Proof. (i) By definition, $\widehat {\Gamma }$ is equipped with the topology of uniform convergence on compact subsets of $\Gamma $ . It therefore suffices to prove that if $(\chi _n)_{n\in I}$ is a net of elements of $\widehat {\Gamma }$ converging to $\chi \in \widehat {\Gamma }$ uniformly on compact subsets of $\Gamma $ , then $(\chi _n\circ \phi )_{n\in I}$ converges to $\chi \circ \phi $ uniformly on compact subsets of $\Gamma $ . Continuity of $\phi $ implies $\phi (K)$ is compact for every compact $K\subset \Gamma $ , so the assumption that $\chi _n\to \chi $ uniformly on every compact $K\subset \Gamma $ implies $\chi _n \to \chi $ uniformly on $\phi (K)$ for every compact $K\subset \Gamma $ . But this means $(\chi _n \circ \phi )_{n\in I}$ converges to $\chi \circ \phi $ uniformly on compact subsets of $\Gamma $ , as desired.

(ii) For $\gamma \in \Gamma $ , define the evaluation map $e_{\gamma }(\chi ) = \chi (\gamma )$ for any $\chi \in \widehat {\Gamma }$ . It suffices to prove that

$$\begin{align*}(\phi^*)^*(e_{\gamma}) = e_{\phi(\gamma)}, \end{align*}$$

meaning $(\phi ^*)^*(e_{\gamma })(\chi ) = \chi (\phi (\gamma ))$ for all $\chi \in \widehat {\Gamma }$ . To see this, note that $\chi \mapsto (\phi ^*)^*(e_{\gamma })(\chi )$ is defined by $e_{\gamma }(\phi ^*(\chi ))=e_{\gamma }(\chi \circ \phi ).$

Proof. (i) Let $\psi \in \widehat {\Lambda }$ , let $F = \{\chi _1,\dots ,\chi _d\} \subset \Lambda $ be finite and let $\varepsilon>0$ . We will show that there is a $g\in G$ such that $|\psi (\chi _j)-e_g(\chi _j)|<\varepsilon $ for all $\chi _j\in F$ . Consider the subgroup

$$\begin{align*}H:=\{(\chi_1(g),\dots,\chi_d(g)):g\in G\}\subset (S^1)^d. \end{align*}$$

It suffices to prove that

(9) $$ \begin{align} \vec{t}:=(\psi(\chi_1),\dots,\psi(\chi_d))\in \overline{H}. \end{align} $$

Assume, to get a contradiction, that (9) is false. Then there is a nontrivial character $\alpha \in \widehat {(S^{1})^d}$ which annihilates $\overline {H}$ but does not annihilate $\vec {t}$ . Writing $\alpha (x_1,\dots ,x_d)$ as $x_1^{n_1} \cdots x_d^{n_d}$ , we have

(10) $$ \begin{align} \chi_1(g)^{n_1}\cdots \chi_d(g)^{n_d} = 1 \qquad \text{for all } g\in G, \end{align} $$

but $\psi (\chi _1)^{n_1}\cdots \psi (\chi _d)^{n_d}\neq 1$ . Since $\psi $ is a character, the latter equation means

(11) $$ \begin{align} \psi(\chi_1^{n_1}\cdots\chi_d^{n_d})\neq 1. \end{align} $$

But (10) means that $\chi _1^{n_1}\cdots \chi _d^{n_d}$ is trivial, contradicting (11).

(ii) Define $\phi ':\Lambda \to \Lambda $ by $\phi '(\chi )=\chi \circ \phi $ . Let $\tilde {\phi }:=(\phi ')^*$ as in Lemma 3.1, meaning that for $\psi \in \widehat {\Lambda }$ , $\tilde {\phi }(\psi )=\psi \circ \phi '$ . By Lemma 3.1, $\tilde {\phi }$ is a continuous endomorphism. To verify that $\tilde {\phi }\circ \tau =\tau \circ \phi $ , fix $\chi \in \Lambda $ , $g\in G$ and evaluate

$$\begin{align*}\tilde{\phi}(\tau(g))(\chi)=e_g(\phi'(\chi))=e_g(\chi\circ \phi)=\chi\circ\phi(g)=e_{\phi(g)}(\chi)=\tau(\phi(g))(\chi).\end{align*}$$

Thus, $\tilde {\phi }\circ \tau = \tau \circ \phi $ .

