Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 / 1

Let $R_z=(A-zI{)}^{-1}$ be compact and let $w$ be another resolvent point. The resolvent identity givesorThe bracket is bounded and the product of a bounded operator with a compact operator is compact. Thus compactness at one resolvent point implies compactness at every resolvent point.

Fix such a $z$. Spectral mapping for the bounded compact operator $R_z$ givesEvery nonzero spectral point of a compact operator is an isolated eigenvalue of finite multiplicity, and zero is its only possible accumulation point. Hence a compact resolvent operator has only isolated eigenvalues of finite multiplicity, with no finite accumulation point. The spectrum is allowed to be empty.

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For $f\in C_c^{\infty }(\mathbb{R})$, integration by parts givesAnother integration by parts removes the factor $x$ from the real cross term and yields the boundThe Sobolev interpolation estimate $\Vert f^{\prime }{\Vert }^2\le \epsilon \Vert f^{\prime \prime }{\Vert }^2+C_{\epsilon }\Vert f{\Vert }^2$ therefore impliesThus 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 $-f^{\prime \prime }+ixf\in L^2$, and cutoff followed by mollification approximates it in this graph norm. Hence

The operator is accretive becauseConsequently $T+I$ and its adjoint $T^{\ast }+I=-d^2/d{x}^2-ix+I$ 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 $\Vert g\Vert \le 1$, then $\Vert f\Vert \le 1$, and the graph estimate bounds $\Vert f^{\prime \prime }\Vert$ and $\Vert xf\Vert$. 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 $(U_{a}f)(x)=f(x-a)$,Thereforeso 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

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On $L^2({\mathbb{R}}^2)=L^2({\mathbb{R}}_x)\otimes L^2({\mathbb{R}}_y)$, takewith its natural tensor-product domain. For every $z\in \mathbb{C}$,so $\sigma (\mathcal{A})=\varnothing$. It is not compact: for fixed nonzero $f$ and any orthonormal sequence $(e_n)$ in $L^2({\mathbb{R}}_y)$, the resolvent imagesare nonzero, mutually orthogonal, and have equal norm, so no subsequence converges.

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