= Solution
For a <finite-rank operator> $T$, define $|T|=(T^*T)^{1/2}$ and its <trace norm> by
$$
\boxed{\|T\|_1=\operatorname{Tr}|T|=\sum_{j=1}^r s_j,}
$$
where $s_j>0$ are its nonzero <singular values>, counted with multiplicity. The finite-rank spectral theorem gives orthonormal systems $(u_j),(v_j)$ with $T=\sum_js_jR_{u_j,v_j}$, where $R_{x,y}z=\langle z,y\rangle x$.
For every <bounded operator> $B$,
$$
\operatorname{Tr}(BT)=\sum_js_j\langle Bu_j,v_j\rangle,
\qquad |\operatorname{Tr}(BT)|\leq\|B\|\sum_js_j.
$$
Choose the contraction $B$ taking $u_j$ to $v_j$ and vanishing on their <orthogonal complement>. It attains equality, so
$$
\boxed{\|T\|_1=\sup_{\|B\|\leq1}|\operatorname{Tr}(BT)|.}
$$
This formula proves the <triangle inequality> and absolute homogeneity. Also $\|T\|\leq\|T\|_1$, so vanishing of the <trace norm> forces $T=0$. Thus it is a <norm> on the <vector> space of <finite-rank operators>.
The map $B\mapsto F_B$, where $F_B(T)=\operatorname{Tr}(BT)$, is a bounded linear map into the dual of this normed space. Its <norm> is at most $\|B\|$. Rank-one tests give the reverse inequality: $\|R_{x,y}\|_1=\|x\|\|y\|$ and $F_B(R_{x,y})=\langle Bx,y\rangle$, so taking unit <vectors> $x,y$ gives $\|F_B\|=\|B\|$.
Conversely, let $F$ be a <bounded linear functional> for the <trace norm>. The <sesquilinear form> $\beta(x,y)=F(R_{x,y})$ satisfies
$$
|\beta(x,y)|\leq\|F\|\|x\|\|y\|.
$$
The <Riesz representation theorem> gives a unique <bounded operator> $B$ with $\beta(x,y)=\langle Bx,y\rangle$, and $\|B\|\leq\|F\|$. Since every <finite-rank operator> is a finite sum of <rank-one operators>, <linearity> implies $F(T)=\operatorname{Tr}(BT)$ for every $T$. Hence
$$
\boxed{(\mathcal F(H),\|\cdot\|_1)^*\cong B(H)\text{ isometrically}.}
$$
The same identification extends to the completion, the <trace-class operators>. This is <trace-class duality>; completeness of the original finite-rank space is not required to identify its bounded dual.
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