(\[ec-on\]) of the operator *E*-norm.

C\) This assertion follows directly from definition (\[ec-on-b\]) of the operator *E*-norm.

D\) The first inequality is obvious. Let $U$ be the partial isometry from the polar decomposition of $A\rho B^*$, i.e. $A\rho B^*=U|A\rho B^*|$. By using the operator Cauchy-Schwarz inequality we obtain $$\|{\mathrm{Tr}}A\rho B^*\|^2_1=[{\mathrm{Tr}}U^*\!A\rho B^*]^2\leq [{\mathrm{Tr}}UU^*\!A\rho A^*][{\mathrm{Tr}}B\rho B^*]\leq [{\mathrm{Tr}}A\rho A^*][{\mathrm{Tr}}B\rho B^*],$$ where that last inequality is due to the fact that $UU^*\leq I_{\H}$. By Proposition \[en-def\]A the right hand side of this inequality does not exceed $\left[\|A\|^{G}_{E_{\rho}}\|B\|^{G}_{E_{\rho}}\right]^2$.

E\) By Kadison’s inequality and Proposition \[en-def\]A we have $${\mathrm{Tr}}[\Phi(A)]^*\Phi(A)\rho\leq\|\Phi(I_{\H})\|{\mathrm{Tr}}\Phi(A^*A)\rho=\|\Phi(I_{\H})\|{\mathrm{Tr}}A^*A\Phi_*(\rho)\leq \|\Phi(I_{\H})\| \left[\|A\|^{G}_{Y_{\Phi}(E)}\right]^2$$ for any $A\in\B(\H)$ and any $\rho\in\S(\H)$ such that ${\mathrm{Tr}}{G}\rho\leq E$ (since the condition $\,\Phi(I_{\H})\leq I_{\H}\,$ guarantees that $\Phi_*(\rho)\in\T_{+,1}(\H)$). This implies the first inequality. The second inequality follows from (\[E-n-eq\]). $\square$

Properties of the *E*-norms related to tensor products
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If ${G}_1$ and ${G}_2$ are positive densely defined operators on Hilbert spaces $\H_1$ and $\H_2$ satisfying condition (\[H-cond\]) then ${G}_{12}={G}_{1}\otimes I_{2}+ I_{1}\otimes {G}_{2}$ is an operator on the Hilbert space $\H_{12}=\H_1\otimes\H_2$ with the same properties.[^11][^12] The following proposition contains several estimates for the operator *E*-norms of product operators used in Section 4.

\[pen-tp\] *Let ${G}_1$, ${G}_2$ and ${G}_{12}$ be the operators described above.*

A\) *For arbitrary 