Sup Path Attaining Representations

We first note that for a sup path attaining representation $T$ of $G$ we have

$$\Vert \mu\Vert \leq C\sup \left\{ {\rm ess\ sup}_{t\in G} \left\v... ... h {\rm\ is\ a\ simple\ function\ with\ }\ \Vert h\Vert _\infty\leq 1 \right\}.$$ (11)

To see this, note that for any bounded measurable function $h$, there exist a sequence of simple functions $s_n$ that converge uniformly to $h$. Then it is easy to see that

$${\rm ess\ sup}_{t\in G} \left\vert \int_\O s_n d (T_t\mu) \right\vert \to {\rm ess\ sup}_{t\in G} \left\vert \int_\O h d (T_t\mu) \right\vert ,$$

and from this (11) follows easily.

Our first example is related to the setting of Forelli [9].

Example 2.1   Let $G$ be a locally compact abelian group. Suppose that $\Omega$ is a topological space and $\left( T_t\right)_{t\in G}$ is a group of homeomorphisms of $\Omega$ onto itself such that the mapping

$$(t,\omega)\mapsto T_t\omega$$

is jointly continuous. Suppose that ${\cal A}$ is an algebra of bounded continuous complex valued functions on $\Omega$ such that if $h \in {\cal A}$, and if $\varphi:\mathbb{C}\to\mathbb{C}$ is any bounded continuous function, and if $t\in G$, then $\varphi\circ h\circ T_t\in {\cal A}$. Let $\sigma({\cal A})$ denote the minimal sigma-algebra so that functions from ${\cal A}$ are measurable. For any measure $\mu\in M(\sigma({\cal A}))$, and $A\in \sigma({\cal A})$, define $T_t \mu(A) = \mu(T_t(A))$, where $T_t(A)=\{T_t\omega:\ \omega\in A\}$. Note that $T$ satisfies (1) and (3). To discuss the weak measurability of $\mu$, and the sup path attaining property of $T$, we need that for each $h \in {\cal A}$ that the map $t \mapsto \int_\Omega h \, d(T_t\mu)$ is continuous. This crucial property follows for any measure $\mu\in M(\sigma({\cal A}))$, by the dominated convergence theorem, if, for example, $G$ is metrizable. For arbitrary locally compact abelian groups, it is enough to require that $\mu$ has sigma-compact support.

Henceforth, we assume that $G$ is metrizable, or that $\mu\in M(\sigma({\cal A}))$ has sigma-compact support, and proceed to show that $\mu$ is weakly measurable in the sense of Definition 1.1, and that the representation $T$ is sup path attaining. Let

$${\cal R}=\{A\in \sigma({\cal A}) :\ 1_A=\lim_n f_n,\ f_n \in {\cal A},\ \Vert f_n\Vert _\infty\leq 1\},$$

and let

$${\cal C}=\{A\in \sigma({\cal A}): \ t\mapsto T_t\mu(A)\ {\rm is\ Borel\ measurable}\}.$$

Clearly, ${\cal C}$ is a monotone class, closed under nested unions and intersections. Also, ${\cal R}$ is an algebra of sets, closed under finite unions and set complementation. Furthermore, it is clear that ${\cal R}\subset {\cal C}$. Hence, by the monotone class theorem, it follows that ${\cal C}$ contains the sigma algebra generated by ${\cal R}$. Now, if $f\in{\cal A}$ and $V \subset \mathbb{C}$ is open, then $f^{-1}(V) \in {\cal R}$, because

$$1_{f^{-1}(V)}(\omega)= \lim_{n\rightarrow \infty} \min \{ n\ {\rm dist}(f(\omega),\mathbb{C}\setminus V) ,1\}.$$

Consequently, we have that ${\cal C}= \sigma({\cal A})$, that is, $\mu$ is weakly measurable.

Next, let us show that ${\cal A}$ is dense in $L^1(\vert\mu\vert)$. Let $g:\Omega\to\mathbb{C}$ be a bounded $\sigma({\cal A})$-measurable function such that $\int_\Omega hg\,d\mu = 0$ for all $h \in {\cal A}$. Define ${\cal R}$ as above, and let

$${\cal C}= \left\{ A \in \sigma({\cal A}): \int_A g \, d\mu = 0 \right\} .$$

Then ${\cal C}$ is a monotone class containing ${\cal R}$, and so arguing as before, ${\cal C}= \sigma({\cal A})$, that is, $g = 0$ almost everywhere with respect to $\vert\mu\vert$. Thus it follows by the Hahn-Banach Theorem that ${\cal A}$ is dense in $L^1(\vert\mu\vert)$. Hence

$$\Vert\mu\Vert = \sup\left\{\left\vert\int_\Omega h \, d\mu \right... ...\, d(T_t\mu)\right\vert : h \in {\cal A},\ \Vert h\Vert _\infty \le 1\right\} ,$$

where the last inequality follows from the fact that the map $t \mapsto \int_\Omega h \, d(T_t\mu)$ is continuous. Hence, $T$ is sup path attaining with $C = 1$.

