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We have that $\lambda, \overline{\lambda} \in \rho(T)$ and $\lambda \in \mathbb{C}$. Now, I want to show that a symmetric operator and closed operator $T: \operatorname{dom(T)} \rightarrow H$ must be self-adjoint. Notice, that $T$ is not necessarily densily defined. Does anybody here have any ideas? Actually, I concluded the closedness of this operator from the fact that the resolvent is not empty by myself, so this may be somehow a tautology in this exercise.

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    $T$ must be densely defined, or it doesn't have an adjoint. – Daniel Fischer Nov 04 '14 at 22:19
  • the point is that you should probably show that this includes that $T$ is densely defined. –  Nov 04 '14 at 22:20
  • If $T$ is symmetric, then $T\subset T^\ast$ (the definition of being symmetric), which implies that $T^\ast$ exists. Which implies that $T$ is densely defined. – Daniel Fischer Nov 04 '14 at 22:23
  • Don't get it. Don't you need the fact that $T$ is densely defined a priori, in order to know that $T^*$ exists uniquely? ( I also don't quite see the duplicate) –  Nov 04 '14 at 22:24
  • One of your premises is that $T$ is symmetric. That implies that $T$ must be densely defined, since otherwise $T^\ast$ doesn't exist. So the denseness of $\operatorname{dom} T$ is part of the premises, though implicit. – Daniel Fischer Nov 04 '14 at 22:27
  • could you explain, how this follows? I mean symmetric means $\forall x,y \in \operatorname{dom}(T): \langle Tx,y\rangle = \langle x,Ty\rangle$. How does this imply the existence of $T^*$? In my functional analysis textbook it says that you can only define the adjoint, if you know that your operator is densely defined, so your argument sounds strange to me. –  Nov 04 '14 at 22:29
  • Sigh. A different definition of symmetric :( But T.A.E.'s answer at the duplicate handles that. – Daniel Fischer Nov 04 '14 at 22:32
  • could you first try to explain how my assumption is related to $R(A+i)=H$? ( There should be one, as you insist on the fact that my question is a duplicate ;-) )... all I get is $(T-\lambda)^{-1}$ has only zero in the nullspace, thus the image of $((T-\lambda)^{-1})^{*}$is dense. –  Nov 04 '14 at 22:35
  • You have $\lambda, \overline{\lambda} \in \rho(T)$ - I assume that indicates $\lambda \notin \mathbb{R}$, otherwise we have a problem. Reducing that to the case $\lambda = i$ is done by adding a real multiple of the identity, and scaling. – Daniel Fischer Nov 04 '14 at 22:37
  • no, $\lambda , \overline{\lambda} \in \mathbb{C}$ is what the exercise wants me to do. Thus, $\lambda \in \mathbb{R}$ is not forbidden I think. –  Nov 04 '14 at 22:41
  • I have the inkling that it doesn't follow for $\lambda\in\mathbb{R}$, but I'm not sure. – Daniel Fischer Nov 04 '14 at 22:52
  • counterexamples would even be better, as this would complete my exercise even faster ;-) –  Nov 04 '14 at 22:54
  • I'm trying to think of one. – Daniel Fischer Nov 04 '14 at 22:54

1 Answers1

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If $A$ is symmetric and $(A-\lambda I)$ is surjective for for some $\lambda\notin\mathbb{R}$, then the domain of $A$ is automatically dense. To see this, assume $A-\lambda I$ is surjective for some non-real $\lambda$ and suppose that $y \perp \mathcal{D}(A)$. Then $y=(A-\lambda I)x$ for some $x \in \mathcal{D}(A)$, which gives $$ 0=(y,x)=((A-\lambda I)x,x)=((A-\Re\lambda I)x,x)+i\Im\lambda(x,x). $$ Because $A$ is symmetric, then $(Ax,x)=(x,Ax)$ must be real. Therefore, $$ 0 = \Im ((A-\lambda I)x,x) = \Im\lambda\|x\|^{2} \implies x = 0\implies y = 0. $$ The conclusion is that $A$ must be densely-defined if $A$ is symmetric and if $A-\lambda I$ is surjective for some $\lambda \notin \mathcal{R}$. Therefore $A^{\star}$ is closed and densely-defined with $A\preceq A^{\star}$, meaning that $A$ is a restriction of $A^{\star}$ to $\mathcal{D}(A)$.

Now assume $A$ a densely-defined symmetric operator. Then $A$ is closable, which implies that $A^{\star}$ is closed and densely-defined, even if $A$ is not closed. Assuming $A-\lambda I$ and $A-\overline{\lambda}I$ are surjective for some $\lambda$, I'll now show that $A=A^{\star}$ (here $\lambda$ can be real.) We know that $A\preceq A^{\star}$ because $A$ is symmetric; so it is enough to show that $y \in \mathcal{D}(A^{\star})$ implies that $y \in \mathcal{D}(A)$. To prove this, assume $y\in\mathcal{D}(A^{\star})$ and choose $z \in \mathcal{D}(A)$ such that $$ (A-\overline{\lambda}I)z = (A^{\star}-\overline{\lambda})y. $$ Then, for all $x\in\mathcal{D}(A)$ one has $$ ((A-\lambda I)x,y)=(x,(A^{\star}-\overline{\lambda}I)y)=(x,(A-\overline{\lambda})z)=((A-\lambda I)x,z),\\ ((A-\lambda I)x,y-z) = 0,\;\;\; x \in \mathcal{D}(A). $$ Because $A-\lambda I$ is also surjective, then $y = z \in \mathcal{D}(A)$, as was to be shown.

Added: If you assume that $A$ is densely-defined and symmetric with $A-\lambda I$ surjective for some real $\lambda$, then $A=A^{\star}$. That's because the arguments of the second paragraph above remain valid. Alternatively, if $A-\lambda I$ is surjective for some $\lambda$ for which $((A-\lambda)x,x) > 0$ for all non-zero $x \in\mathcal{D}(A)$, then $A=A^{\star}$ is a closed densely-defined selfadjoint operator.

Disintegrating By Parts
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  • what do you think about the condition $\lambda \notin \mathbb{R}$. is this necessary? –  Nov 04 '14 at 23:19
  • The first part of my argument does not work if $\lambda \in \mathcal{R}$; I'm not sure if the result holds either in that case. However, if $A$ is non-negative and $(A+\lambda I)$ is surjective for some real $\lambda$, then it is true that $A=A^{\star}$ because $((A+\lambda I)x,x)\ge \lambda|x|^{2}$ holds, and the first part of the argument is okay. Then the second part also holds, too, because that argument works whether or not $\lambda$ is real (I changed the answer to reflect that in the second part of the argument.) – Disintegrating By Parts Nov 04 '14 at 23:28