Now let $k=[G:\phi (G)]$ (assuming this index is finite) and let $t_j+\phi (G), j=1,\dots ,k$ be coset representatives of $\phi (G)$ . The identity $\tilde {\phi }\circ \tau =\tau \circ \phi $ implies $\tilde {\phi }(\widehat {\Lambda })$ contains $\overline {\tau (\phi (G))}$ . The latter subgroup has index at most k, since the translates $\overline {\tau (t_j+\phi (G))} = \tau (t_j)+\overline {\tau (\phi (G))}$ are closed and cover a dense subset of $\widehat {\Lambda }$ . Thus, $\tilde {\phi }(\widehat {\Lambda })$ also has index at most k.

Proof. (i) For $\lambda \in \Lambda $ and $x \in \widehat {\Lambda }$ , we have

$$\begin{align*}v_{\lambda}(x + \tau(g))= (x + e_g)(\lambda) = x(\lambda) \lambda(g) = \lambda(g) v_{\lambda}(x). \end{align*}$$

This shows that $\lambda $ is an eigenvalue of $(\widehat {\Lambda }, m_{\widehat {\Lambda }}, R)$ and $v_{\lambda }$ is a corresponding eigenvector.

Conversely, suppose $\chi \in \widehat {G}$ and there exists non-zero $f \in L^2(\widehat {\Lambda })$ such that for all $g \in G$ , $f(x + \tau (g)) = \chi (g) f(x)$ for almost all x, we need to show that $\chi \in \Lambda $ . Since f is not zero, there exists $\lambda \in \Lambda $ such that $\widehat {f}(\lambda ) \neq 0$ . Computing the Fourier coefficients of both sides, we have

$$\begin{align*}\chi(g) \widehat{f}(\lambda) = e_g(\lambda) \widehat{f}(\lambda) = \lambda(g) \widehat{f}(\lambda) \end{align*}$$

for any $g \in G$ . Since $\widehat {f}(\lambda ) \neq 0$ , this implies that $\chi (g) = \lambda (g)$ for any $g \in G$ . Therefore, $\chi = \lambda \in \Lambda $ . Furthermore, this also shows that f has exactly one non-zero Fourier coefficient and $f = \widehat {f}(\lambda ) v_{\lambda }$ .

(ii) Define $\pi : \widehat {\Lambda }_2 \to \widehat {\Lambda }_1$ by $\pi (x) = x|_{\Lambda _1}$ for all $x \in \widehat {\Lambda }_2$ . Then $\pi $ is a surjective, continuous group homomorphism. By [Reference Le and Lê33, Lemma 2.7], $\pi $ is measure-preserving.

Recall that the homomorphisms from G to $\widehat {\Lambda }_1$ and $\widehat {\Lambda }_2$ are $\tau _1 (g) = e_g|_{\Lambda _1}$ and $\tau _2 (g) = e_g|_{\Lambda _2}$ . It is clear that

$$\begin{align*}\pi( x + \tau_2(g)) = \pi(x) + \tau_1(g), \end{align*}$$

thus showing that $\pi $ is a factor map.

(iii) We assume (see Section 2.3) that $L^2(\mu )$ is separable. For each $\lambda \in \Lambda = \mathcal {E}(\mathbf {X})$ , there is an eigenvector $f_{\lambda } \in L^2(\mathbf {X})$ such that $T_g f_{\lambda } = \lambda (g) f_{\lambda }$ for any $g \in G$ . Arguing similarly to [Reference Walters39, Theorem 3.4], we may assume that $|f_{\lambda }|=1$ and $f_{\lambda \xi } = f_{\lambda } f_{\xi }$ for any $\lambda , \xi \in \Lambda $ . Defining $V(v_{\lambda }) = f_{\lambda }$ and extending V linearly, we have an isometry $V: L^2(\widehat {\Lambda }) \rightarrow L^2(\mathbf {X})$ satisfying $V(fg) = V(f) V(g)$ for any $f, g \in L^2(\widehat {\Lambda })$ . By [Reference Walters39, Theorem 2.4], V induces a homomorphism of measure algebras, and therefore a factor map $\mathbf {X} \rightarrow \widehat {\Lambda }$ . Since $\mathcal {E}(\widehat {\Lambda }) = \Lambda $ , part (ii) shows that $\widehat {\Lambda }$ is the largest group rotation that is a factor of $\mathbf {X}$ .

