2.1 CAT(0) spaces

Recall that a CAT(0) space, also known as Hadamard space, is a complete metric space $(M,\delta )$ that satisfies the following semiparallelogram law: given $x, y \in M$ , there exists $m \in M$ satisfying

(2.1) $$ \begin{align} \delta^2(m, z) \leq \displaystyle\tfrac{1}{2}\delta^2(x, z) + \displaystyle\tfrac{1}{2}\delta^2(y, z) - \displaystyle\tfrac{1}{4}\delta^2(x, y) \end{align} $$

for all $z \in M$ . The point m is unique, and it is called midpoint between x and y, since $ \delta (x,m)=\delta (m,y)=\tfrac 12 \delta (x,y)$ . Recall that, given a continuous curve $\gamma :[a,b]\to M$ , its length is computed as

$$ \begin{align*}\inf \sum_{n=1}^N \delta(\gamma(t_{n+1}),\gamma(t_n)) \end{align*} $$

where the infimum is taken over all the partitions $\{t_1,\ldots ,t_N\}$ of the interval $[a,b]$ . The existence and uniqueness of midpoints give rise to a unique (continuous) geodesic $\gamma _{x,y} : [0, 1] \rightarrow M$ connecting any given two points x and y. Indeed, we first define $\gamma _{x,y}(1/2)$ to be the midpoint of x and y. Then, using an inductive argument, we define the geodesic for all dyadic rational numbers in $[0, 1]$ . Finally, by completeness, it can be extended to all $t \in [0, 1]$ . It can be proved that this curve is the shortest path connecting x and y. As we mentioned in the introduction, we will use the notation $x \#_t y$ instead of $\gamma _{x,y}(t)$ . It is not difficult to see that the points of this geodesic satisfy the following generalized semiparallelogram inequality:

(2.2) $$ \begin{align} \delta^2(x \#_t y, z) \leq (1-t)\delta^2(x, z) + t\delta^2(y, z) - t(1-t)\delta^2(x, y). \end{align} $$

As consequence of this inequality the next result on the convexity of the metric is obtained (see, for example, [Reference Sturm, Auscher, Coulhon and Grigor’yan30, Corollary 2.5]).

Proposition 2.1. Given four points $a, a^{\prime }, b, b^{\prime } \in M$ , let

$$ \begin{align*}f(t) = \delta(a \#_t a^{\prime}, b \#_t b^{\prime}). \end{align*} $$

Then f is convex on $[0, 1]$ ; that is,

(2.3) $$ \begin{align} \delta(a \#_t a^{\prime}, b \#_t b^{\prime}) \leq (1-t)\delta(a, b) + t\delta(a^{\prime}, b^{\prime}). \end{align} $$

Another very important result in CAT(0) spaces is the so-called Reshetnyak quadruple comparison theorem (see, for example, [Reference Sturm, Auscher, Coulhon and Grigor’yan30, Proposition 2.4]).

Theorem 2.2. Let $(M, \delta )$ be a Hadamard space. For all $x_1, x_2, x_3, x_4 \in M$ ,

(2.4) $$ \begin{align} \delta^2(x_1, x_3) + \delta^2(x_2, x_4) \leq \delta^2(x_2, x_3) + \delta^2(x_1, x_4) + 2\delta(x_1, x_2)\delta(x_3, x_4). \end{align} $$

2.2 Barycenters in CAT(0) spaces

Let $\mathcal {B}(M)$ be the $\sigma $ -algebra of Borel sets (that is, the smallest $\sigma $ -algebra that contains the open sets). Denote by $\mathcal {P}(M)$ the set of all probability measures on $\mathcal {B}(M)$ with separable support, and for $1 \leq \theta < \infty $ , let $\mathcal {P}^{\theta }(M)$ be the set of those measures $\mu \in \mathcal {P}(M)$ such that

$$ \begin{align*}\int_M \delta^{\theta}(x, y)\,d{\kern-0.7pt}\mu(y) < \infty, \end{align*} $$

for some (and hence for all) $x \in M$ . By means of $\mathcal {P}^{\infty }(M)$ we will denote the set of all measures in $\mathcal {P}(M)$ with bounded support. Finally, given a measure space $(X,\mathfrak {X},\mu )$ and a measurable function $f:X\to M$ , we say that f belongs to $L^p(X,M)$ if the pushforward of $\mu $ by f belongs to $\mathcal {P}^p(M)$ ( $1\leq p\leq \infty $ ).

