I need to find the spectrum of the following operator
$A: C[a, b] \rightarrow C[a, b], f \mapsto Af(x) = e^x f(x)$.
- What is the spectral set of $A$?
I know that all the $\lambda$ for which the resolvent $R_{\lambda} = (A – \lambda I)^{-1}$ is defined and continuous are the regular points and all other values of $\lambda$ is the spectrum of $A$.
I also know the following proposition. Prop. If A is a bounded linear operator mapping a Banach space into itself and $| \lambda | > ||A||$, then $\lambda$ is a regular point.
So I think that I can somehow use this proposition to show for which values of $\lambda$ we have regular points and this might help in eliminating which are not in the spectrum.
- Classify the spectrum points into point and continuous.
I know that the set of all eigenvalues i.e. for which $(A – \lambda I) = 0$ fails to exist for some $x \neq 0$ and the eigenvalues are called the point spectrum and the rest of the spectrum is the continuous spectrum.
- Is the operator compact?
I just know the definition that a linear operator A mapping a Banach space into itself is completely continuous iff. it maps every bounded set into a relatively compact set. But how to I prove that the given operator is compact?
Best Answer
(3) $A$ is continuously invertible with inverse $A^{-1}f = e^{-x}f$. If $A$ were compact, then $A^{-1}A=I$ would be compact as well, which is not the case because the unit ball in $C[a,b]$ is not compact.
(1) The spectrum $\sigma(A)$ of $A$ is contained in $S=\{ e^x : a \le x \le b \}$ because, for any $\lambda\notin S$, it is clear that $A-\lambda I$ is invertible with inverse $$ (A-\lambda I)^{-1}f = \frac{1}{e^x-\lambda}f(x) $$ To prove the converse, note that, for $\lambda\in S$, there exists $c\in [a,b]$ such that $\lambda=e^c$; so it is not hard to construct a sequence $\{ f_n \} \subseteq C[a,b]$ such that $\|f_n\|_{C[a,b]}=1$ and $(A-e^c I)f_n\rightarrow 0$. So $A-e^c I$ does not have a bounded inverse for any $c\in [a,b]$, which proves that $S\subseteq\sigma(A)$. Hence $S=\sigma(A)$.
(2) The spectrum is all continuous. There is no point spectrum.