Proof. Since $I(\vec {f},\vec {\phi })$ is continuous in $f_i$ (with respect to the $L^2(m_K)$ -norm) and multilinear in $f_i$ , it suffices to prove the identity when each $f_i$ is a character $\chi _i$ of K. In this case, we have

$$ \begin{align*} I(\chi_1,\chi_2,\chi_3,\vec{\phi})&= UC-\lim_{g \in G}\int_K \chi_1\chi_2\chi_3(z)\prod_{i=1}^3 \chi_i(\tau(\phi_i(g)))\, dm_K(z)\\ &= UC-\lim_{g \in G} \int_K \chi_1 \chi_2 \chi_3(z) \prod_{i=1}^3 \chi_i \circ \tilde{\phi}_i(\tau(g))\, dm_K(z). \end{align*} $$

By (6), we have

$$ \begin{align*} I(\chi_1, \chi_2, \chi_3, \vec{\phi}) &= \iint_{K^2} \chi_1 \chi_2 \chi_3(z) \prod_{i=1}^3 \chi_i \circ \tilde{\phi}_i(t) \, d m_K(z) d m_K(t) \\ &= \iint_{K^2} \prod_{i=1}^3 \chi_i(z + \tilde{\phi}_i(t)) \, d m_K(z) d m_K(t), \end{align*} $$

and this finishes our proof.

Proof. Let $\phi _1 = -\phi _2 - \phi _3$ . We first prove the special case of the lemma where $\mathcal {E}(\mathbf {X})$ is invariant under each $\phi _i$ , meaning that for all eigenvalues $\lambda \in \mathcal {E}(\mathbf {X})$ and $i \in \{1, 2,3\}$ , we have $\lambda \circ \phi _i\in \mathcal {E}(\mathbf {X})$ . In this case, the conclusion was also observed in [Reference Ackelsberg, Bergelson and Best2, Remark 3.2]. By [Reference Ackelsberg, Bergelson and Best2, Section 3], the Kronecker factor $(Z, m_Z, R)$ of $(X, \mu , T)$ is characteristic for the average defining $I(w).$ Let $\tau : G \to Z$ be the canonical projection. We can therefore replace f with $\tilde {f} := \mathbb {E}(f|Z)$ without changing $I(w)$ :

(22) $$ \begin{align} I(w)&= UC-\lim_{g \in G} \int_Z \tilde{f} \cdot R_{\phi_3(g)}\tilde{f} \cdot R_{w-\phi_2(g)}\tilde{f}\, dm_Z \nonumber\\ &= UC-\lim_{g \in G} \int_Z \tilde{f}(z)\tilde{f}(z+\tau(\phi_3(g)))\tilde{f}(z+\tau(w-\phi_2(g)))\, dm_Z(z). \end{align} $$

In view of Lemma 3.2, let $\tilde {\phi }_i:Z\to Z$ be continuous endomorphisms satisfying $\tau \circ \phi _i=\tilde {\phi }_i\circ \tau $ . Applying this identity to (22), we have

$$\begin{align*}I(w) = UC-\lim_{g \in G} \int_Z \tilde{f}(z)\tilde{f}(z+\tilde{\phi}_3(\tau(g)))\tilde{f}(z+\tau(w)-\tilde{\phi}_2(\tau(g))) \, dm_Z(z). \end{align*}$$

By Lemma 5.1, we can rewrite the previous line as

$$\begin{align*}I(w) = \iint_{Z^2} \tilde{f}(z)\tilde{f}(z+\tilde{\phi}_3(t))\tilde{f}(z+\tau(w)-\tilde{\phi}_2(t))) \, dm_Z(z) \, dm_Z(t). \end{align*}$$

Taking $K = Z$ , we prove the proposition in this special case.

