Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 5

The Weyl-von Neumann theorem states that on a separable Hilbert space a self-adjoint operator is a diagonal self-adjoint operator plus a compact self-adjoint operator. For a bounded operator $T$, the perturbation can in fact be chosen Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. The separable-space formulation is also recorded in Kadison's Corollary B.

Here is a constructive proof. The bounded self-adjoint spectral theorem decomposes a separable $H$ into countably many cyclic reducing spaces, on each of which $T$ is multiplication by $x$ in $L^2(\mu )$ for a finite measure supported in a bounded real interval. First work on one such space. Partition the interval into $N$ intervals of length at most $\delta$, and repeatedly bisect them. Let $E_j$ be the finite-dimensional space of functions constant on the level-$j$ intervals. These spaces are nested and their union is dense, since dyadic step functions approximate continuous functions and continuous functions are dense in $L^2(\mu )$.

Choose normalized interval indicators as a basis of $E_0$, and choose orthonormal bases of $E_j\ominus E_{j-1}$ within each parent interval. This is possible because the refinement splits into orthogonal parent blocks. Each new basis vector at level $j\ge 1$ is supported in an interval of length at most $\delta 2^{-(j-1)}$, and there are at most $N{2}^{j-1}$ such vectors. For every basis vector $e$ choose a real ${\lambda }_e$ in its supporting interval and define the diagonal operator $De={\lambda }_{e}e$. ThensoOne may take $N=O({\delta }^{-1})$, so this bound tends to zero as $\delta \to 0$. The error is Hilbert-Schmidt, hence compact; it is self-adjoint because both $T$ and $D$ are.

On the countably many cyclic summands choose error bounds ${\epsilon }_j$ with ${\sum }_j{\epsilon }_j^2<{\epsilon }^2$. Combining their orthonormal bases and diagonal operators gives $T=D+K$ with $\Vert K{\Vert }_{\mathrm{HS}}<\epsilon$, proving the assertion. For an unbounded self-adjoint operator, first split by spectral intervals $[j,j+1)$, apply the bounded construction on each, and choose summable error bounds. The resulting total $K$ is bounded compact, and $D=T-K$ has the same domain as $T$; the diagonal domain condition follows from the boundedness of $K$.

Separability matters. Take an uncountable orthogonal sum of copies of multiplication by $x$ on $L^2[0,1]$. The range closure of a compact operator is separable, and a separable subspace of this direct sum is supported in only countably many summands: choose a countable dense set and unite its countable supports. A self-adjoint compact error therefore vanishes on every remaining summand, each of which reduces the proposed diagonal operator $D=T-K$. The restriction of a diagonal self-adjoint operator to a reducing subspace has pure point spectrum, whereas multiplication by $x$ on $L^2[0,1]$ has no eigenvectors. This is a contradiction, qualifying the theorem if the unspoken separability convention is dropped.

For the Fredholm convention, defineThe equality follows from the Atkinson theorem. The quotient is a unital complex Banach algebra when $H$ is infinite-dimensional. Its invertible set is open by the Neumann series, so the spectrum is closed, and it lies in the disk $|\lambda |\le \Vert [T]\Vert$.

For nonemptiness, suppose the quotient resolvent $R(\lambda )=(\lambda 1-[T]{)}^{-1}$ existed for every complex $\lambda$. It is analytic everywhere, and outside a sufficiently large disk its Neumann expansion gives $\Vert R(\lambda )\Vert \le 1/(|\lambda |-\Vert [T]\Vert )$. Continuity bounds it on the remaining compact disk. Every continuous linear functional applied to $R$ is therefore a bounded entire scalar function, constant by Liouville's theorem and zero because it tends to zero at infinity. The Hahn-Banach theorem, or separation by the dual, forces $R(\lambda )=0$, contradicting its inverse identity. The essential spectrum is nonempty and closed.

Infinite dimension is necessary: every operator on a finite-dimensional Hilbert space is Fredholm, so its essential spectrum in this convention is empty. The proof works on nonseparable infinite-dimensional spaces as well; it needs neither self-adjointness nor a sequence basis.

Interpret limit points as subsequential limits, including a value repeated infinitely often. Let $L$ be the common nonempty compact cluster set. Boundedness implies $\mathrm{dist}(a_n,L)\to 0$ and likewise for $b_n$: otherwise a subsequence staying a fixed distance away would have a convergent further subsequence whose limit both belongs to $L$ and stays away from it.

Construct a matching of indices in alternating steps. At step $2j-1$ choose the least unused index $m$ in the first sequence. Choose a nearest point $\ell \in L$ to $a_m$ and an unused index $n$ in the second sequence with $|b_n-\ell |<1/j$. Such indices exist infinitely often because $\ell$ is a subsequential limit. Pair $m$ with $n$; their values differ by at most $\mathrm{dist}(a_m,L)+1/j$. At step $2j$ choose the least unused second-sequence index and match it to an unused first-sequence index within $1/j$ of a nearest point of $L$. This gives the analogous bound with $b_n$.

Every index is used exactly once: no step reuses an index, and by step $2j$ at least the first $j$ indices in each sequence have been used. Hence the produced lists are permutations satisfyingThe indices actively chosen as least unused tend to infinity on their respective odd and even steps. Their distances to $L$ therefore tend to zero, and $1/j\to 0$. The two matching estimates prove $\boxed{|a_{m_k}-b_{n_k}|\to 0}$. The cluster-set convention is important here; accumulation points of the set of distinct values would ignore infinite repetitions.

Work with bounded self-adjoint operators on separable infinite-dimensional Hilbert spaces, the setting in which the classification holds. If $UT{U}^{\ast }-S$ is compact for a unitary $U$, their images in the Calkin algebra have identical spectra, by compact perturbation invariance and unitary conjugation. Thus their essential spectra coincide.

Conversely apply the Weyl-von Neumann theorem to write $T=D_a+K_a$ and $S=D_b+K_b$ in orthonormal bases, with bounded real diagonal sequences and compact errors. The essential spectrum of a diagonal operator is exactly its sequence cluster set. Indeed outside that cluster set all but finitely many entries stay a positive distance from $\lambda$, so the diagonal has an inverse on a finite-codimensional subspace and is Fredholm. At a cluster point, infinitely many distinct basis vectors have $(D-\lambda )e_j\to 0$; these are weakly null unit vectors, contradicting any Fredholm parametrix. This also handles a value with infinite multiplicity.

The two diagonal sequences therefore have the same cluster set. Part (c) provides permutations with $a_{m_k}-b_{n_k}\to 0$. Send the $m_k$th basis vector for $T$ to the $n_k$th basis vector for $S$. Under this unitary, the diagonal difference has entries tending to zero and is compact, by finite-rank truncation. Adding the two conjugated compact errors gives $\boxed{UT{U}^{\ast }-S\text{ compact}}$.

Without separability, equality of essential spectra need not suffice, even on the same Hilbert space. Let $H$ have uncountable dimension and let $T,S$ be projections with infinite-dimensional kernels, but with range dimensions respectively countable and uncountable. Both essential spectra are $\{0,1\}$. After any unitary conjugation the range of $T$ is still separable. If its difference from $S$ were compact, the range of $S$ would lie in the sum of that separable range and the separable closure of the compact error's range, which is impossible. Finite-dimensional spaces also need a separate dimension convention, since essential spectra are then empty. These are genuine qualifications to a statement made without the standard separable infinite-dimensional setting.