Let $V$ be a Banach space and $B(V)$ be the bounded operators on $V$. Now, the space $B(V)$ has a norm, the strong topology (which is the topology of pointwise convergence) and it has a stronger topology called ultrastrong or $\sigma$-strong topology. It is known that these two topologies coincide on norm-bounded sets. I want to know if the $\sigma$-strong topology is the strongest topology which coincides with the strong topology (respectively itself) on bounded sets.


Edit:
The $\sigma$-strong topology is a topology generated by certain seminorms. Namely for all sequences $b\_n$ with $\sum\_n \|b\_n\|^2<\infty$, the seminorm
$B(V)\times B(V) \to \mathbf{R}$
given by the square root of $(f,g)\mapsto \sum\_n \|f(b\_n)-g(b\_n)\|^2$ should be continuous and these seminorms generate the topology.


Edit:
The answer seems to be "no" for this general case. I have no example though.
To simplify this question, I removed a second part. Thank you for the helpful comments on both parts.



Is the sigma-strong topology generated by bounded sets?
The finest topology that coincides with $\tau\*$ ($\sigma$-strong in this case) on $\tau$-bounded (norm-bounded in this case) subsets is the mixed topology $\gamma(\tau,\tau^\*)$, introduced by [A. Wiweger, Linear spaces with mixed topology. Studia Mathematica 20 (1961), 47--68](https://eudml.org/doc/217004); see 2.2.2. For the Hilbert space $\ell\_2$ (or perhaps any separable infinite dimensional Banach space?), the inequalities
$$\mbox{$\sigma$-strong} \le \mbox{uniform convergence on compact subsets} \le \gamma(\mbox{norm},\mbox{$\sigma$-strong})$$
are strict (Addendum: I'm no longer so sure if the second inequality is strict).




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A note for posterity: I was noticed that there is some confusion in the literature as to the difference of the above topologies. E.g., Proposition I.8.6.3 in [Blackadar's book](https://mathscinet.ams.org/mathscinet-getitem?mr=2188261) erroneously claims that the $\sigma$-strong topology on $B(H)$ coincides with the strict topology arising as the multiplier of $K(H)$. The essentially same claim is seen in Chapter 8 in [Lance's book](https://mathscinet.ams.org/mathscinet-getitem?mr=1325694). In fact, these topologies are different (although they coincide on norm bounded subsets, and that's what's needed in daily life), as the following example shows.


Take a sequence of mutually orthogonal rank one orthogonal projections $p\_n$ on a separable Hilbert space $H$.
Then, the $\sigma$-strong closure of $\{ \sqrt{n} p\_n \}$ contains $0$. Indeed, for any positive linear functional $f$ on $B(H)$, one has $\liminf\_n f(n p\_n)=0$, for otherwise $f(p\_n)>\frac{\epsilon}{n}$ for all $n$ and $\lim\_m f(\sum\_{n=1}^m p\_n)=\infty$, a contradiction. However, $T:=\sum\_{n=1}^\infty \frac{1}{\sqrt{n}}p\_n \in K(H)$ (or the compact subset $\{ Tv : \|v\|\le1\})$ separates $0$ from $\{ \sqrt{n} p\_n \}$ in the strict topology (or the compact open topology).