If $\mu \in \mathcal {P}^2(M)$ , then the usual Cartan definition of barycenter $\beta _\mu $ can be extrapolated to this setting:

$$ \begin{align*}\beta_\mu=\arg\!\min_{z \in M} \int_M \delta^2(z, x)\,d{\kern-0.7pt}\mu(x). \end{align*} $$

The existence of a unique minimizer is guaranteed by the convexity properties of the metric. This definition can be extended to measures in $\mathcal {P}^1(M)$ . Following the ideas of Sturm in [Reference Sturm, Auscher, Coulhon and Grigor’yan30], given any point $y\in M$ , the barycenter of a measure $\mu \in \mathcal {P}^1(M)$ is defined as the unique minimizer of the functional

$$ \begin{align*}z \mapsto \int_M [\delta^2(z, x) - \delta^2(y, x)]\,d{\kern-0.7pt}\mu(x). \end{align*} $$

Although the functional depends on the point y, it is easy to see that the minimizer is independent of it. Hence, the barycenter is well defined. Moreover, if $\mu \in \mathcal {P}^2(M)$ this definition coincides with Cartan’s definition. Note that in this case the quantity

$$ \begin{align*}\int_M \delta^2(z, x)\,d{\kern-0.7pt}\mu(x) \end{align*} $$

can be thought as a variance. Moreover, the barycenters in this case also satisfy the following inequality known as the variance inequality:

(2.5) $$ \begin{align} \int_{M} [\delta^2(z, x) - \delta^2({\beta}_{\mu}, x)]\,d{\kern-0.7pt}\mu(x) \geq \delta^2(z, {\beta}_{\mu}). \end{align} $$

Therefore, sometimes the barycenter is considered as a nonlinear version of the expectations. For instance, this idea was used by Sturm to extend different result from probability theory to this nonlinear setting (see [Reference Sturm, Auscher, Coulhon and Grigor’yan30, Reference Tao29] and the references therein).

Special notation

As we mentioned in the introduction, we will use a special notation in the following two cases. On the one hand, let $(X,\mathfrak {X},\mu )$ be a measure space and let $f:X\to M$ be a measurable function in $L^1(X,M)$ . By means of $\beta _f$ we will denote the barycenter of the pushforward measure $f_*(\mu )$ . On the other hand, given n points $x_1,\ldots ,x_n\in M$ , by means of $\beta (x_1,\ldots ,x_n)$ we will denote the barycenter of the point measure $\mu =\delta _{x_1}+\cdots +\delta _{x_n}$ .

The main issue dealing with barycenters is that they are difficult to compute. The computation of the barycenter of three or more points may be difficult. Although there exists a very rich convex theory in Hadamard space (see, for instance, [Reference Bačák3]), sometimes the approximation of the barycenter using convex optimization is not satisfactory. A good example of this situation is as follows.

Example 2.3. (Positive matrices)

Recall that the set of positive invertible matrices $\mathcal {M}_n(\mathbb {C})^+$ is an open cone in the real vector space of self-adjoint matrices $\mathcal {H}(n)$ . In particular, it is a differentiable manifold and the tangent spaces can be identified for simplicity with $\mathcal {H}(n)$ . The manifold $\mathcal {M}_n(\mathbb {C})^+$ can be endowed with a natural Riemannian structure. With respect to this metric structure, if $\alpha :[a,b]\to \mathcal {M}_n(\mathbb {C})^+$ is a piecewise smooth path, its length is defined by

$$ \begin{align*}L(\alpha)=\int_a^b \|\alpha^{-1/2}(t)\alpha'(t)\alpha^{-1/2}(t)\|_2 \,dt, \end{align*} $$

where $\|\cdot \|_2$ denotes the Frobenius or Hilbert–Schmidt norm. In this way, $\mathcal {M}_n(\mathbb {C})^+$ becomes a Riemannian manifold with non-positive curvature, and in particular a CAT(0) space. The geodesic connecting two positive matrices A and B has the following simple expression:

$$ \begin{align*}\gamma_{AB}(t)=A^{1/2}(A^{-1/2}BA^{-1/2})^{t}A^{1/2}\,. \end{align*} $$

