Proof. For $\( y\in H \)$, consider $\( x\mapsto Φ\mathopen{}\left( x, y\right)\mathclose{} \)$. This is a bounded linear functional on $\( H \)$ of norm at most $\( M\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \)$. Get $\( R\mathopen{}\left( y\right)\mathclose{}\in H \)$ such that $\( Φ\mathopen{}\left( x, y\right)\mathclose{}= \mathopen{}\left\langle{}x, R\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{} \)$ with $\( \mathopen{}\left\lVert{}R\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}\leq M\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \)$. By uniqueness of $\( R\mathopen{}\left( y\right)\mathclose{} \)$ for any given $\( y \)$, we get that $\( R : H \to H \)$ is linear, so $\( R\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$ with $\( \mathopen{}\left\lVert{}R\right\rVert\mathclose{}\leq M \)$. Let $\( S= R^{*} \)$.

Proof. Define $\( Φ : H\times H \to \mathbb{C} \)$ by $\( Φ\mathopen{}\left( x, y\right)\mathclose{}= φ\mathopen{}\left( x\otimes y\right)\mathclose{} \)$. Then $\( Φ \)$ satisfies the hypothesis of Proposition III.35 because $\( \mathopen{}\left\lvert{}Φ\mathopen{}\left( x, y\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\otimes y\right\rVert\mathclose{}= \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \)$. Thus there exists $\( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$ such that $\( φ\mathopen{}\left( x\otimes y\right)\mathclose{}= \mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{} \)$. Let $\( \mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ be an orthonormal basis for $\( H \)$. Polarize $\( T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$. Let $\( {P}_{n}= \sum_{j=1}^{n}{} {e}_{j}\otimes {e}_{j} \)$ (the projection on $\( \operatorname{span}\mathopen{}\left( {e}_{1}, \dotsc, {e}_{n}\right)\mathclose{} \)$). Then $\( {P}_{n} V^{*}= \sum_{j=1}^{n}{} {e}_{j}\otimes V{e}_{j} \)$, $\( \mathopen{}\left\lVert{}{P}_{n} V^{*}\right\rVert\mathclose{}\leq 1 \)$, and $\( {P}_{n} V^{*}\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \)$. So, $$\[ \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\geq \mathopen{}\left\lvert{}φ\mathopen{}\left( {P}_{n} V^{*}\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}φ\mathopen{}\left( \sum_{j=1}^{n}{} {e}_{j}\otimes V\mathopen{}\left( {e}_{j}\right)\mathclose{} \right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\sum_{j=1}^{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{j}\right)\mathclose{}, V\mathopen{}\left( {e}_{j}\right)\mathclose{}\right\rangle\mathclose{} \right\rvert\mathclose{}= \mathopen{}\left\lvert{}\sum_{j=1}^{n}{} \mathopen{}\left\langle{}V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{j}\right)\mathclose{}\right)\mathclose{}, V\mathopen{}\left( {e}_{j}\right)\mathclose{}\right\rangle\mathclose{} \right\rvert\mathclose{}= \mathopen{}\left\lvert{}\sum_{j=1}^{n}{} \mathopen{}\left\langle{} V^{*}\mathopen{}\left( V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{j}\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, {e}_{j}\right\rangle\mathclose{} \right\rvert\mathclose{}= \sum_{j=1}^{n}{} \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{j}\right)\mathclose{}, {e}_{j}\right\rangle\mathclose{} \text{.} \]$$ Let $\( n \)$ approach $\( \infty \)$. We find $\( T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \)$ and $\( \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\leq \mathopen{}\left\lVert{}φ\right\rVert\mathclose{} \)$. Next, $\( φ\mathopen{}\left( x\otimes y\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( Tx\otimes y\right)\mathclose{} \)$ for all $\( x \)$ and $\( y \)$. This implies $\( φ\mathopen{}\left( F\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( TF\right)\mathclose{} \)$ for all $\( F\in \mathcal{F}\mathopen{}\left( H\right)\mathclose{} \)$. For general $\( K\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{} \)$, get $\( \mathopen{}\left({F}_{n}\right)\mathclose{}\in \mathcal{F}\mathopen{}\left( H\right)\mathclose{} \)$ such that $\( {F}_{n} \to K \)$ (in norm). Then $\( φ\mathopen{}\left( {F}_{n}\right)\mathclose{} \to φ\mathopen{}\left( K\right)\mathclose{} \)$, and, by Remark III.28, $\( \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( T\mathopen{}\left({F}_{n}-K\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\mathopen{}\left\lVert{}{F}_{n}-K\right\rVert\mathclose{} \to 0 \)$, i.e. $\( \operatorname{Tr}\mathopen{}\left( T{F}_{n}\right)\mathclose{} \to \operatorname{Tr}\mathopen{}\left( TK\right)\mathclose{} \)$. So $\( φ\mathopen{}\left( K\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( TK\right)\mathclose{} \)$ for all $\( K\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{} \)$. So, $\( \mathopen{}\left\lvert{}φ\mathopen{}\left( K\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( T\mathopen{}\left( K\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\mathopen{}\left\lVert{}K\right\rVert\mathclose{} \)$ for all $\( K \)$, and so $\( \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}\leq \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \)$, and $\( \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}= \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \)$.

