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Let $A:H^n(0, \infty) \subset L^2(0, \infty) \to L^2(0, \infty)$ be the differential operator defined by $$Af:= \sum_{j=0}^na_j f^{(j)}$$ for all $f \in H^n(0, \infty),$ where $a_j \in \mathbb{C}$ and $a_n \neq 0$. I need to show that $$\sigma(A)= \{ P(iz) : \mbox{Im}(z) \geq 0\}$$ if $P(z):= a_0 + a_1z+\cdots+a_nz^n$. I have shown that $S:=\{ P(iz) : \mbox{Im}(z) \geq 0\} \subseteq \sigma(A)$ and that if $\lambda \notin S$ then $(A-\lambda I)^{-1}: R(A-\lambda I) \subseteq L^2(0, \infty) \to L^2(0, \infty)$ is a bounded operator and $R(A-\lambda I)$ is closed, but I am struggling to show that $R(A-\lambda I)=L^2(0, \infty)$ if $\lambda \notin S$.

Thanks for any help.

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  • $\begingroup$ When $\lambda \not\in \sigma(A)$, you expect the image of $A-\lambda I$ to be all of $L^2$. The image of the resolvent would then be all of the domain of $A$ (or more specifically the appropriate closure of $A$), which for unbounded operators is smaller than $L^2$ itself. I hope I didn't misunderstand what you wrote in the question. $\endgroup$
    – Igor Khavkine
    Commented Jul 29, 2019 at 1:51
  • $\begingroup$ @IgorKhavkine: I think that the notation $R(A-\lambda I)$ in the question means the range of $A-\lambda I$. The OP does not seem to use the letter $R$ to denote the resolvent. $\endgroup$
    – Jochen Glueck
    Commented Jul 29, 2019 at 4:44
  • $\begingroup$ @Mainkit: Have you tried to explicitly solve the equation $Au-\lambda u =f$ (where $f\in L^2$), and to show that the solution $u$ is in $H^n$ if $\lambda\not\in S$? $\endgroup$
    – Jochen Glueck
    Commented Jul 29, 2019 at 4:54
  • $\begingroup$ @JochenGlueck: I see, thanks! The notation was a bit confusing. But then, if one already knows that that $(A-\lambda I)^{-1}$ exists and is bounded, it's domain is already all of $L^2$. End of story. In a way, this is rephrasing of Jochen's last comment. $\endgroup$
    – Igor Khavkine
    Commented Jul 29, 2019 at 12:17
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    $\begingroup$ @IgorKhavkine But I don't know if $R(A-\lambda I)$ is dense in $L^2(0, \infty)$ $\endgroup$
    – Mike Van
    Commented Jul 29, 2019 at 16:00

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