So, the barycenter of the measure $\mu =\tfrac 12(\delta _{A}+\delta _{B})$ is given by

$$ \begin{align*}A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}. \end{align*} $$

However, if we add an atom to $\mu $ , there is no longer a closed formula for the barycenter (also called the geometric mean in this setting). As a consequence, simple questions such as the monotonicity of the barycenter with respect to the usual order of matrices become difficult. Using convex optimization, it is possible to construct a sequence that approximates the barycenter of a measure. However, that sequence does not contain enough information in order to prove that the barycenter is monotone. This issue, for instance, motivated intensive research with the aim of finding good ways to approximate the barycenters of more than two matrices [Reference Bini and Iannazzo7, Reference Lawson and Lim17, Reference Lim and Pálfia20]. The barycenters in this setting have attracted much attention in recent years because of their interesting applications in signal processing (see [Reference Bhatia and Karandikar6] and the references therein), and gradient or Newton-like optimization methods (see [Reference Bochi and Navas8, Reference Moakher and Zerai23]).

2.3 The inductive means

Recall that, given $a \in M^{\mathbb {N}}$ , the inductive means are define as:

$$ \begin{align*} S_1(a) & = a_1, \\ S_{n}(a) & = S_{n-1}(a) \#_{{1}/{n}} a_{n} \quad (n \geq 2). \end{align*} $$

As a consequence of (2.3), we directly get the following result.

Corollary 2.4. Given $a,b \in M^{\mathbb {N}}$ , then

(2.6) $$ \begin{align} \delta(S_n(a), S_n(b)) \leq \frac{1}{n}\sum_{i=1}^{n} \delta(a_i, b_i). \end{align} $$

The next lemma follows from (2.2), and it is a special case of a weighted inequality considered by Lim and Pálfia in [Reference Lim and Pálfia21].

Lemma 2.5. Given $a \in M^{\mathbb {N}}$ and $z\in M$ , for every $k,m \in \mathbb {N}$ ,

$$ \begin{align*} \delta^2(S_{k+m}(a), z) &\leq \ \frac{k}{k+m}\ \delta^2(S_{k}(a), z) + \displaystyle\frac{1}{k+m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(a_{k+j+1}, z)\\ &\quad - \displaystyle\frac{k}{(k+m)^2}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(S_{k+j}(a), a_{k+j+1}). \end{align*} $$

Proof. By the inequality (2.2) applied to $S_{n+1}(a)=S_n(a)\,\#_{n+1}\,(a_{n+1})$ we obtain

$$ \begin{align*} (n+1)\ \delta^2(S_{n+1}(a), z) - n\ \delta^2(S_{n}(a), z) & \leq \delta^2(a_{n+1}, z) - \displaystyle\frac{n}{(n + 1)}\delta^2(S_{n}(a), a_{n+1}). \end{align*} $$

Summing these inequalities from $n=k$ to $n=k+m-1$ , we get that the difference

$$ \begin{align*} (k+m)\ \delta^2(S_{k+m}(a), z) - k\ \delta^2(S_{k}(a), z), \end{align*} $$

obtained from the telescopic sum of the left-hand side, is less than or equal to

$$ \begin{align*} \sum_{j=0}^{m-1}\bigg(\delta^2(a_{k+j+1}, z) - \displaystyle\frac{k+j}{(k+j+1)}\delta^2(S_{k+j}(a), a_{k+j+1})\bigg). \end{align*} $$

Finally, using that $({k+j})/({k+j+1})\geq {k}/({k+m})$ for every $j\in \{0,\ldots ,m-1\}$ , this sum is bounded from above by

$$ \begin{align*} \sum_{j=0}^{m-1}\bigg(\delta^2(a_{k+j+1}, z) - \displaystyle\frac{k}{(k+m)}\delta^2(S_{k+j}(a), a_{k+j+1})\bigg), \end{align*} $$

which completes the proof.

Given a sequence $a \in M^{\mathbb {N}}$ , let $\Delta (a)$ denote the diameter of its image, that is,

$$ \begin{align*}\Delta(a) := \sup_{n,m\in\mathbb{N}} \delta(a_n, a_m). \end{align*} $$

Note that, also by (2.2), $\delta (S_n(a), a_k) \leq \Delta (a)$ for all $n, k \in \mathbb {N}$ .

