the following inequalities hold $$m(A)\|B\|^{G}_E\leq \|AB\|^{G}_E\leq \|A\|\|B\|^{G}_E,$$ where $m(A)$ is the infimum of the spectrum of the operator $|A|=\sqrt{A^*A}$.*

C\) *For arbitrary operators $A$ and $B$ in $\B(\H)$ such that $\,\langle A\varphi |B\varphi\rangle=0$ for any $\varphi\in\H$ the following inequalities hold* $$\max\!{\hspace{1pt}}\left\{\|A\|^{G}_{E},\|B\|^{G}_{E}{\hspace{1pt}}\right\}\leq \|A+B\|^{G}_{E}\leq \sqrt{\left[\|A\|^{G}_{E}\right]^2+\left[\|B\|^{G}_{E}\right]^2}.$$

D\) *For an operator $\rho$ in $\T_{+,1}(\H)$ with finite $E_{\rho}\doteq{\mathrm{Tr}}{G}\rho$ and any operators $A$ and $B$ in $\B(\H)$ the following inequalities hold* $$|{\mathrm{Tr}}A\rho B^*|\leq \|A\rho B^*\|_1 \leq \|A\|^{G}_{E_{\rho}}\|B\|^{G}_{E_{\rho}}.$$

E\) *For any 2-positive map $\Phi:\B(\H)\rightarrow\B(\H)$ such that $\,\Phi(I_{\H})\leq I_{\H}\,$ having the predual map[^9] $\Phi_*:\T(\H)\rightarrow\T(\H)$ with finite $\,Y_{\Phi}(E)\doteq \sup\!\left\{{\mathrm{Tr}}{G}\Phi_*(\rho)\,|\,\rho\in\mathfrak{S}(\mathcal{H}),{\mathrm{Tr}}{G}\rho\leq E{\hspace{1pt}}\right\}$ and arbitrary operator $A$ in $\B(\H)$ the following inequalities hold* [^10] $$\|\Phi(A)\|^{G}_E\leq\sqrt{\|\Phi(I_{\H})\|}\,\|A\|^{G}_{Y_{\Phi}(E)}\leq \sqrt{\|\Phi(I_{\H})\|K_{\Phi}}\,\|A\|^{G}_{E},\qquad  K_{\Phi}=\max\{1, Y_{\Phi}(E)/E\}.$$

Proposition \[bp-en-0\] shows that the linear transformations $A\mapsto BA$ and $A\mapsto\Phi(A)$ of $\B(\H)$, where $B\in\B(\H)$ and $\Phi$ is a map with the properties pointed in part E, are bounded operators w.r.t. the norm $\|\!\cdot\!\|^{G}_E$ (in contrast to the transformation $A\mapsto AB$).

*Proof.* A) The equality $\|A\|^{G}_E=\||A|\|^{G}_E$ is obvious. The inequality $\|A\|^{G}_E\leq\sqrt{\|A^*A\|^{G}_E}$ follows from the operator Cauchy-Schwarz inequality $$[{\mathrm{Tr}}A^*A\rho]^2\leq [{\mathrm{Tr}}[A^*A]^2\rho][{\mathrm{Tr}}\rho].$$ To show that $\|A^*\|^{G}_E$ may not coincide with $\|A\|^{G}_E$ take any operator ${G}$ having form (\[H-rep\]). It is easy to see that $\||\tau_0\rangle\langle\tau_n|\|^{G}_E=\sqrt{E/E_n}\,$ while $\||\tau_n\rangle\langle\tau_0|\|^{G}_E=1\,$ for all $E>0$.

B\) This assertion follows directly from definition 