Analyticity

We continue with the notation of the previous section. Using the order structure on $\Gamma$ we define some classes of analytic functions on $G$:

$$H^1(G) = \left\{ f\in L^1(G): \widehat{f}=0\ {\rm on}\ (-P)\setminus\{0\} \right\};$$ (14)

$$H^1_0(G) = \left\{ f\in L^1(G): \widehat{f}=0\ {\rm on}\ -P \right\};$$ (15)

and

$$H^\infty(G)=\left\{ f\in L^\infty(G): \int_G f(x)g(x)dx=0 \ {\rm for\ all}\ g\in H^1_0(G) \right\}.$$ (16)

We clearly have

$$H^1(G)=\left\{ f\in L^1(G): \widehat{f}=0\ {\rm on}\ \overline{(-P)\setminus\{0\}} \right\}.$$

We can now give the definition of analytic measures in ${\cal M}_T(\Sigma)$.

Definition 3.1   Let $T$ be a sup path attaining representation of $G$ by isomorphisms of $M(\Sigma)$. A measure $\mu\in{\cal M}_T(\Sigma)$ is called weakly analytic if the mapping $t\mapsto T_t\mu(A)$ is in $H^\infty(G)$ for every $A\in \Sigma$.

Definition 3.2   Recall the $T$-spectrum of a weakly measurable $\mu\in{\cal M}_T(\Sigma)$,

$${\rm spec}_T (\mu)= \bigcap_{f\in {\cal I}(\mu)} \left\{ \chi\in\Gamma:\ \ \widehat{f}(\chi)=0 \right\}.$$ (17)

A measure $\mu$ in ${\cal M}_T(\Sigma)$ is called $T$-analytic if ${\rm spec}_T (\mu)\subset \overline{P}$.

That the two definitions of analyticity are equivalent will be shown later in this section.

Since ${\cal I}(\mu)$ is translation-invariant, it follows readily that for all $t\in G$,

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

and hence

$${\rm spec}_T (T_t(\mu))={\rm spec}_T (\mu).$$ (18)

We now recall several basic results from spectral theory of bounded functions that will be needed in the sequel. Our reference is [21, Section 40]. 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

$$\sigma \left[\phi\right]=Z({\cal I}(\phi)).$$ (19)

(See [21, Theorem (40.5)].)

Recall that a closed subset $E$ of $\Gamma$ is a set of spectral synthesis for $L^1(G)$, or an $S$-set, if and only if ${\cal I}([E])$ is the only ideal in $L^1(G)$ whose zero set is $E$.

There are various equivalent definitions of $S$-sets. Here is one that we will use at several occasions.
A set $E\subset \Gamma$ is an $S$-set if and only if every essentially bounded function $g$ in $L^\infty(G)$ with $\sigma[g]\subset E$ is the weak-* limit of linear combinations of characters from $E$.
(See [21, (40.23) (a)].) This has the following immediate consequence.

Proposition 3.3   Suppose that $B$ is an $S$-set, $g\in L^\infty(G)$, and ${\rm spec}(g)\subset B$. (i) If $f$ is in $L^1(G)$ and $\widehat{f}=0$ on $B$, then $f*g(x)=0$ for all $x$ in $G$. In particular,

$$\int_G f(x) g(-x)\, d\, x =0.$$

(ii) If $\mu$ is a measure in $M(G)$ with $\widehat{\mu}=0$ on $B$, then $\mu*g(x)=0$ for almost all $x$ in $G$.

Proof.    Part (i) is a simple consequence of [21, Theorems (40.8) and (40.10)]. We give a proof for the sake of completeness. Write $g$ as the weak-* limit of trigonometric polynomials, $\sum_{\chi\in E} a_\chi \chi(x)$, with characters in $E$. Then

$$\int_G f(x)g(y-x)\, d\,x = \lim \int_G \sum_{\chi\in E} a_\chi \chi(y)f(x)\chi(-x)\, d\,x$$

$$= \lim \sum_{\chi\in E} a_\chi \chi(y) \widehat{f}(\chi)=0$$

since $\widehat{f}$ vanishes on $E$.

To prove (ii), assume that $\mu*g$ is not 0 a.e.. Then, there is $f$ in $L^1(G)$ such that $f*(\mu*g)$ is not 0 a.e.. But this contradicts (i), since $f*(\mu*g)=(f*\mu)*g$, $f*\mu$ is in $L^1(G)$, and $\widehat{f*\mu}=0$ on $B$.

The following is a converse of sorts of Proposition 3.3 and follows easily from definitions.