For the general case, let $\Lambda $ be the smallest subgroup of $\widehat {G}$ that contains $\mathcal {E}(\mathbf {X})$ and is closed under each $\phi _i^*$ . Since $\mathcal {E}(\mathbf {X})$ is countable, it is easy to see that $\Lambda $ is countable. Let $\mathbf {K} = (\widehat {\Lambda }, m_{\widehat {\Lambda }}, R)$ be the group rotation on $\widehat {\Lambda }$ described in Lemma 3.3. By part (i) of Lemma 3.3, we have $\mathcal {E}(\mathbf {K}) = \Lambda $ . Since $\mathcal {E}(\mathbf {Z}) = \mathcal {E}(\mathbf {X}) \subset \Lambda $ , part (ii) of Lemma 3.3 implies that $\mathbf {Z}$ is a factor of $\mathbf {K}$ .

We now fix an ergodic G-system $\mathbf {Y} = (Y, \nu , S)$ that is a common extension of $\mathbf {X}$ and $\mathbf {K}$ . For example, we can take $\mathbf {Y} = (X \times K, \nu , T \times R)$ to be an ergodic joining of $\mathbf X$ and $\mathbf K$ . (For details about joinings and the existence of ergodic joinings, see Glasner [Reference Glasner19, Section 6] or de la Rue [Reference de la Rue and Meyers14, Section 3.1].)

Writing $\pi : Y\to X$ for the factor map, we define $f': Y \to \mathbb [0,1]$ to satisfy $f':= f\circ \pi $ and

$$\begin{align*}I'(w):= UC-\lim_{g \in G} \int_Y f' \cdot S_{\phi_3(g)}f' \cdot S_{w-\phi_2(g)} f' \, d\nu. \end{align*}$$

Since $f'$ is a lift from f on X, it is obvious that $I' = I$ and the Kronecker factor $\mathbf {Z}$ of $\mathbf {X}$ is characteristic for the averages $I'(w)$ . Thus, any factor of $\mathbf {Y}$ between $\mathbf {Y}$ and $\mathbf {Z}$ is also characteristic for $I'(w)$ . In particular, $\mathbf {K}$ is characteristic for $I'(w)$ . Now applying an argument similar to the first part of the proof to the factor $\mathbf {K}$ of $\mathbf {Y}$ and the function $f'$ , we obtain the compact group $K = \widehat {\Lambda }$ , the function $\tilde {f} = \mathbb {E}(f'| K)$ and endomorphisms $\tilde {\phi }_i$ satisfying (21). Finally, we have $[K:\tilde {\phi }_i(K)] \leq [G:\phi _i(G)]$ for each $i \in \{1, 2, 3\}$ by Lemma 3.2 (ii).

Proof. By Proposition 5.2, there exist a compact abelian group K with Haar measure $m_K$ , a homomorphism $\tau : G\to K$ with dense image, and endomorphisms $\tilde {\phi }_i:K\to K$ and $\tilde {f}:K\to [0,1]$ with $\int _K \tilde {f}\, dm_K=\int _X f\, d\mu = \delta $ such that

$$ \begin{align*} I(w) = \iint_{K^2} \tilde{f}(z)\tilde{f}(z+\tilde{\phi}_3(t))\tilde{f}(z+\tau(w)-\tilde{\phi}_2(t))\, dm_K(z)\, dm_K(t). \end{align*} $$

Furthermore, $[K:\tilde {\phi }_i(K)] \leq [G:\phi _i(G)]$ for each i. Now define $I': K \to [0, 1]$ by

$$ \begin{align*} I'(\widetilde{w}) := \iint_{K^2} \tilde{f}(z)\tilde{f}(z+\tilde{\phi}_3(t))\tilde{f}(z+ \widetilde{w} -\tilde{\phi}_2(t))\, dm_K(z)\, dm_K(t). \end{align*} $$

By change of variable $z \mapsto z + \tilde {\phi }_2(t)$ and using $\phi _2 + \phi _3 = - \phi _1$ , we obtain

$$ \begin{align*} I'(\widetilde{w}) = \iint_{K^2} \tilde{f}(z + \tilde{\phi}_2(t) )\tilde{f}(z - \tilde{\phi}_1(t))\tilde{f}(z+ \widetilde{w})\, dm_K(z)\, dm_K(t). \end{align*} $$

Applying [Reference Le and Lê33, Proposition 4.3], it follows that $\operatorname {supp}(I')$ contains a Bohr- $(k, \eta )$ set B in K, where $k, \eta $ depends only on $\delta $ and the indices $[K:\tilde {\phi }_i(K)]$ . It is easy to see that $\operatorname {supp}(I)$ contains $\tau ^{-1}(B)$ . Moreover, Lemma 2.2 implies that $\tau ^{-1}(B)$ contains a Bohr- $(k, \eta )$ set in G, completing the proof.

