Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 (Spectral Computations in Infinite Dimensions)

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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Choose normalized eigenvectors $u_n$ of the finite compressions and embed them as $v_n=P_n^{\ast }{u}_n\in H$. ThenThe bounded sequence $(v_n)$ has weakly convergent subsequences. If $v_{n_j}\rightharpoonup v$, then for every $y\in H$, strong convergence $P_n^{\ast }{P}_{n}y\to y$ and the compressed eigenvalue equation giveBecause $\lambda \notin \sigma (A)$, this forces $v=0$. Every weak cluster point is zero, so $v_n\rightharpoonup 0$. Since $⟨A{v}_n,v_n⟩={\lambda }_n\to \lambda$, the weak-null characterization givesThus finite-section spectral pollution of a bounded operator can occur only in its essential numerical range.

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LetThe tail spaces decrease, so $C_{n+1}\subseteq C_n$. If $\lambda \in {\bigcap }_n{C}_n$, choose a unit vector $v_n$ supported after coordinate $n$ with $|⟨A{v}_n,v_n⟩-\lambda |<1/n$. Such vectors converge weakly to zero, hence $\lambda \in W_e(A)$.

Conversely, if $v_j\rightharpoonup 0$ and $⟨A{v}_j,v_j⟩\to \lambda$, then for every fixed $n$ the first $n$ coordinates of $v_j$ tend to zero. After normalizing $Q_n{v}_j$, its numerical values still tend to $\lambda$, so $\lambda \in C_n$. Therefore

When the intersection is nonempty, decreasing closed sets have distance functions increasing pointwise to the distance from their intersection; on each compact set this convergence is uniform. This is preciselyIf $W_e(A)=\varnothing$ and a compact $K$ met every $C_n$, nestedness and compactness would supply a convergent sequence whose limit belongs to all $C_n$, a contradiction. Hence $K\cap C_n=\varnothing$ for all sufficiently large $n$.

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Let $H_0=-d^2/d{x}^2+x^2$. The Hermite expansion givesSince $T=-d^2/d{x}^2+i{x}^2$,and therefore

Write $X=\Vert xf{\Vert }^2$ and $S={\sum }_{m=n+1}^{\infty }(2m+1)|c_m{|}^2$. Then $\mathrm{Re}⟨Tf,f⟩=S-X$, $\mathrm{Im}⟨Tf,f⟩=X$, andIf $|z|\le n$, then $|\mathrm{Re}z+\mathrm{Im}z|\le \sqrt{2}n<2n+3$, soEvery fixed compact set is eventually excluded from the tail numerical ranges. Part (b) therefore implies

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The spectral theorem for normal operators on a separable Hilbert space states that a normal operator $A$ has a unique projection-valued measure $E$ on its spectrum such thatFor bounded $A$ this integral acts on all of $H$. For an unbounded normal operator,Equivalently, $A$ is unitarily equivalent to multiplication by a measurable function on a direct sum of $L^2$ spaces.

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By the spectral theorem, the operator in parentheses acts at spectral value $t\in \mathbb{R}$ byThe Poisson kernel converges to $1$ for $t\in (a,b)$, to $1/2$ at $t=a,b$, and to $0$ outside $[a,b]$. It is uniformly bounded, so dominated convergence in the spectral measure of a normal operator gives the strong limitApplying this operator to $v$ proves the claimed Stone formula.

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Let $p(x)=x^3-x$. Since $A$ is multiplication by $p$, its spectral projection is multiplication by $1_{p^{-1}(S)}$. HenceSplit $[-1,1]$ at the two critical points $\pm 1/\sqrt{3}$. On each resulting interval, $p$ is monotone, so one-dimensional change of variables shows that the measure is absolutely continuous. For almost every $t$,The density vanishes outsideIts inverse-square-root singularities at the two critical values are locally integrable, so they do not create singular spectral measure.

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