Spectrum of symmetric, non-selfadjoint operator on Hilbert space I heard that any (unbounded) densely defined and symmetric operator $A: \text{dom}(A)\subset \mathcal{H}  \to \mathcal{H}$, which is not selfadjoint, has $\text{spec}( A )= \mathbb{C}$. $\mathcal{H}$ is here a Hilbert space.
Is this true? I didn't find anything in the internet.
Cheers
 A: Suppose $A$ is symmetric. Equivalently, assume $(Ax,x)$ is real for all $x\in\mathcal{D}(A)$.

Claim: If $A-\lambda I$ is surjective for some $\lambda\notin\mathbb{R}$, then $A$ is densely-defined.

Proof: Suppose $x \perp \mathcal{D}(A)$. If $(A-\lambda I)$ is surjective, then $(A-\lambda I)y=x$ for some $y\in\mathcal{D}(A)$, which gives
$$
         0 = \Im (x,y)=\Im((A-\lambda I)y,y)=-\Im\lambda\|y\|^{2}.
$$
Therefore $y=0$ and $x=(A-\lambda I)y=0$. So $\mathcal{D}(A)^{\perp}=\{0\}$. $\;\;\Box$

Claim: $|\Im\lambda|\|x\| \le \|(A-\lambda I)x\|$ for all $x \in \mathcal{D}(A)$ and $\lambda\in\mathbb{C}$.

Proof: This is just a mild extension of the above argument. For $x\in\mathcal{D}(A)$:
$$
       \Im ((A-\lambda I)x,x)= -\Im\lambda \|x\|^{2},\\
     |\Im\lambda|\|x\|^{2} \le \|(A-\lambda I)x\|\|x\|.
$$
The result follows by considering cases $x=0$ and $x\ne 0$ separately. $\;\;\Box$

Theorem: Let $A$ be closed and symmetric on a complex Hilbert space $X$. If $A-\lambda I$ is surjective for some $\lambda$ for which $\Im\lambda > 0$ (resp. $\Im\lambda < 0$), then the half plane $\Im\mu > 0$ (resp. $\Im\mu < 0$) is in the resolvent set of $A$.

Proof:
Suppose $(A-\lambda I)$ is surjective for some $\lambda\notin\mathbb{R}$. By the previous claim, $A$ is densely-defined and $R(\lambda)=(A-\lambda I)^{-1}$ exists and is bounded in norm by $\|R(\lambda)\| \le 1/|\Im\lambda|$. So $\lambda\in\rho(A)$. Also notice that
$$
        (A-\mu I)R(\lambda) = (A-\lambda I+(\mu-\lambda)I)R(\lambda)=I+(\mu-\lambda)R(\lambda).
$$
The right side is invertible if $|\mu-\lambda| < 1/|\Im\lambda|$ which means that $(A-\mu I)$ is also surjective for such $\mu$ because
$$
           (A-\mu I)R(\lambda)\{I+(\mu-\lambda)R(\lambda)\}^{-1}=I.
$$
Therefore, if $\lambda \in \rho(A)$, it follows that the largest open disk centered at $\lambda$ which does not touch the real axis is also included in $\rho(A)$. In this way, one gets ever expanding disks included in the resolvent until the entire half space containing the original $\lambda$ is found to be in the resolvent set. $\;\;\Box$
If $A$ has both open half spaces in its resolvent set, then $A=A^{\star}$. So, if $A$ is symmetric but not selfadjoint, then the spectrum is the closed upper half plane, the closed lower half plane, or all of $\mathbb{C}$.
Example: The classic example of spectrum in a half plane is $Af = if'$ considered on the domain $\mathcal{D}(A)$ consisting of all aboslutely continuous $f \in L^{2}[0,\infty)$ for which $f'\in L^{2}[0,\infty)$ and $f(0)=0$. We don't have to check that the domain is dense, and that's a relief. Just check that $A$ is symmetric. First notice that for $f \in \mathcal{D}(A)$, one has $(f^{2})'=2ff' \in L^{1}$, which guarantees the existence of $\lim_{x\rightarrow\infty} f^{2}(x)$. That limit must be $0$ in order for $f \in L^{2}$. Therefore, if $f,g\in\mathcal{D}(A)$,
$$
    (Af,g)-(f,Ag)=\left. i\int_{0}^{\infty}(f\overline{g})'\,dx = f\overline{g}\right|_{0}^{\infty} = 0.
$$
Now, we show that $(A-\lambda I)$ is surjective for $\Im \lambda < 0$. So, let $g \in L^{2}[0,\infty)$ be given. We must solve
$$
                     if'-\lambda f = g,\;\;\; f(0)=0,\;\;\; f,f'\in L^{2}[0,\infty).
$$
The ODE $f'+i\lambda f=-ig$, $f(0)=0$, has an integrating factor $e^{i\lambda t}$ leading to the classical solution
$$
                    e^{i\lambda x}f(x) = -i\int_{0}^{x}e^{i\lambda t}g(t),\\
                  f(x) = -i\int_{0}^{x}e^{-i\lambda(x-t)}g(t)\,dt.
$$
What needs to be checked is that $f, f' \in L^{2}[0,\infty)$. This solution has a chance at being in $L^{2}$ because $\Re(-i\lambda)=\Im\lambda < 0$. $f'=-i\lambda f-ig$ is in $L^{2}$ automatically once it is shown that $f\in L^{2}$. So, check:
$$
\begin{align}
    |f(x)|^{2} & \le \left(\int_{0}^{x}e^{\Im\lambda(x-t)}|g(t)|\,dt\right)^{2} \\
        & \le \int_{0}^{x}e^{\Im\lambda(x-t)}\,dt\int_{0}^{x}e^{\Im\lambda(x-t)}|g(t)|^{2}\,dt \\
        & \le \frac{1}{|\Im\lambda|}\int_{0}^{x}e^{\Im\lambda(x-t)}|g(t)|^{2}\,dt.
\end{align}
$$
Now integrate
$$
\begin{align}
        \int_{0}^{\infty}|f(x)|^{2}\,dx & \le \frac{1}{|\Im\lambda|}\int_{0}^{\infty}\int_{0}^{x}e^{\Im\lambda(x-t)}g(t)\,dt\,dx \\
      & = \frac{1}{|\Im\lambda|}\int_{0}^{\infty}\int_{t}^{\infty}e^{\Im(x-t)}\,dx|g(t)|^{2}\,dt \\
      & = \frac{1}{|\Im\lambda|^{2}}\int_{0}^{\infty}|g(t)|^{2}\,dt =\left(\frac{1}{|\Im\lambda|}\|g\|\right)^{2} < \infty.
\end{align}
$$
Therefore $(A-\lambda I)$ is surjective for $\Im\lambda < 0$, which guarantees that the domain of $A$ is dense in $L^{2}[0,\infty)$. And it guarantees that the open lower half plane is in the resolvent set $\rho(A)$. You cannot get a similar solution in $L^{2}$ for $\Im\lambda > 0$ because you can show that the adjoint of $A$ is $if'$ without the restriction $f(0)=0$. So $A\ne A^{\star}$, which forces
$$     \sigma(A)=\{ \lambda : \Im\lambda \ge 0 \}. $$
