= Solution
For $f\in C_c^\infty(\mathbb R)$, integration by parts gives
$$
\|Tf\|^2=\|f''\|^2+\|xf\|^2
+2\operatorname{Re}\langle-f'',ixf\rangle.
$$
Another integration by parts removes the factor $x$ from the real cross term and yields the bound
$$
|2\operatorname{Re}\langle-f'',ixf\rangle|
\leq2\|f'\|\|f\|.
$$
The Sobolev interpolation estimate $\|f'\|^2\leq\varepsilon\|f''\|^2+C_\varepsilon\|f\|^2$ therefore implies
$$
\boxed{\|f''\|^2+\|xf\|^2
\leq C(\|Tf\|^2+\|f\|^2)}.
$$
Thus convergence in the graph norm of the closure forces convergence in $H^2$ and of $xf$ in $L^2$. Conversely, $f\in H^2$ and $xf\in L^2$ clearly makes $-f''+ixf\in L^2$, and cutoff followed by mollification approximates it in this graph norm. Hence
$$
\boxed{D(T)=\{f\in H^2(\mathbb R):xf\in L^2(\mathbb R)\}}.
$$
The operator is accretive because
$$
\operatorname{Re}\langle Tf,f\rangle=\|f'\|^2.
$$
Consequently $T+I$ and its adjoint $T^*+I=-d^2/dx^2-ix+I$ are bounded below by one. The range of $T+I$ is both closed and dense, hence all of $L^2$, so $-1$ is a resolvent point. If $f=(T+I)^{-1}g$ with $\|g\|\leq1$, then $\|f\|\leq1$, and the graph estimate bounds $\|f''\|$ and $\|xf\|$. The compactness criterion in the question shows that $(T+I)^{-1}$ is compact. Thus the <Imaginary Airy operator> has compact resolvent.
For the unitary translation $(U_af)(x)=f(x-a)$,
$$
U_a^{-1}TU_a=T+iaI.
$$
Therefore
$$
\boxed{\|(T-(z+ia)I)^{-1}\|
=\|(T-zI)^{-1}\|},
$$
so the inverse resolvent norm is constant on every vertical line. The same unitary equivalence gives $\sigma(T)=\sigma(T)+ia$ for every real $a$. If the spectrum contained one point, it would contain its entire vertical line, contradicting the isolated-point spectrum forced by compact resolvent. Hence
$$
\boxed{\sigma(T)=\varnothing}.
$$
Solved by gpt-5.6-sol high.
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