Let $H$ be a complex Hilbert space and let $A:\mathrm{dom}(A)\to H$ be an unbounded symmetric operator with dense domain. Prove that $A$ is self-adjoint if and only if there is a $\lambda\in\mathbb{C}$ s.t. $\lambda I-A:\mathrm{dom}(A)\to H$ is surjective.

This is an excise, but now I doubt whether the other direction can be achieved. For $A$ self-adjoint, any nonreal $\lambda$ will make $\lambda I-A$ surjective since the spectrum of $A$ is a subset of $\mathbb{R}$. However for the other direction, I can only prove the case $\lambda I-A$ and $\overline{\lambda} I-A$ are both surjective (or at least one surjective one with dense image). I found many other books and they only have similar statement for the case in which both hold. Therefore I am wondering is it really possible that only $\lambda I-A$ surjective is sufficient?

Proof for the both-hold case: Since $A$ is densely defined and symmetric, $A^*$ is an closed extension of $A$. Therefore if we want to show $A$ is self-adjoint, we just need to show the inclusion from the other direction that $\mathrm{dom}(A^*)\subset\mathrm{dom}(A)$. Let $x\in \mathrm{dom}(A^*)$, since $\mathrm{Im}(\lambda I-A)=H$, there is a $y\in \mathrm{dom}(A)$ such that $$(\lambda I-A)y =(\lambda I- A^*)x\in H$$ Then for any $z\in\mathrm{dom}(A)$ we have $$\left<(\overline{\lambda} I-A)z,x\right>=\left<z,(\lambda I-A^*)x)\right> =\left<z,(\lambda I-A)y\right>=\left<(\overline{\lambda} I-A)z,y\right>$$ If $\overline{\lambda} I-A$ has dense image we will have $x=y$ and so $x\in\mathrm{dom}(A)$.

I can't think up any amelioration if we remove the condition. Any help will be appreciated.

  • $\begingroup$ You are right, surjectivity of $\lambda-A$ for only one $\lambda\in \mathbb{C}\setminus\mathbb{R}$ is not sufficient. A counterexample can be found here: math.stackexchange.com/questions/893899/… $\endgroup$ – MaoWao Dec 6 '18 at 19:00
  • $\begingroup$ Thank you so much for that link! $\endgroup$ – Apocalypse Dec 7 '18 at 12:20

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