Lemma 2.6. Given $a\in M^{\mathbb {N}}$ such that $\Delta (a) < \infty $ , we have for all $k,m \in \mathbb {N}$ that

$$ \begin{align*} \frac{1}{m} \sum_{j = 0}^{m-1} \delta^2(S_{k}(a), a_{k+j+1}) \leq \tilde{R}_{m,k} + \frac{1}{m} \sum_{j = 0}^{m-1}\delta^2(S_{k+j}(a), a_{k+j+1}), \end{align*} $$

where $\tilde {R}_{m,k}=(({m^2}/{(k+1)^2}) + 2({m}/({k+1}))) \Delta ^2(a)$ .

Proof. Note that by (2.6) and for all k,

$$ \begin{align*}\delta(S_{k+j}(a), S_{k+j+1}(a)) \leq \frac{1}{k+j+1}\Delta(a). \end{align*} $$

Hence

$$ \begin{align*} \delta(S_{k}(a), a_{k+j+1}) & \leq \delta(S_{k}(a), S_{k+j}(a)) + \delta(S_{k+j}(a), a_{k+j+1}) \\[3pt] & \leq \displaystyle\sum_{h=1}^{j} \displaystyle\frac{1}{k+h} \Delta(a) + \delta(S_{k+j}(a), a_{k+j+1}) \\[3pt] & \leq \displaystyle\frac{j}{k+1} \Delta(a) + \delta(S_{k+j}(a), a_{k+j+1}). \end{align*} $$

Therefore, for every $j\leq m$ ,

$$ \begin{align*} \delta^2(S_{k}(a), a_{k+j+1}) & \leq \bigg(\displaystyle\frac{m^2}{(k+1)^2} + 2\displaystyle\frac{m}{k+1}\bigg) \Delta^2(a) + \delta^2(S_{k+j}(a), a_{k+j+1}), \end{align*} $$

where we have used that $\delta (S_{k+j}(a), a_{k+j+1}) \leq \Delta (a)$ for every $k,j\in \mathbb {N}$ . Summing up these inequalities and dividing by m, we get the desired result.

3.1 Continuous case

In this section we will prove Theorem 1.3. Recall that, given a function $A: G \rightarrow M$ , we define $a^{\tau } : G \rightarrow M^{\mathbb {N}}$ by

$$ \begin{align*}a^{\tau}(x) :=\{a^{\tau}_j(x)\}_{n\in\mathbb{N}}\quad \textrm{where } a^{\tau}_j(x)=A(\tau^j(x)). \end{align*} $$

The proof is rather long and technical, so we split it into several lemmas and a technical result, which will be combined at the end of the section to provide the proof of Theorem 1.3.

Lemma 3.1. Let $A: G\rightarrow M$ be a continuous function, and let K be any compact subset of M. For each $n\in \mathbb {N}$ , define $F_n :G \times K \rightarrow \mathbb {R}$ by

$$ \begin{align*}F_n(g, x) = \frac{1}{n} \sum_{j = 0}^{n-1} \delta^{2}(a_j^{\tau}(g), x). \end{align*} $$

Then the family $\{F_n\}_{n \in \mathbb {N}}$ is equicontinuous.

Proof. By the triangular inequality, the map $y\mapsto \delta ^2(A(\cdot ),y)$ is continuous from $(K,\delta )$ into the set of real-valued continuous functions defined on G endowed with the uniform norm. Since K is compact, the family $\{\delta ^2(A(\cdot ),x)\}_{x\in K}$ is (uniformly) equicontinuous. Hence, given $\varepsilon>0$ , there exists $\delta>0$ such that if $d_G(g_1,g_2) < \delta $ then

$$ \begin{align*}|\delta^2(A(g_1),x)-\delta^2(A(g_2),x)|<\frac{\varepsilon}{2}, \end{align*} $$

for every $x\in K$ . Since $\tau $ is isometric and $d_G(g_1,g_2) < \delta $ , we get that

$$ \begin{align*}|F_n(g_1, x)-F_n(g_2, x)| = \bigg|\frac{1}{n} \sum_{j = 0}^{n-1} \delta^{2}(a_j^{\tau}(g_1), x)-\delta^{2}(a_j^{\tau}(g_2), x)\bigg|<\frac{\varepsilon}{2}. \end{align*} $$

