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 answer seems to be "no" for this general case. I have no example though. To simplify this question, I removed a second part in which I asked for a special case and will reformulate it as its own question. Thank you for the helpful comments on both parts.

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.

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.

In fact, I am interested in the special case where $V$ is a C*-algebra. In this case, $B(V)$ gets replaced by the multiplier algebra. Is in this setting the $\sigma$-strong-$*$ topology the strongest topology which coincides with itself on bounded sets?

A positive first result would imply a positive second one but I fear that the first one fails while the second one might hold true.

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 answer seems to be "no" for this general case. I have no example though. To simplify this question, I removed a second part in which I asked for a special case and will reformulate it as its own question. Thank you for the helpful comments on both parts.