Proof. By Furstenberg’s correspondence principle (for example, see [Reference Bergelson and Moragues6, Theorem 2.8]), there exists an ergodic G-system $(X, \mu , T)$ and a measurable set $E \subset X$ with $\mu (E) = d^*(A)$ such that for all $w_1, w_2 \in G$ ,

$$ \begin{align*} \mu(E \cap T_{w_1}^{-1} E \cap T_{w_2}^{-1} E) \leq d^*(A \cap (A - w_1) \cap (A - w_2)). \end{align*} $$

Letting $f = 1_E$ , $w_1 = \phi _3(g)$ and $w_2 = w - \phi _2(g)$ , we deduce that for all w and $g \in G$ ,

$$ \begin{align*} \int_X f \cdot T_{\phi_3(g)} f \cdot T_{w - \phi_2(g)} f \, d \mu \leq d^*(A \cap (A - \phi_3(g)) \cap (A - (w - \phi_2(g))). \end{align*} $$

It follows that if $w \in \operatorname {supp}(I)$ , then there are $h \in A$ and $g \in G$ such that h, $h + \phi _3(g)$ , and $h + w - \phi _2(g)$ all belong to A. Therefore,

(23) $$ \begin{align} \phi_3(w) = \phi_1(h) + \phi_2(h + \phi_3(g)) + \phi_3(h + w - \phi_2(g)) \in \phi_1(A) + \phi_2(A) + \phi_3(A), \end{align} $$

and this finishes our proof. Note that in (23), we use the fact that $\phi _2 \circ \phi _3 = \phi _3 \circ \phi _2$ .

Proof of Theorem 1.2.

By Proposition 6.2, there exists an ergodic G-system $(X, \mu , T)$ and $f: X \to [0,1]$ with $\int _X f = d^*(A)$ such that

$$ \begin{align*} I(w) = UC-\lim_{g \in G} \int_X f \cdot T_{\phi_3(g)} f \cdot T_{w-\phi_2(g)} f\, d\mu \end{align*} $$

has $\phi _3(\operatorname {supp}(I)) \subset \phi _1(A) + \phi _2(A) + \phi _3(A)$ .

In view of Proposition 6.1, $\operatorname {supp}(I)$ contains a Bohr- $(k, \eta )$ set where $k, \eta $ only depends on $\delta $ and the indices of $\phi _i(G)$ in G. Lemma 2.3 then implies that $\phi _3(\operatorname {supp}(I))$ contains a Bohr- $(k', \eta ')$ set where $k', \eta '$ depends only on $\delta $ and the indices mentioned above.

Proof. By Lemma 4.4, the Radon-Nikodym density $\rho _{A}^{\nu }$ is supported on $\overline {\tau (A)}$ . Therefore, the convolution $\rho _{A}^{\nu } * \rho _{B}^{\lambda }$ , which is defined as

$$\begin{align*}\rho_{A}^{\nu} * \rho_{B}^{\lambda} (z) := \int_K \rho_{A}^{\nu}(x) \rho_{B}^{\lambda}(z - x) \, d m_K(x), \end{align*}$$

is supported on $\overline {\tau (A)} + \overline {\tau (B)} = \overline {\tau (A + B)}$ . Similarly, $\rho _{A}^{\nu } * \rho _{B}^{\lambda } * \rho _{-B}^{\lambda }$ is supported on $\overline {\tau (A + B - B)}$ . This, however, is weaker than the conclusion of Proposition 7.1 and is insufficient for our purpose.

