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.

  • $\begingroup$ For the general case, please search the literature/textbooks for the Mackey topology. For the second case, the multiplier algebra is a unital $C^*$-algebra. On unital $C^*$-algebras, the Mackey topology and the $\sigma$-strong$^*$ topologies coincide on bounded sets. $\endgroup$
    – Onur Oktay
    Mar 29 at 20:03
  • $\begingroup$ I do not know the answer for exactly which $V$ the Mackey topology and $\sigma$-strong$^*$ topology coincide on bounded sets on $B(V)$ generally. They do however when $V$ is a Hilbert space, when $V=\ell^p$ $1<p<\infty$. Let's note that Mackey top. is stronger than $\sigma$-strong$^*$ top., and $\sigma$-strong$^*$ top. is stronger than $\sigma$-strong top. on $B(V)$. $\endgroup$
    – Onur Oktay
    Mar 29 at 20:44
  • $\begingroup$ A reference: for the Hilbert space case, you might want to consult the fourth chapter of the monograph "Saks Spaces and Applications to Functional Analysis". $\endgroup$
    – terceira
    Apr 20 at 15:56
  • $\begingroup$ I don't know what ultrastrong topology means in the case $V$ is not a Hilbert space. However, the answer is clearly NO if that topology admits an unbounded convergent net, because one can strengthen the topology by insisting that every convergent net has to be bounded. Perhaps, in the present case, this topology would be given by the uniform convergence on compact subsets. $\endgroup$ Apr 21 at 7:20
  • $\begingroup$ @Narutaka OZAWA Sorry for the late respond. When I am not mistaken, then the description of bounded strictly converging nets is not inherited by a topology as the product $B(V) \times V \to V$ (take the norm topology on $V$) would be continuous and this implies that we already have the operator norm topology on $B(V)$. However, there is a convergent, unbounded net in $l^2(\mathbb{N})$ if that helps you. $\endgroup$ May 5 at 10:34

1 Answer 1


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; 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).

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 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. 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).

  • $\begingroup$ Thank you for the reference. The topology $\gamma $ ($\sigma$-strong, norm-topology) was indeed the topology I had in mind. When I understood you correctly, then you claim that the answer is "no" and $V=l_2$ is a counterexample. You moreover claim the lower chain of strict inclusions of topologies on the operators on $l_2$. Why are they true and why are they strict? $\endgroup$ May 11 at 14:20
  • $\begingroup$ For the sequence of mutually orthogonal projections $\{ p_n\}$, the ultrastrong-closure of the countable subset $\{ n p_n \}$ contains $0$, but it's not true for the compact open topology. $\endgroup$ May 12 at 1:02
  • $\begingroup$ Lance points to Taylor, "The strict topology for double centralizer algebras" zbMath as a source for this erroneous claim. I think Corollary 2.7 in Taylor's paper might be to blame: it claims that if $\beta$ is the strict topology on $M(A)$ and $\beta'$ the strongest l.c. top which agrees with $\beta$ on norm bounded sets, then $\beta=\beta'$. That seems to contradict the strict topology on $B(H)=M(K(H))$ being different to the $\sigma$-strong$^*$-topology (which do agree on bounded sets). $\endgroup$ Jun 9 at 7:51
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    $\begingroup$ @Matthew Daws: In fact, Lance misquoted. Taylor's Corollary 2.7 says that the strict topology is same as the mixed topology (Mackey topology), which sounds right. The Mackey topology is strictly stronger than $\sigma$-strong*. $\endgroup$ Jun 9 at 9:05

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