Proof of the Main Theorem

Throughout this section, $G$ denotes a locally compact abelian group with dual group $\Gamma$; $M(\Sigma)$ is a space of measures on a set $\O$; and $T=(T_t)_{t\in G}$ is a sup path attaining representation of $G$ by isomorphisms of $M(\Sigma)$.

If $\phi$ is in $L^\infty(G)$, write $\left[ \phi\right]$ for the smallest weak-* closed translation-invariant subspace of $L^\infty(G)$ containing $\phi$, and let ${\cal I}([\phi])={\cal I}(\phi)$ denote the closed translation-invariant ideal in $L^1(G)$:

$${\cal I}(\phi)=\{f\in L^1(G): f*\phi=0\}.$$

It is clear that ${\cal I}(\phi)=\{f\in L^1(G): f*g=0, \forall g\in \left[\phi\right]\}$. The spectrum of $\phi$, denoted by $\sigma \left[\phi\right]$, is the set of all continuous characters of $G$ that belong to $\left[ \phi\right]$. This closed subset of $\Gamma$ is also given by

$$\sigma \left[\phi\right]=Z({\cal I}(\phi)).$$ (13)

(See [14, Chapter 7, Theorem 7.8.2, (b), (c), and (d)].)

Remarks 3.1   (a) For a weakly measurable $\mu\in M(\Sigma)$, since ${\cal I}(\mu)$ is a closed ideal in $L^1(G)$, it is translation-invariant, by [14, Theorem 7.1.2]. It follows readily that for all $t\in G$,

$${\cal I}(T_t\mu)={\cal I}(\mu),$$

and hence

$${\rm spec}_T (T_t(\mu))={\rm spec}_T (\mu).$$ (14)

(b) Let $\mu\in M(\Sigma)$ be weakly measurable and let $E\in \Sigma$. It is clear that ${\cal I}(\mu)\supset{\cal I}(t\mapsto (T_t\mu)(E))$. Thus $\sigma\left[ t\mapsto (T_t\mu)(E)\right]\subset {\rm spec}_T(\mu)$.

Lemma 3.2   Let $\mu\in{\cal M}_T(\Sigma)$.
(a) If $g\in L^1(G)$ and $E\in \Sigma$, then

$$\int_G g(t)T_t\mu(E)\overline{\chi(t)}\,dt=0,$$

for all $\chi$ not in ${\rm supp}(\widehat{g})+{\rm spec}_T(\mu)$.
(b) If $\nu \in M(G)$, then

$${\rm spec}_T(\nu*_T\mu)\subset {\rm supp}(\widehat{\nu})\cap {\rm spec}_T(\mu)\,.$$

(c) If $(k_\alpha)$ is an approximate identity for $L^1(G)$, then

$$\Vert\mu\Vert _{M(\Sigma)}/C\leq \mathop{\hbox{\rm lim inf}}\limi... ...mapsto \Vert k_\alpha *_T T_t\mu\Vert _{M(\Sigma)}\right \Vert _{L^\infty(G)},$$

where $C$ is as in (7).

Proof. (a) Fix $E\in \Sigma$ and $\chi\in\Gamma$, and let $f(t)=(T_{-t}\mu)(E)\chi(t)$. Then $\sigma\left[ f\right]\subset \chi -{\rm spec}_T(\mu)$. Hence, for $g\in L^1(G)$ with supp$(\widehat{g})\cap (\chi-{\rm spec}_T\mu)=\emptyset$, we have

$$\int_Gg(t)(T_t\mu)\overline{\chi}(t)dt=(g*f)(0)=0\,.$$

(b) Immediate from ${\cal I}(\nu*_T\mu)\supset {\cal I}(\nu)\cup{\cal I}(\mu)\,.$

(c) Fix $h$ on $\Omega$ with $\vert h\vert\leq 1$ and let $f(t)=\int_\Omega h(x)d(T_t\mu(x))$. Then $f\in L^\infty(G)$ and so, if $(k_\alpha)$ is an approximate identity for $L^1(G)$, then $k_\alpha *f\rightarrow f$ in the weak $*$-topology of $L^\infty(G)$. Hence

$$\Vert f\Vert _\infty\leq \mathop{\hbox{\rm lim inf}}\limits \Vert k_\alpha *f\Vert _\infty = \mathop{\hbox{\rm lim inf}}\limits \left\Vert t\rightarrow \int_\Omega h(x)d(k_\alpha*_T T_t\mu) \right\Vert _\infty$$

$$\leq \mathop{\hbox{\rm lim inf}}\limits \left\Vert t\rightarrow \Vert k_\alpha*_T T_t\mu\Vert \right\Vert _\infty\,.$$

The proof is completed by taking the sup of $\Vert f\Vert _\infty$ over all $h$ and using (7).

Lemma 3.3   Let $f:\ G\rightarrow M(\Sigma)$ be bounded and continuous, and let $\nu \in M(G)$. Suppose that
(i) for all $E\in \Sigma$, $t\mapsto f(t)E$ is in $L^1_S(G)$;
(ii) for all $g\in L^1_S(G),\ \Vert\nu*g\Vert _1\leq \Vert g\Vert _1\,.$
Then

$$\int_G\Vert(\nu*f)(t)\Vert _{M(\Sigma)}dt\leq \int_G \Vert f (t)\Vert _{M(\Sigma)}dt\,.$$ (15)

