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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?

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  • $\begingroup$ 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. $\endgroup$ – Matthias Klupsch Jan 20 '20 at 8:16
  • $\begingroup$ 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? $\endgroup$ – user672245 Jan 20 '20 at 8:19
  • $\begingroup$ Hint: Don't forget $[a, b]$. $\endgroup$ – Jacky Chong Jan 20 '20 at 8:21
  • $\begingroup$ @JackyChong The range of $e^x$ for some $x \in [a, b]$? $\endgroup$ – user672245 Jan 20 '20 at 8:26
  • $\begingroup$ For example, suppose $[a, b]=[0, 1]$, then is $e^x-\pi$ invertible on $[0, 1]$? $\endgroup$ – Jacky Chong Jan 20 '20 at 8:29
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(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.

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  • $\begingroup$ I am stuck on the "it is not hard to construct a sequence ... " part. What sequence? and why does this suffice to show that $S \subset \sigma(A)$? $\endgroup$ – user672245 Jan 20 '20 at 9:06
  • $\begingroup$ 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 $\epsilon\downarrow 0$. $\endgroup$ – Disintegrating By Parts Jan 20 '20 at 16:05
  • $\begingroup$ @Crimenation : I forgot to put your name into my comment. $\endgroup$ – Disintegrating By Parts Jan 20 '20 at 16:17

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