Proposition 3.4   Let $B$ be a nonvoid closed subset of $\Gamma$. Suppose that $f$ is in $L^\infty(G)$ and

$$\int_G f(x)g(x)dx=0$$ (20)

for all $g$ in $L^1(G)$ such that $\widehat{g}= 0$ on $-B$. Then $\sigma[f]\subset B$.

Proof.    Let $\chi_0$ be any element in $\Gamma\setminus B$. We will show that $\chi_0$ is not in the spectrum of $f$ by constructing a function $h$ in $L^1(G)$ with $\widehat{h}(\chi_0)\neq 0$ and $h*f= 0$. Let $U$ be an open neighborhood of $\chi_0$ not intersecting $B$, and let $h$ be in $L^1(G)$ such that $\widehat{h}$ is equal to 1 at $\chi_0$ and to 0 outside $U$. Direct computations show that the Fourier transform of the function $g:\ \ t\mapsto h(x-t)$, when evaluated at $\chi\in\Gamma$, gives $\overline{\chi(x)}\widehat{h}(-\chi)$, and hence it vanishes on $-B$. It follows from (20) that $h*f= 0$, which completes the proof.

A certain class of $S$-sets, known as the Calderón sets, or $C$-sets, is particularly useful to us. These are defined as follows. A subset $E$ of $\Gamma$ is called a $C$-set if every $f$ in $L^1(G)$ with Fourier transform vanishing on $E$ can be approximated in the $L^1$-norm by functions of the form $h*f$ where $h\in L^1(G)$ and $\widehat{h}$ vanishes on an open set containing $E$.
$C$-sets enjoy the following properties (see [21, (39.39)] or [22, Section 7.5]).

• Every $C$-set is an $S$-set.
• Every closed subgroup of $\Gamma$ is a $C$-set.
• The empty set is a $C$-set.
• If the boundary of a set $A$ is a $C$-set, then $A$ is a $C$-set.
• Finite unions of $C$-sets are $C$-sets.

Since closed subgroups are $C$-sets, we conclude that $\overline{P} \cap \overline{(-P)}$, and $C_\alpha$, for all $\alpha$, are $C$-sets. >From the definition of $S_{\alpha_0}$, (13), and the fact that $C_{\alpha_0}$ is open and closed, it follows that the boundary of $S_{\alpha_0}$ is the closed subgroup $\psi_{\alpha_0}^{-1}(0)\cap C_{\alpha_0}$. Hence $S_{\alpha_0}$ is a $C$-set. For $\alpha\neq\alpha_0$, the set $S_\alpha$ is open and closed, and so it has empty boundary, and thus it is a $C$-set. Likewise $C_\alpha\setminus D_\alpha$ is a $C$-set for all $\alpha\neq\alpha_0$. >From this we conclude that arbitrary unions of $S_\alpha$ and $C_\alpha\setminus D_\alpha$ are $C$-sets, because an arbitrary union of such sets, not including the index $\alpha_0$, is open and closed, and so it is a $C$-set.
We summarize our findings as follows.

Proposition 3.5   Suppose that $P$ is a measurable order on $\Gamma$. We have:
(i) $\overline{P}$ and $\overline{(-P)}$ are $C$-sets;
(ii) $S_\alpha$ is a $C$-set for all $\alpha$;
(iii) arbitrary unions of $S_\alpha$ and $C_\alpha\setminus D_\alpha$ are $C$-sets.

As an immediate application, we have the following characterizations.

Corollary 3.6   Suppose that $f$ is in $L^\infty(G)$, then
(i) $\sigma[f] \subset S_\alpha$ if and only if $\int_G f(x) g(x) d x =0$ for all $g\in L^1(G)$ such that $\widehat{g}= 0$ on $-S_\alpha$;
(ii) $\sigma[f] \subset \Gamma\setminus C_\alpha$ if and only if $\mu_\alpha *f=0$;
(iii) $\sigma[f] \subset \overline{P}$ if and only if $f\in H^\infty(G)$.

Proof. Assertions (i) and (iii) are clear from Propositions 3.5 and 3.4. To prove (ii), use Fubini's Theorem to first establish that for any $g\in L^1(G)$, and any $\mu\in M(G)$, we have

$$\int_G (\mu*f)(t) g(t)dt=\int_G f(t)(\mu*g)(t)dt.$$

Now suppose that $\sigma[f] \subset \Gamma\setminus C_\alpha$, and let $g$ be any function in $L^1(G)$. From Propositions 3.5 and 3.4, we have that $\int_G f g dt = 0$ for all $g$ with Fourier transform vanishing on $\Gamma\setminus C_\alpha$, equivalently, for all $g=\mu_\alpha*g$. Hence, $\int_G f (\mu_\alpha*g) dt = \int_G (\mu_\alpha*f) g dt=0$ for all $g$ in $L^1(G)$, from which it follows that $\mu_\alpha *f=0$. The converse is proved similarly, and we omit the details.

