# spectrum of an operator and its point and continuous spectrum

I need to find the spectrum of the following operator

$$A: C[a, b] \rightarrow C[a, b], f \mapsto Af(x) = e^x f(x)$$.

1. 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.

1. 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.

1. 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?

• I do not think that you need any special result to calculate the spectrum, just find out what $A - \lambda I$ is. It does not seem to me that difficult to see for which $\lambda$ this operator is not bijective. – Matthias Klupsch Jan 20 '20 at 8:16
• I just wanted to edit the post with what I found. I found that $(A - \lambda I)f(x) = (e^x - \lambda)f(x)$ and then $(A - \lambda I)^{-1} = \frac{1}{e^x - \lambda}f(x)$. Hence the spectrum consists of all $\lambda$ for which $e^{x} - \lambda$ vanishes? – user672245 Jan 20 '20 at 8:19
• Hint: Don't forget $[a, b]$. – Jacky Chong Jan 20 '20 at 8:21
• @JackyChong The range of $e^x$ for some $x \in [a, b]$? – user672245 Jan 20 '20 at 8:26
• For example, suppose $[a, b]=[0, 1]$, then is $e^x-\pi$ invertible on $[0, 1]$? – Jacky Chong Jan 20 '20 at 8:29

(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 $$S=\sigma(A)$$.
• I am stuck on the "it is not hard to construct a sequence ... " part. What sequence? and why does this suffice to show that $S \subset \sigma(A)$? – user672245 Jan 20 '20 at 9:06
• Create a hat function $f_{c,\epsilon}$ centered at $e^c=\lambda$ and supported in $[c-\epsilon,c+\epsilon]$ and let $\epsilon \downarrow 0$. Then $(A-e^c)f_{c,\epsilon}\rightarrow 0$ as $\epsilon\downarrow 0$. – Disintegrating By Parts Jan 20 '20 at 16:05