Let $\Delta $ be the diameter of the set $\,(\mbox {Image}(A)\times K)$ in $M^2$ . Since both sets are compact, $\Delta <\infty $ . So, take $(g_1,x_1)$ and $(g_2,x_2)$ such that $d_G(g_1,g_2)<\delta $ and $\delta (x_1,x_2)<{\varepsilon }/{4\Delta }$ . Then

$$ \begin{align*} |F_n(g_1, x_1)-F_n(g_2, x_2)|&\leq |F_n(g_1, x_1)-F_n(g_1, x_2)|+|F_n(g_1, x_2)-F_n(g_2, x_2)|\\[3pt] &\leq \frac{2\Delta}{n}\sum_{k=0}^{n-1} \delta(x_1,x_2) \ +\ \frac{\varepsilon}{2}<\varepsilon.\\[-3.8pc] \end{align*} $$

Now, as a consequence of the Arzelà–Ascoli and Birkhoff theorems, we get the following proposition.

Proposition 3.2. Let $A:G\to M$ be a continuous function, and K a compact subset of M. Then

$$ \begin{align*}\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{j = 0}^{n-1} \delta^{2}(a_j^{\tau}(g), x) = \int_{G} \delta^2(A(\gamma), x)\,d{\kern-0.6pt}m(\gamma), \end{align*} $$

and the convergence is uniform in $(g, x) \in G \times K$ .

From now on we will fix the continuous function $A:G\to M$ . Let

$$ \begin{align*}\alpha := \min_{x\in M} \int_{G} \delta^2(A(g), x)\,d{\kern-0.6pt}m(g), \end{align*} $$

and ${\beta }_{\mbox{A}}$ is the point where this minimum is attained, that is, ${\beta }_{\mbox{A}}$ is the barycenter of the pushforward by A of the Haar measure in G. Then we obtain the following upper estimate.

Lemma 3.3. For every $\varepsilon> 0$ , there exists $m_0 \in \mathbb {N}$ such that, for all $m \geq m_0$ and for all $k \in \mathbb {N}$ ,

$$ \begin{align*} \delta^2(S_{k+m}(a^{\tau}(g)), {\beta}_{\mbox{A}}) & \leq \frac{k}{k+m}\delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{m}{k+m}(\alpha+\varepsilon) \\[3pt] &\quad - \displaystyle\frac{km}{(k+m)^2}\bigg(\frac{1}{m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(S_{k+j}(a^{\tau}(g)), a^{\tau}_{k+j+1}(g))\bigg). \end{align*} $$

Proof. For every $\varepsilon> 0$ , there exists $m_0 \in \mathbb {N}$ such that for all $m \geq m_0$ ,

$$ \begin{align*}\bigg|\frac{1}{m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(a_{k+j+1}^{\tau}(g), {\beta}_{\mbox{A}}) - \alpha\bigg| < \varepsilon. \end{align*} $$

Note that $m_0$ is independent of k by Proposition 3.2. Now, by Lemma 2.5,

$$ \begin{align*} \delta^2(S_{k+m}(a^{\tau}(g)),{\beta}_{\mbox{A}}) &\leq \frac{k}{k+m}\ \delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{1}{k+m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(a_{k+j+1}^{\tau}(g), {\beta}_{\mbox{A}})\\ & \quad - \displaystyle\frac{k}{(k+m)^2}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(S_{k+j}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g)) \nonumber\\& = \frac{k}{k+m} \ \delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{m}{k+m}\bigg(\frac{1}{m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(a_{k+j+1}^{\tau}(g), {\beta}_{\mbox{A}})\bigg) \\ & \quad - \displaystyle\frac{km}{(k+m)^2}\bigg(\frac{1}{m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(S_{k+j}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g))\bigg) \\ & \leq \frac{k}{k+m}\ \delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{m}{k+m}(\alpha+\varepsilon) \\ & \quad - \displaystyle\frac{km}{(k+m)^2}\bigg(\frac{1}{m}\displaystyle\sum_{j = 0}^{m - 1}\delta^2(S_{k+j}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g))\bigg).\\[-3.8pc] \end{align*} $$

Since $A:G\to M$ is continuous, note that

$$ \begin{align*}C_a:= \sup_{g\in G} \Delta(a^{\tau}(g))<\infty, \end{align*} $$

where, as we have defined before Lemma 2.6, $\Delta (a^{\tau }(g))$ denotes the diameter of the image of the sequence $a^{\tau }(g)$ .

