Homomorphism theorems

We continue with the notation of the previous section: $G$ is a locally compact abelian group, $\Gamma$ the dual group of $G$, $P$ is a measurable order on $\Gamma$, $T$ is a sup path attaining representation of $G$ acting on $M(\Sigma)$. Associated with $P$ is a collection of homomorphisms $\psi_\alpha$, as described by Theorem 2.1. Let $\phi_\alpha$ denote the adjoint of $\psi_\alpha$. Thus, $\phi_\alpha$ is a continuous homomorphism of $\Bbb R$ into $G$. By composing the representation $T$ with the $\phi_\alpha$, we define a new representation $T_{\phi_\alpha}$ of $\Bbb R$ acting on $M(\Sigma)$ by: $t\in \Bbb R\mapsto T_{\phi_\alpha(t)}$. If $\mu$ in $M(\Sigma)$ is weakly measurable with respect to $T$ then $\mu$ is also weakly measurable with respect to $T_{\phi_\alpha}$. We will further suppose that $T_{\phi_\alpha}$ is sup path attaining for each $\alpha$. This is the case with the representations of Example 1.7.

Our goal in this section is to relate the notion of analyticity with respect to $T$ to the notion of analyticity with respect to $T_{\phi_\alpha}$. More generally, suppose that $G_1$ and $G_2$ are two locally compact abelian groups with dual groups $\Gamma_1$ and $\Gamma_2$, respectively. Let

$$\psi: \Gamma_1\rightarrow \Gamma_2$$

be a continuous homomorphism, and let $\phi:\ G_2\rightarrow G_1$ denote its adjoint homomorphism. Suppose $\nu$ is in $M(G_2)$. We define a Borel measure $\Phi(\nu)$ in $M(G_1)$ on the Borel subsets $A$ of $G_1$ by:

$$\Phi(\nu)(A)=\int_{G_2} 1_A\circ \phi(t)\,d\nu(t)= \int_{G_1} 1_A d\Phi(\nu),$$ (21)

where $1_A$ is the indicator function of $A$. We have $\Vert\Phi(g)\Vert _{M(G_1)}=\Vert\nu\Vert _{M(G_2)}$ and, for every Borel measurable bounded function $f$ on $G_1$, we have

$$\int_{G_1} f d\Phi(\nu)= \int_{G_2} f\circ \phi(t)\,d \nu (t).$$ (22)

In particular, if $f=\chi$, a character in $\Gamma$, then

$$\widehat{\Phi(\nu)}(\chi)=\int_{G_1} \overline{\chi} d\Phi(\nu)=\... ...nu(t) =\int_{G_2} \overline{\psi(\chi)}(t)\, d\nu(t)=\widehat{\nu}(\psi(\chi)),$$ (23)

where $\psi$ is the adjoint homomorphism of $\phi$. So,

$$\widehat{\Phi(\nu)}=\widehat{\nu}\circ \psi.$$ (24)

Our first result is a very useful fact from spectral synthesis of bounded functions. The proof uses in a crucial way the fact that the representation is sup path attaining, or, more precisely, satisfies the property in Proposition 1.4.

Lemma 4.1   Suppose that $T$ is a sup path attaining representation of $G_1$ acting on $M(\Sigma)$, $\phi$ is a continuous homomorphism of $G_2$ into $G_1$ such that $T_\phi$ is a sup path attaining representation of $G_2$. Suppose that $B$ is a nonempty closed $S$-subset of $\Gamma_1$ and that $\mu$ is in $M(\Sigma)$ with ${\rm spec}_T\mu\subset B$. Suppose that $C$ is an $S$-subset of $\Gamma_2$ and $\psi(B)\subset C$. Then ${\rm spec}_{T_\phi}\mu\subset C$.

