Divergence of a series involving an operator [Part II]

Question

Let $T$ be an operator on an infinite dimensional Hilbert space. Assume $r$ is the spectral radius of $T$ and $r$ is also an isolated point spectrum of $T$, $\phi$ is a finite faithful normal trace on the operator algebra generated by $T$. Can we deduce that $$\sum_{n=1}^\infty \dfrac{\phi(T^n)}{nr^n}\to\infty?$$

Related Question: Part I

Answer 1

You didn't require $\phi$ to be faithful, but I assume that's the spirit of the question.

Take $H=L^2[0,1]$, $T=M_x$, i.e. multiplication by the identity function. Then $T$ is selfadjoint, $\|T\|=1$, $r=1\in\sigma(T)$. The von Neumann algebra generated by $T$ is $L^\infty[0,1]$. Let $\phi$ be integration. Then $$ \phi(T^n)=\int_{[0,1]} x^n dx=\frac1{n+1}, $$ so $$ \sum_{n=1}^\infty\frac{\phi(T^n)}{nr^n}=\sum_{n=1}^\infty\frac1{n(n+1)}=1 $$

If $r $ is an isolated point of the spectrum, and the only one with $|\lambda|=r $; and if $T $ is selfadjoint: we may assume $r=1$. Then $T=P+B $, with $P $ a projection and $\|B\|<1$. Also, $T^n=P+B^n $. Then $$ \sum_n\frac {\phi (T^n)}{n}=\sum_n\frac {\phi (B^n)}{n}+\sum_n\frac {\phi (P)}{n}. $$ The first sum converges absolutely because $|\phi (B^n)|\leq\|\phi\|\,\|B\|^n $. The second one is harmonic. So the series diverges.