0
$\begingroup$

If‎ ‎$T\in B(H)$ is ‎invertible ‎and ‎for ‎all $n\geq 1$,‎ ‎$\|T^n\|\leq M$, ‎then ‎‎‎$\sigma(T)\subset\{\lambda\in\mathbb{C}:|\lambda|\leq 1\}$.

‎Here, $\sigma (‎ T‎ )$‎ ‎‎‎is the spectrum of $T$ and $B(H)$ is the set of bounded operators on the Hilbert space $H$.

Do you think there is ‎equality?

  • 0
    Without loss of generality, assume $M > 0$. The spectral radius is $r_{\sigma}(T) =\lim_n \|T^n\|^{1/n} \le\lim_n M^{1/n}=1$. – 2017-01-24

2 Answers 2

2

Let $\lambda \in \mathbb{C}$ with $|\lambda| > 1$. Let $S=T/\lambda$. Then for all $k$ we have that $\|S^k\| \leq \frac{M}{\lambda^k}$ and since $|\lambda|>1$ we have that the geometric series $\sum\frac{M}{\lambda^k}$ converges. Therefore the Neumann series $\sum S^k$ converges in the operator norm.

This gives that $(I - S)=(I-T/\lambda)$ is invertible $\Rightarrow (\lambda I-T)$ is invertible and thus $\lambda\not\in\sigma(T)$.

What do you mean by equality? If you mean $\sigma(T) = \{\lambda\in\mathbb{C}:|\lambda| \leq 1\}$ this is definitely not true. In general $\sigma(T)$ can be any compact subset of $\mathbb{C}$. So for example consider the fact that $\sigma(\lambda I) = \{\lambda\}.$

1

Let $r(A)$ denote the spectral radius of $A\in B(H)$. Note that $\sigma(T^n)=\sigma(T)^n$, hence $r(T^n)=r(T)^n$ for all $n$. As spectral radius is always bounded above by operator norm, $r(T)^n=r(T^n)\leq \|T^n\|$ for all $n$. The hypothesis thus implies that the set $\{r(T)^n:n\in\mathbb N\}$ is bounded, which implies that $r(T)\leq 1$.

Invertibility isn't relevant for this result. Having invertibility means $0$ is not in the spectrum, and the possible spectra of $T$ are the compact subsets of $\{z\in\mathbb C:0<|z|\leq 1\}$.

If you had $\{\|T^n\|:n\in\mathbb Z\}$ bounded (instead of only nonnegative powers), it would imply that $\sigma(T)\subseteq\{z\in\mathbb C:|z|=1\}$. This follows from the above result combined with $\sigma(T^{-1})=\sigma(T)^{-1}$.