Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 358 (Spectral Computation in Infinite Dimensions)

Let $M={\max }_{S}u$ and suppose $u(z_0)=M$. The sub-mean inequality on every closed disk contained in $S$ givesEquality holds throughout. If $u$ were strictly below $M$ at one point of the circle, upper semicontinuity would make it uniformly below $M$ on a small arc, contradicting equality of the average. Thus $u=M$ on every sufficiently small circle centered at $z_0$, and hence throughout a neighborhood of $z_0$.

The set $\{u=M\}$ is therefore open. It is also closed because upper semicontinuity makes $\{u<M\}$ open. Since the domain is connected and $\{u=M\}$ is nonempty, it is all of $S$. This proves the maximum principle for subharmonic functions.

LetThe resolvent set is open, and the resolvent of an element is operator-valued holomorphic on each of its components. Fix $z_0$ in the resolvent set. For any $\delta >0$, choose unit vectors $x,y$ such thatThe scalar function $f(z)=⟨R(z)x,y⟩$ is holomorphic. Its modulus is subharmonic, soLetting $\delta \downarrow 0$ proves that the resolvent norm is subharmonic on the resolvent component containing $z_0$.

Use the convention that the reciprocal resolvent norm is zero on the spectrum. Suppose a bounded component $U$ ofcontained no spectral point. A spectral point in its boundary would belong to the same pseudospectral component, so $\overline{U}$ is contained in the resolvent set. On $\partial U$ one has $\Vert R(z)\Vert \le {ϵ}^{-1}$, whereas inside $U$ one has $\Vert R(z)\Vert >{ϵ}^{-1}$. Continuity on the compact set $\overline{U}$ makes the resolvent norm attain a maximum at an interior point. The subharmonic maximum principle would make it constant, contradicting its boundary values. Hence every bounded component of the pseudospectrum contains spectrum:

Factor the perturbed operator on $D(A)$ asSincethe Neumann series makes $I+B{A}^{-1}$ invertible. Therefore $0\notin \mathrm{Sp}(A+B)$ andThe geometric-series bound gives

Now choose a bounded open neighborhood $U$ of the isolated spectral component $X$ such thatand $\overline{U}$ meets no other component of the spectrum. Compactness of $\partial U$ givesFor all sufficiently large $n$, $\Vert B_n\Vert M<1$. Applying the first part to $A-zI$ shows uniformly that $\partial U\subset \rho (A+B_n)$.

The corresponding Riesz projections areThe resolvent identity and the uniform Neumann bound imply $\Vert P_n-P\Vert \to 0$. The projection $P$ is nonzero because $U$ contains the nonempty spectral component $X$. Projections at distance less than one have isomorphic ranges, so $P_n\ne 0$ for large $n$. Therefore $A+B_n$ has spectrum inside $U$, and any such point $z$ satisfies $\mathrm{dist}(z,X)<ϵ$. Thusfor every sufficiently large $n$.

If $A$ has finite rank, the image of its unit ball is bounded in the finite-dimensional space $\mathrm{ran}A$. The closure of a bounded set in a finite-dimensional normed space is compact. Hence every bounded finite-rank operator is a compact operator.

Let $Q_n=P_n^{\ast }{P}_n$ be the orthogonal projection ontoThen $Q_n\to I$ strongly. Strong convergence is uniform on every compact subset: if $K$ is compact, cover it by finitely many small balls and use $\Vert I-Q_n\Vert \le 1$ at their centers. Since the closure of $A$ applied to the unit ball is compact,The adjoint of a compact operator is compact, so the same argument for $A^{\ast }$ givesSince $A_n=Q_{n}A{Q}_n$,ThereforeThis is the finite-section approximation of a compact operator.

Write $Q_n=P_n=P_n^{\ast }$ for the orthogonal projections and regard the compression $A_n=Q_{n}A{Q}_n$ as acting on $\mathrm{ran}{Q}_n$. The compactness argument from part iii applies to any strongly convergent sequence of orthogonal projections, so

We first rule out spectral pollution. Suppose ${\lambda }_n\in \mathrm{Sp}(A_n)$ and, after taking a subsequence, ${\lambda }_n\to \lambda \ne 0$. Choose unit eigenvectors $x_n\in \mathrm{ran}{Q}_n$:Compactness gives a convergent subsequence of $A{x}_n$. Becausethe relation $x_n={\lambda }_n^{-1}{A}_n{x}_n$ then makes $x_n$ converge to a nonzero vector $x$, and passage to the limit gives $Ax=\lambda x$. Thus every nonzero limit of finite-section spectral points belongs to $\mathrm{Sp}(A)$. The only remaining possible limit is zero, which belongs to the spectrum of a compact operator on an infinite-dimensional space.

Conversely, the Riesz–Schauder theorem says that every nonzero $\lambda \in \mathrm{Sp}(A)$ is an isolated eigenvalue of finite algebraic multiplicity. Put a small contour around $\lambda$ containing no other point of $\mathrm{Sp}(A)$. Norm convergence of $A_n$ gives uniform resolvent convergence on the contour, so the associated Riesz projections converge in norm and eventually have the same positive rank. Hence $\mathrm{Sp}(A_n)$ meets every neighborhood of $\lambda$.

