= Solution
The estimate
$$
|\mu_{x,y}(T)|=|\langle Tx,y\rangle|
\leq\|T\|\,\|x\|\,\|y\|
$$
shows $\|\mu_{x,y}\|\leq\|x\|\|y\|$. For nonzero $x,y$, the rank-one operator
$$
Tz=\left\langle z,\frac{x}{\|x\|}\right\rangle\frac{y}{\|y\|}
$$
has norm one and attains equality. The zero cases are immediate.
Embed the operator unit ball into
$$
\prod_{(x,y)\in H\times H}
[-\|x\|\|y\|,\|x\|\|y\|]
$$
by $T\mapsto(\langle Tx,y\rangle)_{x,y}$. The product is compact by <Tychonoff theorem>. A pointwise limit of these coordinates is a bilinear form $\theta$ satisfying
$$
|\theta(x,y)|\leq\|x\|\|y\|.
$$
By the stated representation theorem for bounded bilinear forms, $\theta(x,y)=\langle Tx,y\rangle$ for a unique operator with $\|T\|\leq1$. The image is therefore closed and compact. Its product topology is precisely the <weak operator topology> $\sigma(\mathcal B(H),M)$.
The linear span $Z=\operatorname{span}M$ separates operators, and the preceding compactness lets part (a) identify $\mathcal B(H)$ isometrically with $Z^*$. Thus $\mathcal B(H)$ is a dual Banach space. This is the <operator predual from matrix coefficients>.
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