Proof.    Since $C$ is an $S$-subset of $\Gamma_2$, it is enough to show that for every ` $A\in \Sigma$, ${\rm spec}_{T_\phi}(x\mapsto T_{\phi(x)}\mu(A))\subset C$, by Proposition 3.7. For this purpose, it is enough by [21, Theorem (40.8)], to show that

$$g*T_{\phi(\cdot)}\mu(A)=0$$

for every $g$ in $L^1(G_2)$ with $\widehat{g}= 0$ on $C$. For $r\in G_2$ and $x\in G_1$, consider the measure

$$T_x(g*_{T_\phi}T_{\phi(r)}\mu)= g*_{T_\phi}T_{x+\phi(r)}\mu.$$

For $A\in \Sigma$, we have

$$g*_{T_\phi} T_{x+\phi(r)}\mu(A) = \int_GT_{-t+x}(T_{\phi(r)}\mu)(A)\,d\Phi(g)(t)$$

$$= \Phi(g)*[t\mapsto T_t(T_{\phi(r)}\mu)(A)](x)$$

$$=$$ 0

for almost all $x\in G_1$. To justify the last equality, we appeal to Proposition 3.3 and note that $\widehat{\Phi(g)}=\widehat{g}\circ \psi$ and so $\widehat{\Phi(g)}=0$ on $B\subset \psi^{-1}(C)$. Moreover, $\sigma[t\mapsto T_t(T_{\phi(r)}\mu)(A)] \subset {\rm spec}_T(\mu)\subset B$. Now, using Proposition 1.4 and the fact that, for every $A\in \Sigma$,

$$T_x[g*_{T_\phi} T_{\phi(r)}]\mu(A) =g*_{T_\phi}T_{x+\phi(r)}\mu(A)=0,$$

for almost all $x\in G_1$, we conclude that the measure $g*_{T_\phi} T_{\phi(r)}\mu$ is the zero measure, which completes the proof.

Given ${\cal C}$, a collection of elements in $L^1(G_1)$ or $M(G_1)$, let

$$Z({\cal C})=\bigcap_{\delta\in {\cal C}} \left\{ \chi:\ \widehat{\delta}(\chi)=0\right\}.$$

This is the same notation for the zero set of an ideal in $L^1(G)$ that we introduced in Section 1. Given a set of measures ${\cal S}$ in $M(G_2)$, let

$$\Phi({\cal S})=\left\{ \Phi(\nu):\ \nu \in {\cal S}\right\}\subset M(G_1).$$

Lemma 4.2   In the above notation, if $\mu\in M(\Sigma)$ is weakly measurable, then

$$Z\left( \Phi({\cal I}_{T_\phi}\mu)\right)= \psi^{-1}\left( Z ({\cal I}_{T_\phi}\mu)\right)= \psi^{-1}\left( {\rm spec}_{T_\phi}\mu\right).$$

Proof. It is enough to establish the first equality; the second one follows from definitions. We have

$$Z\left( \Phi({\cal I}_{T_\phi}\mu)\right) = \bigcap_{\delta\in \Phi({\cal I}_{T_\phi}(\mu)) } \left\{ \chi\in \Gamma:\ \widehat{\delta}(\chi)=0\right\}$$

$$= \bigcap_{g \in {\cal I}_{T_\phi}(\mu) } \left\{ \chi\in \Gamma:\ \widehat{\Phi(g)}(\chi)=0\right\}$$

$$= \bigcap_{g \in {\cal I}_{T_\phi}(\mu) } \left\{ \chi\in \Gamma:\ \widehat{g }(\psi(\chi))=0\right\}$$

$$= \bigcap_{g \in {\cal I}_{T_\phi}(\mu) } \psi^{-1}\left( Z(g) \right)$$

$$= \psi^{-1}\left( \bigcap_{g \in {\cal I}_{T_\phi}(\mu) } \left( Z(g) \right)\right)$$

$$= \psi^{-1}\left( Z \left( {\cal I}_{T_\phi}(\mu) \right)\right) =\psi^{-1}\left({\rm spec}_{T_\phi}\mu\right).$$

Lemma 4.3   Suppose that $C$ is a nonempty closed $S$-subset of $\Gamma_2$ and that $\psi^{-1}(C)$ is an $S$-subset of $\Gamma_1$. Suppose that $\mu$ is in $M(\Sigma)$ and ${\rm spec}_{T_\phi}(\mu)\subset C$. Then ${\rm spec}_T\mu\subset \psi^{-1}(C)$.

