Trace-class duality and finite-rank density

Let $J=\{1,\dots ,r\}$ if $T$ has finite rank $r$, and $J={\mathbb{N}}_{>0}$ otherwise, as in [F1]. Write its singular expansion $T={\sum }_{j\in J}{s}_j⟨\cdot ,e_j⟩f_j$ and let $T_N={\sum }_{j\in J,j\le N}{s}_j⟨\cdot ,e_j⟩f_j$. The SVD gives $T-T_N={\sum }_{j\in J,j>N}{s}_j⟨\cdot ,e_j⟩f_j$ as a norm-convergent nuclear representation whose coefficient products sum to the tail ${\sum }_{j\in J,j>N}{s}_j\to 0$. The nuclear characterization in [F1] therefore gives $\mathrm{\parallel }T-T_N{\mathrm{\parallel }}_1\le {\sum }_{j\in J,j>N}{s}_j\to 0$. In finite rank the tail is empty once $N\ge r$. Thus finite-rank operators are trace-norm dense.

For $\xi ,\eta \in H$, define the rank-one operator $R_{\xi ,\eta }z=⟨z,\eta ⟩\xi$. Its only possible nonzero singular value is $\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$, so $\mathrm{\parallel }{R}_{\xi ,\eta }{\mathrm{\parallel }}_1=\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$. The form $b(\xi ,\eta ):=\mathrm{\Phi }(R_{\xi ,\eta })$ is linear in $\xi$, conjugate-linear in $\eta$, and satisfies $\mathrm{\mid }b(\xi ,\eta )\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$. For fixed $\xi$, the map $g_{\xi }(\eta ):=\overline{b(\xi ,\eta )}$ is bounded and linear in $\eta$. Riesz representation [F4] gives a unique vector $X\xi$ such that $g_{\xi }(\eta )=⟨\eta ,X\xi ⟩$; conjugating gives $b(\xi ,\eta )=⟨X\xi ,\eta ⟩$. Uniqueness makes $X$ linear, and the bound gives $\mathrm{\parallel }X\xi \mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\mathrm{\parallel }\xi \mathrm{\parallel }$, so $X\in \mathcal{B}(H)$ and $\mathrm{\parallel }X\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }$.

The nuclear representation $R_{\xi ,\eta }X=R_{\xi ,X^{\ast }\eta }$ and [F3] give $\mathrm{Tr}(R_{\xi ,\eta }X)=⟨\xi ,X^{\ast }\eta ⟩=⟨X\xi ,\eta ⟩$. Every finite-rank operator is a finite sum of rank-one operators, so linearity of $\mathrm{\Phi }$ and of the trace gives $\mathrm{\Phi }(T)=\mathrm{Tr}(TX)$ for every finite-rank $T$.

For arbitrary $T\in {\mathcal{S}}_1(H)$, use the finite-rank approximants $T_N$ from step 1.1. Boundedness of $\mathrm{\Phi }$ gives $\mathrm{\Phi }(T_N)\to \mathrm{\Phi }(T)$. Also $\mathrm{\mid }\mathrm{Tr}((T-T_N)X)\mathrm{\mid }\le \mathrm{\parallel }(T-T_N)X{\mathrm{\parallel }}_1\le \mathrm{\parallel }T-T_N{\mathrm{\parallel }}_1\mathrm{\parallel }X\mathrm{\parallel }\to 0$ by [F2] and [F3]. Hence the same trace identity holds for every trace-class $T$.

If $H=\{0\}$, then ${\mathcal{S}}_1(H)=\mathcal{B}(H)=\{0\}$, so the only bounded functional and representing operator are zero and the isometry assertion holds. Otherwise the trace identity and [F2]–[F3] imply $\mathrm{\mid }\mathrm{\Phi }(T)\mathrm{\mid }\le \mathrm{\parallel }T{\mathrm{\parallel }}_1\mathrm{\parallel }X\mathrm{\parallel }$, so $\mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }\le \mathrm{\parallel }X\mathrm{\parallel }$. For unit $\xi ,\eta$, step 2.1 gives $\mathrm{\mid }⟨X\xi ,\eta ⟩\mathrm{\mid }=\mathrm{\mid }\mathrm{\Phi }(R_{\xi ,\eta })\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }$, since $\mathrm{\parallel }{R}_{\xi ,\eta }{\mathrm{\parallel }}_1=1$. Taking suprema over unit $\xi ,\eta$ gives $\mathrm{\parallel }X\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }$. If $Y$ also represents $\mathrm{\Phi }$, applying the identity to each $R_{\xi ,\eta }$ gives $⟨(X-Y)\xi ,\eta ⟩=0$ for all $\xi ,\eta$, hence $X=Y$.

Steps 1.1–4.1 prove finite-rank trace-norm density and establish a unique operator $X$ with the stated pairing and norm equality for every bounded functional $\mathrm{\Phi }$.