Proof. We suppose throughout the proof that the right side of (15) is finite; otherwise there is nothing to prove. Write $f_E(t)=f(t)E$. Thus

$$(\nu*f)(t)E=(\nu*f_E)(t)$$

is continuous on $G$. By (i) and (ii),

$$\int_G\vert(\nu*f)(t)E\vert dt\leq \int_G\vert f(t)E\vert dt,$$ (16)

for all $E\in \Sigma$. Now let $P=\{E_j\}$ be a finite measurable partition of $\Omega$, and define

$$h(P,t)= \sum_j\vert(\nu*f)(t)E_j\vert\,.$$

Then $h(P,\cdot)$ is continuous on $G$ and, by (16),

$$\int_G h(P,t)dt=\sum_j\int_G\vert(\nu*f)(t)E_j\vert dt \leq \int_G\Vert f(t)\Vert _{M(\Sigma)}dt\,.$$ (17)

If $R$ is a common refinement of $P$ and $Q$, then, for all $t$,

$$\max\{h(P,t),h(Q,t)\}\leq h(R,t)\,.$$

It follows from [12, (11.13)] that

$$\int_G\Vert\nu*f\Vert _{M(\Sigma)}dt = \int_G \sup_Ph(P,t)dt$$

$$= \sup_P\int_G h(P,t)dt \leq \int_G\Vert f(t)\Vert _{M(\Sigma)}dt$$

by (17).

Proof of Theorem 1.6. Let $f(t)=T_t\mu$ and suppose first that $f$ is continuous and that there is a neighborhood $V$ of $0\in\Gamma$ such that $V+{\rm spec}_T(\mu)\subset S$. Choose a continuous $g\in L^1(G)$ such that

$$g\geq 0;$$

$$\widehat{g}(0)= \int_G g(t)dt=1;$$

and

$${\rm supp}(\widehat{g})\subset V\,.$$

Then, for all $E\in \Sigma$, the mapping $t\mapsto g(t)f(t)E$ is bounded and continuous on $G$, and, by Lemma 3.2, belongs to

$$L^1_{{\rm supp}(\widehat{g})+{\rm spec}_T(\mu)}\subset L^1_{V+{\rm spec}_T(\mu)}\subset L^1_S\,.$$

Hence Lemma 3.3 implies that

$$\Vert\nu*(gf)\Vert _{L^1(G,M(\Sigma))}\leq \Vert gf\Vert _{L^1(G,M(\Sigma))}\,.$$ (18)

We now proceed to show that (10) is a consequence of (18). We start with the right side of (18):

$$\Vert gf\Vert _1 = \int_Gg(t)\Vert f(t)\Vert dt$$

$$\leq c \Vert\mu\Vert \int_Gg(t) dt =c \Vert\mu\Vert.$$ (19)

Consider now the left side of (18). We have

$$\Vert\nu *(gf)\Vert _1 = \int_G\Vert\nu *(gf)(t)\Vert dt$$

$$= \int_G\Vert\int_G (gf) (t-s)d\nu(s)\Vert dt$$

$$= \int_G\Vert\int_G g (t-s) T_{t-s} \mu d \nu (s)\Vert dt$$

$$\geq \frac{1}{c} \int_G\Vert\int_G T_{-t}\left[ g (t-s) T_{t-s} \mu \right] d \nu (s)\Vert dt$$

$$= \frac{1}{c} \int_G\Vert\int_G g (t-s) T_{-s} \mu d \nu (s)\Vert dt$$

$$\geq \frac{1}{c} \Vert \int_G \int_G g (t-s) d t T_{-s} \mu d \nu (s)\Vert$$

$$= \frac{1}{c} \widehat{g}(0) \Vert \int_G T_{-s} \mu d \nu (s)\Vert$$

$$= \frac{1}{c} \Vert \nu*_T \mu \Vert.$$ (20)

Inequalities (18), (19), and (20) imply that

$$\Vert\nu*_T\mu\Vert\leq c^2 \Vert\mu\Vert\,.$$ (21)

Now fix $\epsilon>0$ and let $k\in L^1(G)$ be such that $\widehat{k}$ has compact support in $\Gamma$. We have by Lemma 3.2

$${\rm spec}_T (k*_T \mu)\subset$$   supp$$(\widehat{k}) \cap {\rm spec}_T(\mu)\,.$$

Moreover, the set $\{\gamma\in\Gamma:\ \int_G\vert 1-\gamma\vert d\vert\nu\vert<\epsilon\}$ is a neighborhood of $0\in\Gamma$. Hence, by the hypothesis on $S$, there is a neighborhood $V$ of 0 in $\Gamma$ and $\gamma\in\Gamma$ such that

$$\int_G\vert 1-\gamma\vert d\vert\nu\vert<\epsilon$$   and$$\quad V+{\rm spec}_T(k*_T\mu)\subset S-\gamma\,.$$

If $h\in L^1_{S-\gamma}$, then $\gamma h\in L^1_S$; hence

$$\Vert(\overline{\gamma}\nu)*h\Vert _1=\Vert\nu*(\gamma h)\Vert _1\leq \Vert\gamma h\Vert _1=\Vert h\Vert _1,$$

for all $h\in L^1_{S-\gamma}$. Since translation in $L^1(G)$ is continuous, it follows that the map $t\mapsto T_t(k*_T\mu)$ is continuous, and hence we obtain from (21),

$$\Vert(\overline{\gamma}\nu)*_T(k*_T\mu)\Vert\leq c^2\Vert k*_T\mu\Vert\,.$$

As $\Vert\overline{\gamma}\nu-\nu\Vert<\epsilon$ and $\epsilon$ is arbitrary, we get

$$\Vert k*_T(\nu*_T\mu)\Vert=\Vert\nu*_T(k*_T\mu)\Vert\leq c^2\Vert k*_T\mu\Vert\leq c^3\Vert\mu\Vert\,.$$

Letting $k$ run through an approximate identity of $L^1(G)$ and using Lemma 3.2, we obtain $\Vert \nu*_T\mu \Vert\leq Cc^3\Vert\mu\Vert$.