Closed operator with trivial resolvent set For a separable Hilbert space $H$, and 
$$
D:{\frak dom}(D) \subseteq H \to H
$$
a closed operator, it is possible that $D$ has trivial resolvent. In the other words, is it possible that $T - \lambda$id$_H$ does not admit a bounded inverse, for all $\lambda \in \mathbb{C}$.
Moreover, we know that if $T$ is self-adjoint, then it has non-empty resolvent since, for example, $i$ cannot be in the spectrum. However, if $T$ is merely closed and symmetric, or just symmetric, will the resolvent still be empty?
 A: Another example on $L^2[a,b]$ is $Lf=if'$ defined on the domain $\mathcal{D}(L)$ consisting of all absolutely continuous functions $f$ on $[a,b]$ such that $f'\in L^2$ and $f(a)=f(b)=0$. It is easy to check that the operator $L$ is symmetric by using integration by parts.
Claim: $\sigma(L)=\mathbb{C}$:
Proof: The resolvent equation $(L-\lambda I)f=g$ is equivalent to solving for $f\in L^2[a,b]$ such that
$$
              if'-\lambda f=g,\;\;\; f(a)=f(b)=0.
$$
The ODE $if'-\lambda f=g$ or $f'+i\lambda f=-ig$ or $(e^{i\lambda x}f)'=-ie^{i\lambda x}g$ has a unique solution $f$ for which $f(a)=0$, and it is given by
$$
             f(x)=-e^{-i\lambda x}\int_{a}^{x}ie^{i\lambda t}g(t)dt.
$$
However, this is not a full solution unless $\int_{a}^{b}e^{i\lambda t}g(t)dt=0$, which does not happen for all $g$. In fact, this last condition is equivalent to
$$
        \mathcal{R}(L-\lambda I)= [\{ e^{-i\overline{\lambda}t} \}]^{\perp}.
$$
So the range of $L-\lambda I$ is of co-dimension $1$ for all $\lambda\in\mathbb{C}$, which proves the claim. $\;\;\blacksquare$
A: Unbounded operators can have empty resolvent set. Consider the unbounded operator on $L^2\left(\mathbb{R}^2\right)$ given by multiplication by $x_1+ix_2$, with domain given by all $u\in L^2(\mathbb{R^2})$ such that $(x_1+ix_2)u\in L^2\left(\mathbb{R}^2\right).$ I'll leave it to you to verify that is an operator with empty resolvent.