Proof.     We will use the notation of Lemma 4.2. If $f\in {\cal I}_{T_\phi}(\mu)$ and $t\in G_1$, then $f\in {\cal I}_{T_\phi}(T_t \mu)$. So, for $A\in \Sigma$, we have $f*_{T_\phi}(T_t\mu)(A)=0$. But

$$f*_{T_\phi}(T_t\mu)(A) = \int_\Bbb RT_{t-\phi(x)}\mu(A)f(x)\, dx$$

$$= \int_G T_{t- x }\mu(A)\, d\Phi(f),$$

where $\Phi(f)$ is the homomorphic image of the measure $f(x)\,dx$. Hence, $\Phi(f) \in {\cal I}_T ([t\mapsto T_t\mu(A)])$, and so $\Phi({\cal I}_{T_\phi}(\mu))\subset {\cal I}_T([t\mapsto T_t\mu(A)])$, which implies that

$$Z\left(\Phi({\cal I}_{T_\phi}(\mu))\right)\supset Z\left({\cal I}_T([t\mapsto T_t\mu(A)])\right) ={\rm spec}_T(t\mapsto T_t\mu(A)).$$

By Lemma 4.2,

$$Z\left(\Phi({\cal I}_{T_\phi}(\mu))\right)= \psi^{-1}\left( {\rm spec}_{T_\phi}\mu \right)\subset \psi^{-1}(C).$$

Hence, ${\rm spec}_T(t\mapsto T_t\mu(A))\subset \psi^{-1}(C)$ for all $A\in \Sigma$, which by Proposition 3.7 implies that ${\rm spec}_T(\mu)\subset \psi^{-1}(C)$.

Taking $G_1=G,\ G_2=\Bbb R$ and $\psi=\psi_\alpha$ to be one of the homomorphisms in Theorem 2.1, and using the fact that $[0,\infty[$, $S_\alpha$, $C_\alpha\setminus D_\alpha$ are all $S$-sets, we obtain useful relationships between different types of analyticity.

Theorem 4.4   Let $G$ be a locally compact abelian group with ordered dual group $\Gamma$, and let $P$ denote a measurable order on $\Gamma$. Suppose that $T$ is a sup path attaining representation of $G$ by isomorphisms of $M(\Sigma)$, such that $T_{\phi_\alpha}$ is sup path attaining, where $\phi_\alpha$ is the adjoint of $\psi_\alpha$ in Theorem 2.1.
(i) If $\mu\in M(\Sigma)$ and ${\rm spec}_T(\mu)\subset C_\alpha\setminus D_\alpha$. Then

$${\rm spec}_T(\mu)\subset S_\alpha\Leftrightarrow {\rm spec}_{T_{\phi_\alpha}}(\mu)\subset [0,\infty[.$$

(ii) If $\mu\in M(\Sigma)$ and ${\rm spec}_T(\mu)\subset C_{\alpha_0}$. Then

$${\rm spec}_T(\mu)\subset S_{\alpha_0}\Leftrightarrow {\rm spec}_{T_{\phi_{\alpha_0}}}(\mu)\subset [0,\infty[.$$

We can use the representation $T_\phi$ to convolve a measure $\nu\in M(G_2)$ with $\mu\in M(G_1)$:

$$\nu*_{T_\phi}\mu(A)=\int_{G_2}T_{-\phi(x)}\mu(A)d\nu(x)=\int_{G_2}\mu(A-\phi(x))\,d\nu(x) ,$$

for all Borel $A\subset G_1$.

Alternatively, we can convolve $\Phi(\nu)$ in the usual sense of [20, Definition 19.8] with $\mu$ to yield another measure in $M(G_1)$, defined on the Borel subsets of $G_1$ by

$$\Phi(\nu)*\mu(A)=\int_{G_1}\int_{G_1}1_A(x+y)d\Phi(\nu)(x)d\mu(y).$$

Using (22), we find that

$$\Phi(\nu)*\mu(A) = \int_{G_1}\int_{G_2}1_A(\phi(t)+y)d\nu(t)d\mu(y)$$

$$= \int_{G_2}\mu(A-\phi(t))d\nu(t) =\nu *_{T_\phi}\mu(A).$$

Thus,

$$\Phi(\nu)*\mu=\nu*_{T_\phi}\mu.$$ (25)

