Operator topologies on $L^{\infty}(X,\mu )$

Let $(X,\mu )$ be a measure space. Then, $L^2(X):=L^2(X,\mu )$ is a Hilbert space in the usual way and we may view $L^{\infty}(X):=L^{\infty}(X,\mu )$ as a subalgebra of bounded operators on $L^2(X)$ via $L^{\infty}(X)\ni f\mapsto M_f$, where $M_f\colon L^2(X)\rightarrow L^2(X)$ is the multiplication operator defined by $[M_f(g)](x):=f(x)g(x)$.

Regarding $L^{\infty}(X)$ as an algebra of bounded operators in this way, the question can be stated as

How can we describe explicitly convergence in the various operator topologies on $L^{\infty}(X)$?

At least one is relatively easy. For example, (i) $\lambda \mapsto M_{f_{\lambda}}$ converges to $M_f$ in the operator norm topology iff $\lambda \mapsto f_{\lambda}$ converges to $f$ in the $L^{\infty}$ norm. Furthermore, it seems that (ii) convergence in measure corresponds to ultra-weak convergence, (iii) pointwise almost-everywhere convergence can't correspond to any topology, and (iv) the weak-$^*$ topology corresponds to the weak operator topology (at least when $(X,\mu )$ is localizable so that $L^{\infty}(X)\cong L^1(X)^*$)---sketch of argument below. Unfortunately, this still leaves quite a few of the operator topologies unaccounted for. Perhaps most of them just don't have a very nice description?

Weak operator topology corresponds to weak-$^*$ topology:

$\lambda \mapsto f_{\lambda}\in L^{\infty}(X)$ converges to $f\in L^{\infty}(X)$ in the weak-$^*$ topology iff $\lambda \mapsto \int _Xgf_{\lambda}$ converges to $\int _Xgf$ for all $g\in L^1(X)$, and on the other hand, $\lambda \mapsto M_{f_{\lambda}}$ converges to $M_f$ in the weak operator topology iff $\lambda \mapsto \int _Xgf_{\lambda}h$ converges to $\int _Xgfh$ for all $g,h\in L^2(X)$. As $gh\in L^1(X)$ for $g,h\in L^2(X)$, we obtain the $(\Rightarrow )$ direction immediately. For the $(\Leftarrow )$ write $g=u|g|$ for some Borel function $u$ with $|u|=1$, so that we have $\lambda \mapsto \int _Xgf_{\lambda}=\int _X(u|g|^{1/2})f_{\lambda}|g|^{1/2}$ converges to $\int _X(u|g|^{1/2})f|g|^{1/2}=\int _Xgf$.

Disclaimer: This is essentially a migration of the original question on math.stackexchange.

Summary

The final answer is due to Cameron Zwarich, but I thought I'd post a quick summary here for easy reference. (If nothing else, I will find likely find it useful in the future.)

• The operator norm topology corresponds to the norm topology.
• The ultra-strong-$^*$, ultra-strong, strong-$^*$, and strong operator topologies all coincide, and the Arens-Mackey topology agrees with these on bounded subsets of $L^{\infty}(X)$. Furthermore, if $\mu (X)<\infty$, the all of these topology correspond to convergence in measure on bounded subsets of $L^{\infty}(X)$.
• The ultra-weak and weak operator topologies coincide and correspond to weak-$^*$ convergence (regarding $L^{\infty}(X)$ as the dual of $L^1(X)$).

The argument you give for the equality of the weak operator and weak-$^*$ (or $\sigma$-weak) topologies also shows that the strong and $\sigma$-strong topologies are equal, and similarly for the strong-$^*$ and $\sigma$ strong-$^*$. This equality of topologies holds for all von Neumann algebras in standard form.

It's not hard to see that the adjoint on $L^\infty(X)$ is continuous with respect to the strong operator topology and thus the strong operator topology and strong-$^*$ topologies are equal as well. In general, this is equivalent to the finiteness of a von Neumann algebra.

Therefore, you are only left with a few distinct topologies: the weak operator topology, the strong operator topology, and the Mackey topology. It's easiest to describe the behavior of these topologies on the unit ball, and it's a result of Akemann that the Mackey topology is equal to the strong-$^*$ topology on the unit ball, or in our case the strong topology, which brings us down to two topologies: the weak and strong operator topologies.

To most easily describe the behavior of these topologies on bounded sets, it's best to replace them with the finest locally convex topologies that agree with them on bounded sets, the bounded weak-$^*$ (or equicontinuous weak-$^*$) topology and Mackey topology. This identification follows from the Mackey-Arens and Banach-Dieudonné theorems from locally convex space theory. In this case, these topologies are the topologies of uniform convergence on norm compact and weakly compact absolutely convex subsets of $L^1(X)$ respectively.

The norm compact absolutely convex subsets of a Banach space are just the absolute convex hulls of null sequences, so the bounded weak-$^*$ topology is equal to the topology of uniform convergence on null sequences in $L^1(X)$. I don't know if this counts as an explicit description. Maybe it's possible to work out something better.

If $X$ is finite, the Dunford-Pettis Theorem identifies the relatively weakly compact subsets of $L^1(X)$ with the bounded equiintegrable sets. Thus, if $X$ is finite then the Mackey topology and strong topology are equal on bounded sets to the topology of convergence in measure. This contradicts the claim in your link, which states it for the $\sigma$-weak topology. For more general localizable measure spaces, you have to consider the topology of convergence in measure on subsets of $X$ with finite measure. The proof is similar to pretty much any other proof that represents an $L^\infty$ space as a projective limit of $L^\infty$ spaces of finite measure spaces.

The weak and strong operator topologies only agree (even on the unit ball) when $X$ is purely atomic, as this is precisely when every weakly compact subset of $L^1(X)$ is norm compact.

I've never seen a good description of these topologies on unbounded subsets of $L^\infty(X)$, and it seems that it would be difficult to give one. For example, if $X$ is a separable measure space then $L^\infty(X)$ has a separable predual, and thus all of the topologies mentioned here are metrizable on bounded sets, but they are never metrizable when $X$ is infinite. This precludes the Mackey topology from actually being equal to the topology of convergence in measure, because the latter is metrizable.

An interesting book on some of these topologies is Saks Spaces and Applications to Functional Analysis by J.B. Cooper. Since this was migrated from MSE, please let me know if you want more precise references for some of the things I've mentioned.

• I added a quick summary of your results to the question so that I might easily refer to it later. Please let me know if I have incorrectly summarized your answer. Commented Jan 6, 2017 at 21:45
• The only correction I can see is that the Mackey-Arens topology is always equal to the strong-$^*$ and $\sigma$-strong-$^*$ topologies on bounded sets, not just when $X$ is a finite measure space. When $X$ is a finite measure space this further agrees with the topology of convergence in measure. Commented Jan 6, 2017 at 21:51