operator $A$ in $\B(\H_1)$ the following equalities hold* $$\|A\otimes I_{2}\|^{{G}_{12}}_E=\|A\otimes I_{2}\|^{{G}_{1}\otimes {\hspace{1pt}}I_{2}}_E=\|A\|^{{G}_1}_E$$

B\) *For arbitrary operators $A\in\B(\H_1)$ and $B\in\B(\H_2)$ the following inequalities hold* $$\label{pen-tp-1}
\sup_{x\in(0,E)}\|A\|^{{G}_1}_{x}\|B\|^{{G}_2}_{E-x}\leq \|A\otimes B\|^{{G}_{12}}_{E} \leq\sup_{x\in(0,E)} \sqrt{\|A^*A\|^{{G}_1}_{x}\|B^*B\|^{{G}_2}_{E-x}},$$ *and* $$\label{pen-tp-2}
\|A\otimes B\|^{{G}_{12}}_{E} \leq\min\left\{\|A\|^{{G}_1}_{E}\|B\|,\|A\|\|B\|^{{G}_2}_{E}\right\}.$$

Note that the lower and upper bounds in (\[pen-tp-1\]) and the r.h.s. of (\[pen-tp-2\]) tend to $\|A\|\|B\|=\|A\otimes B\|=\lim_{E\to+\infty}\|A\otimes B\|^{{G}_{12}}_{E}$ as $E\to+\infty$.

*Proof.* A) It suffices to note that ${\mathrm{Tr}}[|A|^2\otimes I_{2}]\rho_{12}={\mathrm{Tr}}|A|^2\rho_{1}$ and that ${\mathrm{Tr}}{G}_{12}\rho_{12}={\mathrm{Tr}}{G}_{1}\rho_{1}+{\mathrm{Tr}}{G}_{2}\rho_{2}$ for any state $\rho_{12}\in\S(\H_{12})$, where $\rho_1={\mathrm{Tr}}_{\H_2}\rho_{12}$ and $\rho_2={\mathrm{Tr}}_{\H_1}\rho_{12}$ are the partial states of $\rho_{12}$.

B\) For each $x\in(0,E)$ and any $\varepsilon>0$ there exist states $\rho_1$ in $\S(\H_1)$ and $\rho_2$ in $\S(\H_2)$ such that $\sqrt{{\mathrm{Tr}}|A|^2\rho_1}>\|A\|^{{G}_1}_x-\varepsilon$, ${\mathrm{Tr}}{G}_1\rho_1\leq x$, $\sqrt{{\mathrm{Tr}}|B|^2 \rho_2}>\|B\|^{{G}_2}_{E-x}-\varepsilon$ and ${\mathrm{Tr}}{G}_2\rho_2\leq E-x$. Then ${\mathrm{Tr}}{G}_{12}[\rho_1\otimes\rho_2]\leq x+E-x=E$ and $\sqrt{{\mathrm{Tr}}[|A|^2\otimes|B|^2][\rho_1\otimes\rho_2]}\geq[\|A\|^{{G}_1}_x-\varepsilon][\|B\|^{{G}_2}_{E-x}-\varepsilon]$. Since $\varepsilon$ is arbitrary, this implies the left inequality in (\[pen-tp-1\]).

By the operator Cauchy-Schwarz inequality for any state $\rho_{12}$ in $\S(\H_{12})$ we have $$\begin{array}{c}
{\mathrm{Tr}}[|A|^2\otimes |B|^2]\rho_{12}\leq \sqrt{{\mathrm{Tr}}[|A|^4\otimes I_{2}]\rho_{12}}\sqrt{{\mathrm{Tr}}[I_{1}\otimes |B|^4]\rho_{12}}\\\\=
\sqrt{{\mathrm{Tr}}|A|^4\rho_{1}}\sqrt{{\mathrm{Tr}}|B|^4\rho_{2}}\leq \||A|^2\|^{{G}_1}_{{\mathrm{Tr}}{G}_1\rho_1}\||B|^2\|^{{G}_2}_{{\mathrm{Tr}}{G}_2\rho_2}.
\end{array}$$ Since ${\mathrm{Tr}}{G}_{12}\rho_{12}={\mathrm{Tr}}{G}_{1}\rho_{1}+{\mathrm{Tr}}{G}_{2}\rho_{2}$, this implies the right inequality in (\[pen-tp-1\]).

To prove inequality (\[pen-tp-2\]) it suffices to note that $$A\otimes B=[A\otimes I_{2}][I_{1}\otimes B]=[I_{1}\otimes B][A\otimes I_{2}]$$ and to apply Proposition \[bp-en-0\]B and part A of this proposition. $\square$

The *E*-version of the Kretschmann-Schlingemann-Werner theorem
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In this section we consider application of the operator $E$-norms to 