Taking ${\cal A}$ to be the uniformly continuous functions on a group, we then derive the following useful example.

Example 2.2   Suppose that $G_1$ and $G_2$ are locally compact abelian groups and that $\phi:\ G_2\rightarrow G_1$ is a continuous homomorphism. Define an action of $G_2$ on $M(G_1)$ (the regular Borel measures on $G_1$) by translation by $\phi$. Hence, for $x\in G_2, \mu\in M(G_1)$, and any Borel subset $A\subset G_1$, let $T_x\mu(A)=\mu(A+\phi(x))$. Then every $\mu\in M(G_1)$ is weakly measurable, the representation is sup path attaining with constants $c = 1$ and $C = 1$.

In the following example no topology is required on the measure space.

Example 2.3   Let $(X,\Sigma)$ be an abstract Lebesgue space, that is, $\Sigma$ is countably generated. Then any uniformly bounded representation $T$ of $G$ by isomorphisms of $M(\Sigma)$ is sup path attaining. To see this, note that since $\Sigma$ is countably generated, there is a countable subset ${\cal A}$ of the unit ball of ${\cal L}^\infty(\Sigma)$ such that for any $\mu\in M(\Sigma)$ we have

$$\Vert\mu\Vert =\sup\left\{ \left\vert \int_X h d \mu\right\vert :\ \ h\in {\cal A} \right\} .$$

If $\mu$ is weakly measurable, then for $g\in {\cal A}$ and for locally almost all $u\in G$, we have

$$\left\vert \int_X g d (T_u \mu) \right\vert \leq {\rm ess\ sup}_{t\in G} \left\vert \int_X g d (T_t \mu) \right\vert.$$ (12)

Since ${\cal A}$ is countable we can find a subset $B$ of $G$ such that the complement of $B$ is locally null, and such that (12) holds for every $u\in B$ and all $g\in {\cal A}$. For $u\in B$, take the sup in (12) over all $g\in {\cal A}$, and get

$$\Vert T_u \mu\Vert \leq \sup_{g\in {\cal A}} {\rm ess\ sup}_{t\in G} \left\vert \int_X g d (T_t \mu) \right\vert.$$

But since $\Vert\mu\Vert\leq c \Vert T_u \mu\Vert$, it follows that $T$ is sup path attaining with $C=c$.

Sup path attaining representations satisfy the following property, which was introduced in [1] and was called hypothesis $(A)$.

Proposition 2.4   Suppose that $T$ is sup path attaining and $\mu$ is weakly measurable such that for every $A\in \Sigma$ we have

$$T_t\mu(A)=0$$

for locally almost all $t\in G$. Then $\mu=0$.

The proof is immediate and follows from (11).

The key in Proposition 2.4 is that the set of $t\in G$ for which $T_t\mu(A)=0$ depends on $A$. If this set were the same for all $A\in \Sigma$, then the conclusion of the proposition would trivially hold for any representation by isomorphisms of $M(\Sigma)$.

For further motivation, we recall the following example from [1].

Example 2.5   (a) Let $\Sigma$ denote the sigma algebra of countable and co-countable subsets of $\mathbb{R}$. Define $\nu\in M(\Sigma)$ by

$$\nu (A)=\left\{ \begin{array}{ll} 1 & \mbox{if $A$ is co-countable,}\\ 0 & \mbox{if $A$ is countable.} \end{array}\right.$$

Let $\delta_t$ denote the point mass at $t \in \mathbb{R}$, and take $\mu=\nu-\delta_0$. Consider the representation $T$ of $\mathbb{R}$ given by translation by $t$. Then:
$\mu$ is weakly measurable;
$\Vert\mu\Vert>0$;
$T_t(\mu)=T_t(\nu-\delta_0)=\nu-\delta_t$;
for every $A\in \Sigma$, $T_t(\mu)(A)=0$ for almost all $t \in \mathbb{R}$.
It now follows from Proposition 2.4 that the representation $T$ is not sup path attaining.
(b) Let $\alpha$ be a real number and let $\Sigma,\ \mu, \nu, \delta_t$, and $T_t$ have the same meanings as in (a). Define a representation $T^\alpha$ by

$$T_t^\alpha = e^{i\alpha t}T_t.$$

Arguing as in (a), it is easy to see that $T^\alpha$ is not sup path attaining.