Finally, zero is also approximated. Otherwise some subsequence would have all its eigenvalues bounded away from zero. Outside any small disk, $\mathrm{Sp}(A)$ has only finitely many eigenvalues, and the preceding Riesz-projection argument fixes the total algebraic multiplicity of nearby finite-section eigenvalues. This cannot account forThus finite-section eigenvalues also approach zero. Both directed spectral distances vanish, proving

Put $T=A-zI$. We use the following closed-range lemma:Indeed,Away from this kernel, zero is separated from the spectrum of the positive self-adjoint operator $T^{\ast }T$ exactly whenfor some $c>0$. This is equivalent to closed range. Zero is then absent from the essential spectrum exactly when $\dim \ker T<\infty$. Similarly,because $\ker T^{\ast }=(\mathrm{ran}T{)}^{\perp }$ measures the cokernel.

Applying the lecture definitions of the three essential spectra of a closed operator now givesand

If $A$ is normal, so is $T$. The spectral theorem givesand maps the spectral mass of $T$ at $0$ exactly to the spectral mass of $T^{\ast }T$ at $0$. Thus zero is isolated with finite multiplicity for $T^{\ast }T$ exactly when $z$ is an isolated eigenvalue of finite multiplicity for $A$. Consequently

Normality is essential. Let $U{e}_j=e_{j+1}$ be the unilateral shift and take $A=U^{\ast }$. Its spectrum is the closed unit disk, so $0$ is not in the discrete spectrum. Butwhose zero eigenvalue is isolated and simple. Hence $0\in {\mathrm{Sp}}_{\mathrm{d}}(A^{\ast }A)$ while $0\notin {\mathrm{Sp}}_{\mathrm{d}}(A)$.

Set $T=A-zI$. The squares of the singular values of $T{P}_n^{\ast }$ are the eigenvalues of the finite-dimensional positive operatorThe Rayleigh-Ritz variational principle and its min-max characterization show that, as the trial space grows, its $(j+1)$-st eigenvalue counted upward cannot increase. Therefore, for each fixed $j\ge 0$,once $n>j$. Being nonnegative, it has a limitThe core assumption ensures that these Ritz limits are the min-max values of $T^{\ast }T$, rather than values for a smaller closed restriction.

For a normal operator, the spectral theorem gives three possibilities.

vanish, while ${\sigma }_{\inf +m}(T)>0$.

For $k\in \mathbb{N}$, defineIn the first case $H_k=0$ for all sufficiently large $k$. In the second case its first $m$ summands tend to one and all remaining summands eventually vanish, so $H_k\to m$. In the third case $H_k=k+1\to +\infty$. These are exactly the three values in the definition of $h$, and hence

Letand, for $w\in \mathbb{C}$, form the finite rectangular matrixEvery entry is available from $\Lambda$. Let its singular values, padded and ordered as in the question, beFor fixed $n_2$, Parseval's identity gives convergence of the finite Gram matrices:as $n_1\to \infty$. Consequently, ifthenSingular values are Lipschitz under scalar shifts, so the subsequent limit $z_{n_2}\to z$ and part ii give

Define the finite-information arithmetic functionsFinite-matrix singular values can be obtained by arithmetic eigenvalue approximation, so these functions form the required ${\Delta }_4^A$ arithmetic tower using only $\Lambda$. The two inner limits recover the limiting singular values, while the outer limit is exactly the multiplicity formula from part ii:

For a densely defined closed operator $S$, define its lower normThe closed-range theorem givesThereforeand

We next show that $\mu (T(z))$ is accessible through the matrix-entry evaluations. Put $L_n=\mathrm{span}\{e_1,\dots ,e_n\}$ and defineBecause the canonical span is a core for both operators,

For fixed $m,n$ and grid point $z$, formIts entries belong to $\Lambda$. The smallest singular value of $M_{m,n}(z)$ increases as $m\to \infty$ to ${\nu }_n(T(z))$. Applying the same construction to the conjugate-transposed entries gives ${\nu }_n(T(z{)}^{\ast })$. Thus two nested finite-matrix limits determine $\mu (T(z))$.

The assumed gap continuity of $T$ and the identitymake the corresponding lower-norm tests stable under movement of $z$. The rational grids $G_n$ become dense on every bounded disk. Hence an arithmetic algorithm can, on $G_n\cap {\overline{B}}_n(0)$, use the two finite-section levels above and a vanishing rational tolerance to output all grid cells certified byTaking their closures and letting the mesh and tolerance vanish converges in the Attouch--Wets topology toThe strict existential inequality requires two nested limits, with approximants entering from the prescribed side. In the notation of the arithmetic hierarchy this proves

For the spectrum, equality to zero is the countable intersectionUse the preceding two-level pseudospectral procedure with threshold $1/q$, and add an outer limit $q\to \infty$. Outer approximants remove every point with positive lower norm, while gap continuity and density of the grids retain every zero. Truncating to expanding disks and using vanishing mesh again gives Attouch–Wets convergence. The universal outer intersection reverses the one-sided classification and adds one level, proving