# Is this statement still true with a weaker condition?

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 $$\left<(\overline{\lambda} I-A)z,x\right>=\left =\left=\left<(\overline{\lambda} I-A)z,y\right>$$ 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.

• 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/… – MaoWao Dec 6 '18 at 19:00
• Thank you so much for that link! – Apocalypse Dec 7 '18 at 12:20