Aiming for a characterization of weakly analytic measures in terms of their spectra, we present one more result.

Proposition 3.7   Let $\mu$ be weakly measurable in $M(\Sigma)$.
(i) Suppose that $B$ is a nonvoid closed subset of $\Gamma$ and ${\rm spec}_T\mu\subset B$. Then $\sigma[t\mapsto T_t\mu(A)]\subset B$ for all $A\in \Sigma$.
(ii) Conversely, suppose that $B$ is an $S$-set in $\Gamma$ and that $\sigma[t\mapsto T_t\mu(A)]\subset B$ for all $A\in \Sigma$, then ${\rm spec}_T\mu\subset B$.

Proof. We clearly have ${\cal I}(\mu)\subset {\cal I}([t\mapsto T_t\mu(A)])$. Hence, ${\rm spec}_T \mu=Z({\cal I}(\mu))\supset Z({\cal I}([t\mapsto T_t\mu(A)]))=\sigma[t\mapsto T_t\mu(A)],$ and (i) follows.
Now suppose that $B$ is an $S$-set and let $g\in L^1(G)$ be such that $\widehat{g}= 0$ on $-B$. Then, for all $A\in \Sigma$, we have from Proposition 3.4:

$$\int_G g (t) T_t\mu (A)d t = 0.$$

Equivalently, we have that

$$\int_G g(-t) T_{-t}\mu(A) dt =0.$$

Since the Fourier transform of the function $t\mapsto g(-t)$ vanishes on $B$, we see that ${\cal I}(\mu) \supset \{f:\ \ \widehat{f}=0\ {\rm on}\ B\}$. Thus $Z({\cal I}(\mu))\subset Z(\{f:\ \ \widehat{f}=0\ {\rm on} \ B\})=B$, which completes the proof.

Straightforward applications of Propositions 3.5 and 3.7 yield the desired characterization of weakly analytic measures.

Corollary 3.8   Suppose that $\mu\in{\cal M}_T(\Sigma)$. Then,
(i) $\mu$ is weakly $T-$analytic if and only if ${\rm spec}_T\mu\subset \overline{P}$ if and only if $\sigma [t\mapsto T_t\mu (A)]\subset \overline{P}$, for every $A\in \Sigma$;
(ii) ${\rm spec}_T\mu\subset S_\alpha$ if and only if $\sigma[t\mapsto T_t\mu(A)]\subset S_\alpha$ for every $A\in \Sigma$.
(iii) ${\rm spec}_T\mu\subset C_\alpha$ if and only if $\sigma[t\mapsto T_t\mu(A)]\subset C_\alpha$ for every $A\in \Sigma$.
(iv) ${\rm spec}_T\mu\subset \Gamma\setminus C_\alpha$ if and only if $\sigma[t\mapsto T_t\mu(A)]\subset \Gamma\setminus C_\alpha$ for every $A\in \Sigma$.

The remaining results of this section are simple properties of measures in ${\cal M}_T(\Sigma)$ that will be needed later. Although the statements are direct analogues of classical facts about measures on groups, these generalization require in some places the sup path attaining property of $T$.

Proposition 3.9   Suppose that $\mu\in{\cal M}_T(\Sigma)$ and $\nu \in M(G)$. Then ${\rm spec}_T \nu*_T\mu$ is contained in the support of $\widehat{\nu}$, and ${\rm spec}_T \nu*_T\mu\subset {\rm spec}_T \mu$.

Proof. Given $\chi_0$ not in the support of $\widehat{\nu}$, to conclude that it is also not in the spectrum of $\nu*_T\mu$ it is enough to find a function $f$ in $L^1(G)$ with $\widehat{f}(\chi_0)=1$ and $f*_T(\nu*_T\mu)=0$. Simply choose $f$ with Fourier transform vanishing on the support of $\widehat{\nu}$ and taking value 1 at $\chi_0$. By Fourier inversion, we have $f*\nu=0$, and since $f*_T(\nu*_T\mu)=(f*\nu)*_T\mu$, the first part of the proposition follows. For the second part, we have ${\cal I}(\mu)\subset {\cal I}(\nu*_T\mu)$, which implies the desired inclusion.