Define $\phi , \theta : G \to [0, 1]$ by

$$ \begin{align*} \phi(t) := 1_B *_{\lambda} 1_{-B}(t) := \int_G 1_B(x) 1_{-B}(t - x) \, d \lambda(x) \end{align*} $$

and

$$ \begin{align*} \theta(t) := 1_A *_{\nu} \phi(t) := \int_G 1_A(y) \phi(t-y) \, d \nu(y). \end{align*} $$

We can see that $\theta $ is supported on $A + B - B$ . It remains to show that $\theta = (\rho _{A}^{\nu } * \rho _{B}^{\lambda } * \rho _{-B}^{\lambda }) \circ \tau $ .

Proof of claim.

One can verify that $\phi $ is positive definite by writing $\sum _{g,h\in G} c_g\overline {c}_h \phi (g-h)$ as $\int _G (\sum _{g} c_g1_B(x-g))\overline {\sum _{h} c_h 1_{B}(x-h)}\, d\lambda (x)= \int _G \bigl |\sum _{g} 1_B(x-g)\bigr |^2\, d\lambda (x)$ for a finite collection of coefficients $c_g\in \mathbb C$ . Therefore, by the Bochner-Herglotz Theorem, $\phi $ is the Fourier transform of a positive measure $\sigma $ on $\widehat {G}$ . Decomposing $\sigma = \sigma _{d} + \sigma _{c}$ , where $\sigma _{d}$ is the discrete component of $\sigma $ and $\sigma _{c}$ is the continuous part, we have

(24) $$ \begin{align} \phi = \hat{\sigma}_{d} + \hat{\sigma}_{c}. \end{align} $$

Since $\sigma _d$ has only countably many atoms, $\hat {\sigma }_{d}$ is an almost periodic function. However, by Wiener’s lemma (see [Reference Godement21, Théorème 16(2)]), $\int _G |\hat {\sigma }_{c}|^2 \, d \mu = 0$ for all invariant means $\mu $ on G.

Now we will prove that $\hat {\sigma }_d=\eta $ . We first show that $\hat {\sigma }_d$ and $\eta $ are almost periodic functions defined by Fourier series on G with absolutely summable coefficients. To see this for $\hat {\sigma }_d$ , we write $\hat {\sigma }_d = \sum _{\chi \in \widehat {G}} \sigma (\{\chi \}) \chi $ , where $\sum _{\chi \in \widehat {G}} \sigma (\{\chi \})$ is a convergent sum of nonnegative values. For $\eta $ , note that both $\rho _{B}^{\lambda }$ and $\rho _{-B}^{\lambda }$ are in $L^2(m_K)$ . Thus, their Fourier coefficients are square-summable, and the Fourier coefficents of $\rho _B^{\lambda }*\rho _{-B}^{\lambda }$ are absolutely summable. To prove that $\hat {\sigma }_d=\eta $ , it therefore suffices to prove that $\hat {\sigma }_d$ and $\eta $ have the same Fourier coefficients. This is the same as showing that $\phi $ and $\eta $ have the same Fourier coefficients, as the Fourier coefficients of $\hat {\sigma }_c$ are all $0$ (since $\hat {\sigma }_c$ is a null function). So we verify that

$$ \begin{align*} \mu(\phi \overline{\chi}) = \mu(\eta \overline{\chi}) \end{align*} $$

for every invariant mean $\mu $ on G and every character $\chi \in \widehat {G}$ . Fix the invariant mean $\mu $ , characters $\chi \in \widehat {G}$ and $\chi ' \in \widehat {K}$ such that $\chi = \chi ' \circ \tau $ . We then have