Lemma 3.4. For every $\varepsilon> 0$ , there exists $m_0 \in \mathbb {N}$ such that for all $m \geq m_0$ and for all $k \in \mathbb {N}$ ,

$$ \begin{align*}\delta^2(S_k(a^{\tau}(g)), {\beta}_{\mbox{A}}) - \varepsilon + \alpha - R_{m,k} \leq \frac{1}{m} \sum_{j = 0}^{m-1}\delta^2(S_{k+j}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g)), \end{align*} $$

where $\displaystyle R_{m,k}= ({m^2}/{(k+1)^2}) + 2({m}/({k+1})) C_a^2$ .

Proof. Consider the compact set

$$ \begin{align*}K := \overline{cc\{S_k(a^{\tau}(x)) : k \in \mathbb{N}\}}, \end{align*} $$

where the convex hull is in the geodesic sense. For every $\varepsilon> 0$ , there exists $m_0 \in \mathbb {N}$ such that, for all $m \geq m_0$ , by the variance inequality (2.5) and Proposition 3.2,

$$ \begin{align*} \delta^2(S_k(a^{\tau}(x)), {\beta}_{\mbox{A}}) & \leq \int_G \delta^2(S_k(a^{\tau}(g)), A(\gamma))\,d{\kern-0.6pt}m(\gamma)\ -\ \alpha \\[3pt] &\leq \varepsilon+ \frac{1}{m} \sum_{j = 0}^{m-1} \delta^2(S_{k}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g)) - \alpha. \end{align*} $$

Finally, by Lemma 2.6,

$$ \begin{align*} \delta^2(S_k(a^{\tau}(x)), {\beta}_{\mbox{A}}) & \leq \varepsilon + \frac{1}{m} \sum_{j = 0}^{m-1}\delta^2(S_{k+j}(a^{\tau}(x)), a_{k+j+1}^{\tau}(x))+R_{m,k}- \alpha, \end{align*} $$

where $\displaystyle R_{m,k}=(({m^2}/{(k+1)^2}) + 2({m}/({k+1}))) C_a$ .

Lemma 3.5. Given $\varepsilon> 0$ , there exists $m_0\geq 1$ such that for every $\ell \in \mathbb {N}$ ,

$$ \begin{align*}\delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) \leq \frac{L}{\ell} + \varepsilon, \end{align*} $$

uniformly in $g\in G$ , where $L = \alpha + 3 C_a^2$ .

Proof. Fix $\varepsilon> 0$ . By Lemmas 3.3 and 3.4, there exists $m_0 \geq 1$ such that for all $k \in \mathbb {N}$ ,

$$ \begin{align*} \delta^2(S_{k+m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) & \leq \frac{k}{k+m_0}\delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{m_0}{k+m_0}(\alpha+\varepsilon) \\[3pt] &\quad - \displaystyle\frac{km_0}{(k+m_0)^2}\bigg(\frac{1}{m_0}\displaystyle\sum_{j = 0}^{m_0 - 1}\delta^2(S_{k+j}(a^{\tau}(x)), a_{k+j+1}^{\tau}(x))\bigg) \end{align*} $$

and

$$ \begin{align*} \frac{1}{m_0} \sum_{j = 0}^{m_0-1}\delta^2(S_{k+j}(a^{\tau}(g)), a_{k+j+1}^{\tau}(g)) \geq \delta^2(S_k(a^{\tau}(g)), {\beta}_{\mbox{A}}) - \varepsilon+ \alpha - R_{m_0,k}. \end{align*} $$

Therefore, combining these two inequalities, we obtain

$$ \begin{align*} \delta^2(S_{k+m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) & \leq \frac{k}{k+m_0}\delta^2(S_{k}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{m_0}{k+m_0}(\alpha+\varepsilon) \\[3pt] & \quad - \displaystyle\frac{km_0}{(k+m_0)^2}(\delta^2(S_k(a^{\tau}(g)), {\beta}_{\mbox{A}}) - \varepsilon + \alpha - R_{m_0,k}). \end{align*} $$

