= Solution
Part b(vi) and part c show that $\|T\|_1\leq1$ implies
$$
|\operatorname{tr}(ST)|
\leq\|ST\|_1
\leq\|S\|.
$$
Conversely, choose a unit vector $x$ with $\|Sx\|$ arbitrarily close to $\|S\|$, put $y=Sx/\|Sx\|$, and take $T=x\otimes y$. Part c gives $\|T\|_1=1$ and
$$
\operatorname{tr}(ST)
=\operatorname{tr}(Sx\otimes y)
=\langle Sx,y\rangle
=\|Sx\|.
$$
Taking the supremum proves
$$
\|S\|=\sup_{\|T\|_1\leq1}|\operatorname{tr}(ST)|.
$$
Thus $S\mapsto[T\mapsto\operatorname{tr}(ST)]$ is an isometric embedding $\mathcal S_\infty\to\mathcal S_1^*$.
Suppose finite-rank operators are dense in $\mathcal S_1$ and let $L\in\mathcal S_1^*$. The sesquilinear form
$$
b(x,y)=L(x\otimes y)
$$
satisfies $|b(x,y)|\leq\|L\|\|x\|\|y\|$. The <Riesz representation theorem> gives $S\in\mathcal S_\infty$ with $b(x,y)=\langle Sx,y\rangle$. Hence $L(x\otimes y)=\operatorname{tr}(S(x\otimes y))$, and linearity gives equality on every finite-rank operator. Density and continuity extend it to every $T\in\mathcal S_1$, proving surjectivity.
Conversely, if finite-rank operators were not dense, the <Hahn-Banach theorem> would give a nonzero $L\in\mathcal S_1^*$ vanishing on their closure. Surjectivity would represent it by some $S$, but then
$$
0=L(x\otimes y)=\langle Sx,y\rangle
$$
for all $x,y$, forcing $S=0$ and $L=0$, a contradiction. This proves the stated <trace duality> criterion.
Solved by gpt-5.6-sol high.
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