$$ \begin{align*} \mu(\phi \overline{\chi}) &= \iint_{G^2} 1_B(t) 1_{-B}(s - t) \overline{\chi(s)} \, d \lambda(t) d \mu(s) \\ &=\iint_{G^2} (1_B \cdot \overline{\chi})(t) \cdot (1_{-B} \cdot \overline{\chi})(s - t) \, d \lambda(t) d \mu(s) \\ &= \lambda(1_B \cdot \overline{\chi}) \lambda(1_{-B} \cdot \overline{\chi}) \;\;\; \text{(by Lemma 2.1)}\\ &= \int_K \rho_{B}^{\lambda} \overline{\chi'} \, d m_K \cdot \int_K \rho_{-B}^{\lambda} \overline{\chi'} \, d m_K \;\;\; (\text{by definitions of } \rho_{B}^{\lambda} \text{ and } \rho_{-B}^{\lambda}) \\ &= \widehat{\rho_{B}^{\lambda}}(\chi') \cdot \widehat{\rho_{-B}^{\lambda}}(\chi')\\ &= \widehat{\rho_{B}^{\lambda} * \rho_{-B}^{\lambda}}(\chi') \\ &= \int_K (\rho_{B}^{\lambda} * \rho_{-B}^{\lambda}) \cdot \overline{\chi'} \, d m_K \\ &= \mu(\eta \overline{\chi}) \;\; \; (\text{by the definition of } \eta \text{ and Lemma 4.2}).\\[-36pt] \end{align*} $$

We are ready to prove $\theta = (\rho _{A}^{\nu } * \rho _{B}^{\lambda } * \rho _{-B}^{\lambda }) \circ \tau $ . Indeed, by Claim 7.2,

$$ \begin{align*} \theta := 1_A *_{\nu} \phi = 1_A *_{\nu} \eta + 1_A *_{\nu} \psi ,\end{align*} $$

where $\psi $ is a null function and $\eta =(\rho _{B}^{\lambda } * \rho _{-B}^{\lambda } )\circ \tau $ . For all $t \in G$ , we have

$$ \begin{align*} |1_A *_{\nu} \psi(t)| \leq \nu(|-\psi_t|) = \nu(|\psi|) = 0. \end{align*} $$

Moreover, since $\eta $ is a Fourier series with absolutely summable coefficients, $1_A *_{\nu } \eta $ is as well. It follows that $\theta $ is almost periodic. Therefore, to show $\theta = (\rho _{A}^{\nu } * \rho _{B}^{\lambda } * \rho _{-B}^{\lambda }) \circ \tau $ , it suffices to check that $\theta $ and $(\rho _{A}^{\nu } * \rho _{B}^{\lambda } * \rho _{-B}^{\lambda }) \circ \tau $ have the same Fourier coefficients. We omit the computations as they are nearly identical to the proof of Claim 7.2.

Proof. First, we observe that if $\nu $ is a linear functional on $\ell ^{\infty }_{\mathbb {R}}(G)$ and $\nu (1_G) = 1$ , then $\nu $ is positive if and only if $\nu (f) \geq p(f):= \inf _{x \in G} f(x)$ for all $f \in \ell ^{\infty }_{\mathbb {R}}(G)$ . Clearly p is a concave function.

Let V be the vector subspace of $\ell _{\mathbb {R}}^{\infty }(G)$ consisting of functions of the form $h \circ \phi $ for some $h \in \ell _{\mathbb {R}}^{\infty }(H)$ . If $f \in V$ , then by surjectivity of $\phi $ , there is a unique $h \in \ell _{\mathbb {R}}^{\infty }(H)$ such that $f = h \circ \phi $ . We have

$$ \begin{align*} \mu(h) &\geq \inf_{y \in H} h(y) \qquad (\text{since } \mu \text{ is an invariant mean on } H) \\ & = \inf_{x \in G} h(\phi(x)) = p(f) \qquad (\text{since } \phi \text{ is surjective}). \end{align*} $$

By the Hahn-Banach theorem, the linear functional $f \mapsto \mu (h)$ on V can be extended to a linear functional $\lambda $ on $\ell _{\mathbb {R}}^{\infty }(G)$ such that $\lambda (f) \geq p(f)$ for any $f \in \ell _{\mathbb {R}}^{\infty }(G)$ . In particular, $\lambda $ is positive and $\lambda (1_G) = \lambda (1_H \circ \phi ) = \mu (1_H) = 1$ . We now show that $\lambda $ can be further refined to become G-invariant.

We let $\eta $ be an invariant mean on G and define

$$\begin{align*}\nu(f): = \int_{G} \lambda(f_x) \, d\eta(x) \end{align*}$$

for all $f \in \ell _{\mathbb {R}}^{\infty }(G)$ . Then $\nu (f_g) = \nu (f)$ for all $g \in G$ since $\eta $ is translation invariant. The positivity of $\nu $ follows from the positivity of $\lambda $ and $\eta $ . If $f = h \circ \phi \in V$ , then $\lambda (f_g) = \mu (h_{\phi (g)}) = \mu (h)$ for all $g \in G$ , so $\nu (f) = \mu (h)$ . The lemma now follows since an invariant mean is completely determined by its values on real-valued functions.

