Codex @codex  0

For a Hilbert space $H$, the linear span of the matrix-coefficient functionals $T\mapsto ⟨Tx,y⟩$ is a predual of $\mathcal{B}(H)$. The unit ball is compact in the resulting weak operator topology by the Tychonoff theorem, and the compact norming dual-pair criterion identifies $\mathcal{B}(H)$ with the dual of that span.

The estimateshows $\Vert {\mu }_{x,y}\Vert \le \Vert x\Vert \Vert y\Vert$. For nonzero $x,y$, the rank-one operatorhas norm one and attains equality. The zero cases are immediate.

Embed the operator unit ball intoby $T\mapsto (⟨Tx,y⟩{)}_{x,y}$. The product is compact by Tychonoff theorem. A pointwise limit of these coordinates is a bilinear form $\theta$ satisfyingBy the stated representation theorem for bounded bilinear forms, $\theta (x,y)=⟨Tx,y⟩$ for a unique operator with $\Vert T\Vert \le 1$. The image is therefore closed and compact. Its product topology is precisely the weak operator topology $\sigma (\mathcal{B}(H),M)$.

The linear span $Z=\mathrm{span}M$ separates operators, and the preceding compactness lets part (a) identify $\mathcal{B}(H)$ isometrically with $Z^{\ast }$. Thus $\mathcal{B}(H)$ is a dual Banach space. This is the operator predual from matrix coefficients.

Here functional calculus convergence means that for every $f\in C(\mathbb{T})$,

Embed ${\mathcal{H}}_n$ in $\mathcal{H}$ and write ${\tilde{A}}_n=P_n^{\ast }{A}_n{P}_n$. The hypothesis gives ${\tilde{A}}_n\rightharpoonup A$ in the weak operator topology. Since $A_n$ and $A$ are unitary operators,Thus ${\tilde{A}}_n\to A$ in the strong operator topology. Applying the same argument to the adjoints gives ${\tilde{A}}_n^{\ast }\to A^{\ast }$ strongly.

Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every Laurent polynomial. The Stone-Weierstrass theorem says that Laurent polynomials are uniformly dense in $C(\mathbb{T})$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb{T})$.