n)$. But $F_f(\lambda):=\begin{cases} 0 &\text{ if } \lambda=0,\\ f((1-\lambda)/\lambda) &\text{ if } \lambda\in (0,1]. \end{cases}$. Hence I got the above problem. I was trying with the property that $\frac{1-\lambda}{\lambda}\in\sigma(Ae_n)\implies f((1-\lambda)/\lambda)\in\sigma(f(Ae_n))$ for any continuous function $f$ on $\sigma(Ae_n)$ but did not get anything. Thanks in advance for any help.
You have $$ e_n=1_{\big(\tfrac1{n+1},\infty\big)}(a)=1_{\big(\tfrac1{n+1},\infty\big)}((1+A)^{-1})=1_{(0,n)}(A). $$ Then $$ Ae_n=A\,1_{(0,n)}(A),$$ $$ae_n=(1+A)^{-1}\,1_{(0,n)}(A)=\Big(\tfrac1{1+t}\,1_{(0,n)}(t)\Big)(A). $$ So, if $\lambda\in\sigma(ae_n)$, then $\lambda=\tfrac1{1+\alpha}$ with $\alpha\in\sigma(A)\cap(0,n)$ and $$ \frac{1-\lambda}{\lambda}=\frac{1}{\lambda}-1=1+\alpha-1=\alpha. $$ Conversely, if $\tfrac{1-\lambda}{\lambda}\in\sigma(Ae_n)$, then $$ \lambda=\frac{1}{1+\frac{1-\lambda}{\lambda}}\in\sigma(ae_n). $$