Let $B[0,1]$ be the Banach space of bounded complex functions on $[0,1]$ endowed with the supremum norm. I've have to show that the spectrum of the multiplication operator $T_q: B[0,1] \rightarrow B[0,1]$

$$ (T_q f)(t) : = q(t)f(t), \,\,\, t \in [0,1] $$ is $ \sigma(T_q)=\overline{\{q(t):t\in[0,1]\}}, \,\,. \text{for each}\,\, q \in B[0,1]. $

My question is why do we need the closure?

  • $\begingroup$ The spectrum needs to be closed, while the range of $q$ isn't necessarily (it is when $q$ is continuous). $\endgroup$ – Davide Giraudo Nov 4 '13 at 14:18

If $T$ is a linear operator on a (complex) Banach space $X$, then the resolvent set $\rho(T)$ is the collection of all complex numbers $\lambda$ such that $\lambda - T$ is a bounded invertible map $X \to \operatorname{Dom}(T)$. An important aspect of linear operators on Banach spaces is that their resolvent set is an open subset of $\mathbb{C}$. In fact, you learn that for $\lambda \in \rho(T)$, then the resolvent map $\lambda \mapsto (\lambda - T)^{-1}$ is analytic in $\lambda$ (that is, can be expressed as a power series within a sufficiently small ball around $\lambda$). This applies to your problem since the resolvent $\rho(T)$ is, in fact, the compliment of the spectrum $\sigma(T)$, implying that $\sigma(T)$ is closed. In your case, if $\lambda \in \{ q(s) : s \in [0,1]\}$ then suppose $q(t) = \lambda$. Now, let $f \in B[0,1]$ and $g \in B[0,1]$ be related by $f(s) = g(s)$ for every $s \neq t$ and $g(t) \neq f(t)$. Then for every $s \in [0,1]$ you have $(\lambda -q(s))f(s) = (\lambda-q(s))g(s)$. This means that $\lambda \in \sigma(T_q)$ since $\lambda - T_q$ is not invertible. Since $\sigma(T_q) = \overline{(\sigma(T_q))}$ (since it is closed) this implies that $\overline{\{q(t):t \in [0,1]\}} \subset \sigma(T_q)$. However, if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda - q(s) \neq 0$ for any $s \in [0,1]$. In fact, there is some $\delta > 0$ such that $\sup_{s\in [0,1]} \| \lambda - q(s) \| > \delta$. It is easy enough now to see that $(\lambda - T_q)^{-1}f(s) = (\lambda - q(s))^{-1} f(s)$ and that $\| (\lambda - T_q)^{-1}\| \leq \delta^{-1} < \infty$. This means that if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda \in \rho(T_q)$ and you have $\sigma(T_q) \subset \overline{\{q(t):t \in [0,1]\}}$ from which your question follows.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.