We end the section with homomorphism theorems, which complement the well-known homomorphism theorems for $L^p$-multipliers (see Edwards and Gaudry [11, Appendix B]). In these theorems, we let $G_1$ act on $M(G_1)$ by translation. That is, if $\mu\in M(G_1)$, $x\in G_1$, and $A$ is a Borel subset of $G_1$, then

$$T_x\mu(A)=\mu(A+x).$$

Let $\phi:\ G_2\rightarrow G_1$ be a continuous homomorphism. By Example 1.3, $T$ and $T_\phi$ are sup path attaining. (Recall that if $t\in G_2$, $\mu\in M(G_1)$, then $T_{\phi(t)}\mu(A)=\mu(A+\phi(t))$.) A simple exercise with definitions shows that for $\mu\in M(G_1)$

$${\rm spec}_T\mu={\rm supp}\widehat{\mu}.$$

Theorem 4.5   Suppose that $\Gamma_1$ and $\Gamma_2$ contain measurable orders $P_1$ and $P_2$, respectively, and $\psi:\ \Gamma_1\rightarrow \Gamma_2$ is a continuous, order-preserving homomorphism (that is, $\psi(\overline{P_1})\subset \overline{P_2}$). Suppose that there is a positive constant $N(\nu)$ such that for all $f\in H^1(G_2)$

$$\Vert\nu*f\Vert _1\leq N(\nu)\Vert f\Vert _1.$$ (26)

Then

$$\Vert\Phi(\nu)*\mu\Vert\leq N(\nu)\Vert\mu\Vert$$ (27)

for all Borel measures in $M(G_1)$ such that $\widehat{\mu}$ is supported in $\overline{P_1}$.

Proof.    We have $\Phi(\nu)*\mu=\nu*_{T_\phi}\mu$. Also $\overline{P_2}$ is a ${\cal T}$-set. So (27) will follow from Theorem 1.8 once we show that ${\rm spec}_{T_\phi}\mu\subset \overline{P_2}$. For that purpose, we use Lemma 4.1. We have

$${\rm spec}_T\mu ={\rm supp}\widehat{\mu}\subset \overline{P_1},$$

and $\psi(\overline{P_1})\subset \overline{P_2}$ is an $S$-set. Hence ${\rm spec}_{T_\phi}\mu\subset \overline{P_2}$ by Lemma 4.1.

The following special case of Theorem 4.5 deserves a separate statement.

Theorem 4.6   With the above notation, suppose that there is a positive constant $N(\nu)$ such that for all $f\in H^1(G_2)$

$$\Vert\nu*f\Vert _1\leq N(\nu)\Vert f\Vert _1.$$ (28)

Then for all $f\in H^1(G_1)$ we have

$$\Vert\Phi(\nu)*f\Vert _1\leq N(\nu)\Vert f\Vert _1.$$ (29)

Theorem 4.7   Suppose that there is a positive constant $N(\nu)$ such that for all $f\in H^1(\Bbb R)$

$$\Vert\nu*f\Vert _1\leq N(\nu)\Vert f\Vert _1.$$ (30)

Then for all $\mu\in M(G)$ with support of $\widehat{\mu}$ contained in $C_\alpha\setminus D_\alpha$, where $\alpha< \alpha_0$, we have

$$\Vert\Phi_\alpha(\nu)*\mu\Vert _1\leq N(\nu)\Vert\mu\Vert.$$ (31)

Proof.    The proof is very much like the proof of Theorem 4.5. We have $\Phi_\alpha (\nu)*\mu=\nu*_{T_\phi}\mu$. Apply Theorem 1.8, taking into consideration that

$${\rm spec}_T\mu={\rm supp}\widehat{\mu}\subset C_\alpha\setminus D_\alpha$$

is an $S$-set and so

$${\rm spec}_{T_{\phi_\alpha}}\mu\subset \psi_\alpha(C_\alpha\setminus D_\alpha)\subset [0,\infty[.$$