We next prove a property of $L^\infty(G)$ functions similar to the characterization of $L^1$ functions which are constant on cosets of a subgroup [21, Theorem (28.55)].

Proposition 3.10   Suppose that $f$ is in $L^\infty(G)$ and that $\Lambda$ is an open subgroup of $\Gamma$. Let $\lambda_0$ denote the normalized Haar measure on the compact group $A(G,\Lambda)$, the annihilator in $G$ of $\Lambda$ (see [20, (23.23)]. Then, $\sigma[f]\subset \Lambda$ if and only if $f=f*\lambda_0$ a. e. This is also the case if and only if $f$ is constant on cosets of $A(G,\Lambda)$.

Proof. Suppose that the spectrum of $f$ is contained in $\Lambda$. Since $\Lambda$ is an $S$-set, it follows that $f$ is the weak-* limit of trigonometric polynomials with spectra contained in $\Lambda$. Let $\{f_\alpha\}$ be a net of such trigonometric polynomials converging to $f$ weak-*. Note that, for any $\alpha$, we have $\lambda_0*f_\alpha=f_\alpha$. For $g$ in $L^1(G)$, we have

$$\lim_\alpha \int_G f_\alpha \overline{g}dx= \int_G f\overline{g}dx.$$

In particular, we have

$$\lim_\alpha \int_G f_\alpha (\lambda_0*\overline{g})dx= \int_G f (\lambda_0*\overline{g})dx,$$

and so

$$\lim_\alpha \int_G (f_\alpha *\lambda_0) \overline{g}dx= \int_G (f*\lambda_0) \overline{g}dx.$$

Since this holds for any $g$ in $L^1(G)$, we conclude that $\lambda_0 *f_\alpha$ converges weak-* to $\lambda_0*f$. But $\lambda_0*f_\alpha=f_\alpha$, and $f_\alpha$ converges weak-* to $f$, hence $f*\lambda_0=f$. The remaining assertions of the lemma are easy to prove. We omit the details.

In what follows, we use the symbol $\mu_\alpha$ to denote the normalized Haar measure on the compact subgroup $A(G,C_\alpha)$, the annihilator in $G$ of $C_\alpha$. This measure is also characterized by its Fourier transform:

$$\widehat{\mu_\alpha}=1_{C_\alpha}$$

(see [20, (23.19)]).

Corollary 3.11   Suppose that $\mu\in{\cal M}_T(\Sigma)$. Then,
(i) ${\rm spec}_T\mu\subset C_\alpha$ if and only if $\mu=\mu_\alpha *_T \mu$;
(ii) ${\rm spec}_T\mu\subset \Gamma\setminus C_\alpha$ if and only if $\mu_\alpha *_T \mu=0$.

Proof. (i) If $\mu=\mu_\alpha *_T \mu$, then, by Proposition 3.10, $\sigma[t\mapsto \mu_\alpha*_T T_t\mu(A)]\subset C_\alpha$. Hence by Corollary 3.8, ${\rm spec}_T\mu\subset C_\alpha$. For the other direction, suppose that ${\rm spec}_T\mu\subset C_\alpha$. Then by Corollary 3.8 we have that the spectrum of the function $t\mapsto T_t\mu(A)$ is contained in $C_\alpha$ for every $A\in \Sigma$. By Proposition 3.10, we have that

$$T_t\mu(A)=\int_{G_\alpha}T_{t-y}\mu (A) d\mu_\alpha= T_t(\mu_\alpha*\mu)(A)$$

for almost all $t\in G$. Since this holds for all $A\in \Sigma$, the desired conclusion follows from Proposition 1.4.
Part (ii) follows from Corollary 3.6 (ii), Proposition 3.7(ii), and the fact that $\Gamma\setminus C_\alpha$ is an $S$-set.

Corollary 3.12   Suppose that $\mu\in{\cal M}_T(\Sigma)$ and ${\rm spec}_T\mu\subset C_\alpha$, and let $y\in G_\alpha=A(G,C_\alpha)$. Then $T_y\mu=\mu$.

Proof. For any $A\in \Sigma$, we have from Corollary 3.11

$$T_y\mu (A) = T_y(\mu_\alpha *\mu)(A)=\mu_\alpha*T_y\mu(A)$$

$$= \int_{G_\alpha} T_{y-x} \mu (A) d\mu_\alpha (y)$$

$$= \int_{G_\alpha}T_{-x}\mu(A) d\mu_\alpha (y) =\mu_\alpha *\mu(A)=\mu(A).$$