Proof.

1. $\( \mathcal{K}\mathopen{}\left( H\right)\mathclose{}^{*} \)$, like any dual space, is a Banach space.
2. We have $$\[ \mathopen{}\left(x\otimes y\right)\mathclose{} ^{*}\mathopen{}\left(x\otimes y\right)\mathclose{}= \mathopen{}\left(y\otimes x\right)\mathclose{}\mathopen{}\left(x\otimes y\right)\mathclose{}= \mathopen{}\left\langle{}x, x\right\rangle\mathclose{}y\otimes y= {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}\frac{y}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}\otimes \frac{y}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}= {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}P \text{} \]$$ where $\( P \)$ is the projection on $\( \mathbb{C}y \)$. So, $\( \mathopen{}\left\lvert{}x\otimes y\right\rvert\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\frac{y}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}\otimes \frac{y}{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}} \)$. Thus $\( \operatorname{Tr}\mathopen{}\left( x\otimes y\right)\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \)$.
3. Take $\( T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \)$. Let $\( \mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ be an orthonormal basis diagonalizing $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$. Then $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}= \sum{} {λ}_{n}{e}_{n}\otimes {e}_{n} \)$, with $\( {λ}_{n}\geq 0 \)$ and $\( \sum{} {λ}_{n} \lt \infty \)$. Write $\( {P}_{j}= \sum_{n=1}^{j}{} {e}_{n}\otimes {e}_{n} \)$ (which is the projection on $\( \operatorname{span}\mathopen{}\left( {e}_{1}, {e}_{2}, \dotsc, {e}_{j}\right)\mathclose{} \)$). Then $$\[ \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left(1-{P}_{j}\right)\mathclose{}= \sum_{n=j+1}^{\infty}{} {λ}_{n}{e}_{n}\otimes {e}_{n} \]$$ so $\( \mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left(1-{P}_{j}\right)\mathclose{}\right\rVert\mathclose{}_{\operatorname{Tr}}= \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left(1-{P}_{j}\right)\mathclose{}\right)\mathclose{}= \sum_{n=j+1}^{\infty}{} {λ}_{n} \to 0 \)$. Finally, the polar decomposition $\( T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$ gives $$\[ \mathopen{}\left\lVert{}T\mathopen{}\left(1-{P}_{j}\right)\mathclose{}\right\rVert\mathclose{}_{\operatorname{Tr}}\leq \mathopen{}\left\lVert{}V\right\rVert\mathclose{}\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left(1-{P}_{j}\right)\mathclose{}\right)\mathclose{} \to 0 \text{.} \]$$ Thus $\( \mathopen{}\left\lVert{}T-T{P}_{j}\right\rVert\mathclose{} \to 0 \)$.
4. $\( \mathopen{}\left\lVert{}T\right\rVert\mathclose{}_{\operatorname{Tr}}= \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \)$, which is the sum of the eigenvalues of $\( T \)$, and is therefore at least $\( ρ\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \)$ (equal to the largest eigenvalue of $\( \mathopen{}\left\lvert{}T\right\rvert\mathclose{} \)$). Thus $$\[ \mathopen{}\left\lVert{}T\right\rVert\mathclose{}_{\operatorname{Tr}}\geq ρ\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{2}\right\rVert\mathclose{}}^{\frac{1}{2}}= {\mathopen{}\left\lVert{} T^{*}T\right\rVert\mathclose{}}^{\frac{1}{2}}= \mathopen{}\left\lVert{}T\right\rVert\mathclose{} \text{.} \]$$
5. $\( \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( STK\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( TKS\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}_{\operatorname{Tr}}\mathopen{}\left\lVert{}KS\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}_{\operatorname{Tr}}\mathopen{}\left\lVert{}S\right\rVert\mathclose{}\mathopen{}\left\lVert{}K\right\rVert\mathclose{} \)$ for all $\( K\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{} \)$.

Proof. Consider the conjugate bilinear (sesquilinear) form $\( \mathopen{}\left(x, y\right)\mathclose{}\mapsto f\mathopen{}\left( x\otimes y\right)\mathclose{} \)$. Since $\( \mathopen{}\left\lvert{}f\mathopen{}\left( x\otimes y\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\otimes y\right\rVert\mathclose{}_{\operatorname{Tr}}= \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{} \)$, by Proposition III.35 there exists $\( S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$ such that $\( f\mathopen{}\left( x\otimes y\right)\mathclose{}= \mathopen{}\left\langle{}S\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{} \)$ and $\( \mathopen{}\left\lVert{}S\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{} \)$. Thus $\( f\mathopen{}\left( x\otimes y\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( S\mathopen{}\left( x\otimes y\right)\mathclose{}\right)\mathclose{} \)$ for all $\( x \)$ and $\( y \)$ in $\( H \)$. So $\( f\mathopen{}\left( F\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( SF\right)\mathclose{} \)$ for all $\( F\in \mathcal{F}\mathopen{}\left( H\right)\mathclose{} \)$. Then $\( f\mathopen{}\left( T\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( ST\right)\mathclose{} \)$ by Proposition III.37 for all $\( T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \)$. Further $\( \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( S\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}S\right\rVert\mathclose{}\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{} \)$ so $\( \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}S\right\rVert\mathclose{} \)$.