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

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.