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Let $X$ be a Banach space and $T: X \rightarrow X$ is a bounded operator. The spectrum of $T$ is defined by $$\sigma(T) = \{\lambda \in \mathbb{C}; \lambda I - T ~~ \mbox{is not invertible}\}$$ It is well known that $\sigma(T)$ is a bounded set.

Now, let $S: X \rightarrow X$ be a bounded operator such that $S \neq 0$ and define the set $$\sigma_{S}(T) = \{\lambda \in \mathbb{C}; \lambda S - T ~~ \mbox{is not invertible}\}$$. My question is: the set $\sigma_{S}(T)$ is it bounded? and why?

salma
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1 Answers1

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Not necessarily. Take $H$ be a Hilbert space with orthonormal basis $\{e_n\}_{n\in\mathbb N}$. Define $T$ to be the identity and $$ Se_k=\frac{1}{k}e_k, \quad k\in\mathbb N. $$ Then for $\lambda=k$, $$ (\lambda S-T)e_k=kSe_k-Te_k=e_k-e_k=0. $$ Thus $\mathbb N\subset\sigma_S(T)$.

Yiorgos S. Smyrlis
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