of $\B(\H)$ is not weaker than the strong operator topology. On the other hand, it is not stronger than the norm topology on $\B(\H)$. The following proposition characterizes these extreme cases.

\[ec-on-p2\] A) *The norm $\,\|\!\cdot\!\|^{G}_E$, $E>0$, is equivalent to the operator norm $\,\|\!\cdot\!\|$ on $\B(\H)$ if and only if the operator ${G}$ is bounded.*

B\) *The norm $\,\|\!\cdot\!\|^{G}_E$, $E>0$, generates the strong operator topology on bounded subsets of $\B(\H)$ if and only if $\,{G}$ is an unbounded discrete operator (Definition \[D-H\]).*

*Proof.* A) If ${G}$ is a bounded operator then $\,\|\!\cdot\!\|^{G}_E=\|\!\cdot\!\|$ for any $E\geq \|G\|$.

If ${G}$ is a unbounded operator and $P_n$ is the spectral projector of ${G}$ corresponding to the interval $[n,+\infty)$ then $\|P_n\|=1$ for all $n$. By noting that ${\mathrm{Tr}}P_n\rho\leq E/n$ for any state $\rho$ such that ${\mathrm{Tr}}{G}\rho\leq E$, it is easy to see that $\,\|P_n\|^{G}_E\rightarrow 0$ as $n\rightarrow+\infty$.

B\) The “if” part of this assertion is proved in [@CSR].

Assume there is a spectral projector of the operator ${G}$ corresponding to a finite interval $[0, E_0]$ with infinite-dimensional range $\H_0$. Since $\langle\varphi|G|\varphi\rangle\leq E_0$ for any unit vector $\varphi$ in $\H_0$, we have $\|A\|^{G}_E=\|A\|$ for any $A\in\B(\H_0)$ and $E>E_0$. So, any of the norms $\,\|\!\cdot\!\|^{G}_E$, $E>0$, 