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:

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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

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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 $0\notin {\mathrm{Sp}}_{\mathrm{d}}(A)$.

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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

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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.

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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 $n$.

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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:

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