When is $\inf_{n\geq0}x^n\neq0$?

Question

Let $M$ be a von Neumann algebra acting on a Hilbert space $H$. Let $x$ be a positive element of $M$ with $\|x\|=1$. So, $(x^n)_{n\in\mathbb N}$ is a decreasing sequence of positive elements and $y:=\inf_{n\in\mathbb N}x^n$ does exist. My question is under what hypothesis we have $y\neq0$. When $H$ is finite dimensional, then obviously $y\neq0$. Does the assumption that $x$ is invertible imply $y\neq0$, in the case $H$ is infinite dimensional?

Answer 1

Basically, Borel functional calculus translates pointwise convergence of functions into convergence in the strong operator topology. Since the functions $(x^n)$ converge to the function $\mathbb{1}_{\{1\}}$, $y=\inf x^n$ is equal to the projection onto the eigenspace of $x$ corresponding to the eigenvalue $1$, i.e. it is non-zero iff $1$ is an eigenvalue of $x$. In particular, one can find invertible elements such that the infimum is equal to zero.