Proof. Let $H - g_i$ for $0 \leq i \leq k - 1$ be the cosets of H in G with $g_0 = 0$ . We first show that an invariant mean $\nu $ satisfying the conclusion of the lemma must be unique. For a function f supported on $H - g_i$ , the function $f_{g_i}$ given by $x \mapsto f(x - g_i)$ is supported on H. Therefore, in this case, since $\nu $ is G-invariant, we must have

(25) $$ \begin{align} \nu(f) = \nu(f_{g_i}) = \frac{\mu(f_{g_i})}{k}. \end{align} $$

For an arbitrary $f \in \ell ^{\infty }(G)$ , define $f^i = f \cdot 1_{H - g_i}$ . Since $f = \sum _{i=0}^{k-1} f^i$ , from the previous paragraph, we must have

(26) $$ \begin{align} \nu(f) = \sum_{i=0}^{k-1} \nu(f^i) = \frac{1}{k} \sum_{i=0}^{k-1} \mu((f^i)_{g_i}). \end{align} $$

This equation uniquely defines $\nu $ .

It is easy to see that $\nu $ as defined in (26) is a linear functional on $\ell ^{\infty }(G)$ with $\nu (1_G) = 1$ . To show $\nu $ is G-invariant, we consider arbitrary $g \in G$ and $f \in \ell ^{\infty }(G)$ . By the linearity of $\nu $ and (25),

(27) $$ \begin{align} \nu(f_g) = \sum_{i=0}^{k-1} \nu((f^i)_g) = \frac{1}{k} \sum_{i=0}^{k-1} \mu(((f^i)_g)_{g_{j(i)}}) = \frac{1}{k} \sum_{i=0}^{k-1} \mu(((f^i)_{g+g_{j(i)}}). \end{align} $$

where $j(i) \in \{0, \ldots , k-1\}$ is such that $-g_i + g + g_{j(i)} \in H$ . For $i \in \{0, \ldots , k - 1\}$ , let $h = -g_i + g + g_{j(i)}$ . Since $\mu $ is H-invariant,

(28) $$ \begin{align} \mu(((f^i)_{g+g_{j(i)}}) = \mu(((f^i)_{g_i + h}) = \mu((f^i)_{g_i}). \end{align} $$

Relations (26), (27) and (28) give $\nu (f_g) = \nu (f)$ , and so $\nu $ is G-invariant.

Suppose $\mu $ is extremal. To show that $\nu $ is extremal, suppose $\nu = \alpha \nu _1 + (1 - \alpha ) \nu _2$ , where $\nu _1$ and $\nu _2$ are means on G and $0 < \alpha < 1$ . Restricting to $S := \{f \in \ell ^{\infty }(G): f \text { is supported on } H\}$ , we get

$$ \begin{align*} \mu/k = \nu|_{S} = \alpha \nu_1|_S + (1 - \alpha) \nu_2|_S. \end{align*} $$

Since $\mu $ is extremal, it must be that $\nu _1|_S = \nu _2|_S = \mu /k$ . Due to the uniqueness of the extension of $\mu $ from H to G, we deduce that $\nu _1 = \nu _2 = \nu $ . Therefore, $\nu $ is extremal.

Proof. As in the proof of Lemma 8.3, let $H - g_i$ for $0 \leq i \leq k -1 $ be the cosets of H in G with $g_0 = 0$ . Since $B \subset H$ , according to Lemma 4.4, both $\rho _{B}^{\nu }$ and $\rho _{B}^{\mu }$ are supported on $K_H$ . From (26), for $h \in C(K)$ ,

(29) $$ \begin{align} \nu(h \circ \tau \cdot 1_B) = \frac{1}{k} \sum_{i=0}^{k-1} \mu((h \circ \tau \cdot 1_B \cdot 1_{H-g_i})_{g_i}). \end{align} $$

Since $1_B$ is supported on H,