Consider now the particular case where $k=\ell m_0$ . Since $\displaystyle R_{m_0,\ell m_0} \leq ({3}/{\ell }) C_a^2$ , we get

(3.1) $$ \begin{align} \delta^2(S_{(\ell+1)m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) & \leq \frac{\ell}{\ell+1}\delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) + \displaystyle\frac{1}{l+1}(\alpha+\varepsilon)\nonumber \\[3pt] & \quad -\frac{\ell}{(\ell+1)^2}(\delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) - \varepsilon + \alpha - R_{m_0,\ell m_0})\nonumber\\[3pt] &\leq \frac{\ell^2\ \delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) +(2\ell+1)\varepsilon+\alpha+3C_a^2}{(\ell+1)^2}. \end{align} $$

Using this recursive inequality, the result follows by induction on $\ell $ . Indeed, if $\ell =1$ then

$$ \begin{align*}\delta^2(S_{m_0}(a^{\tau}(g)),{\beta}_{\mbox{A}})\leq C_a^2 \leq L. \end{align*} $$

On the other hand, if we assume that the result holds for some $\ell \geq 1$ , that is,

$$ \begin{align*}\delta^2(S_{\ell m_0}(a^{\tau}(x)), g) \leq \frac{L}{\ell} + \varepsilon, \end{align*} $$

then, combining this inequality with (3.1), we have that

$$ \begin{align*} \delta^2(S_{(\ell+1)m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) &\leq \frac{\ell L+ \ell^2\varepsilon +(2\ell+1)\varepsilon+\alpha+3C_a^2}{(\ell+1)^2}=\frac{L}{\ell+1}+\varepsilon.\\[-3pc] \end{align*} $$

Now we are ready to prove the ergodic formula for continuous functions.

Proof of Theorem 1.3

Given $\varepsilon> 0$ , by Lemma 3.5, there exists $m_0 \in \mathbb {N}$ such that

$$ \begin{align*}\delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) \leq \frac{L}{\ell} + \frac{\varepsilon^2}{8}, \end{align*} $$

for every $\ell \in \mathbb {N}$ . Take $\ell _0 \in \mathbb {N}$ such that for all $\ell \geq \ell _0$ ,

(3.2) $$ \begin{align} \delta^2(S_{\ell m_0}(a^{\tau}(g)), {\beta}_{\mbox{A}}) \leq \frac{\varepsilon^2}{4}. \end{align} $$

Let $n=\ell m_0+d$ such that $\ell \geq \ell _0$ and $d\in \{1,\ldots , m_0-1\}$ . Since $x \#_{t} x = x$ for all $x \in M$ , using Corollary 2.4 with the sequences

$$ \begin{align*} &(\ a^\tau_1(g),\ldots,a^\tau_{\ell m_0}(g), \underbrace{S_{\ell m_0}(a^{\tau}(g)),\ldots, S_{\ell m_0}(a^{\tau}(g))}_{\mbox{d times}}\ )\\ \text{and}\\ &(\ a^\tau_1(g),\ldots,a^\tau_{\ell m_0}(g), \ a^\tau_{\ell m_0+1}(g)\ ,\ \ldots\ ,\ a^\tau_{\ell m_0+d}(g)\ ) \end{align*} $$

we get

$$ \begin{align*} \delta(S_{\ell m_0}(a^{\tau}(g)), S_{\ell m_0+d}(a^{\tau}(g))) & \leq \frac{1}{\ell m_0 + d}\sum_{j = 1}^{d} \delta(S_{\ell m_0}(a^{\tau}(g)),a^\tau_{\ell m_0+j}(g)). \end{align*} $$

Now, taking into account that $\delta(S_{\ell m_0}(a^{\tau}(g)),a^\tau_{\ell m_0+j}(g))\leq C_a$ for every $j\in\{1,\ldots, m_0-1\}$ , we obtain that

$$ \begin{align*} \delta(S_{\ell m_0}(a^{\tau}(g)), S_{\ell m_0+d}(a^{\tau}(g))) & \leq \frac{d}{\ell m_0 + d} C_a \leq\frac{1}{\ell} C_a\xrightarrow[k \rightarrow \infty]{} 0. \end{align*} $$

Combining this with (3.2) we conclude that, for n big enough, $\delta(S_{n}(a^{\tau}(g)), {\beta }_{\mbox{A}})<\varepsilon$ .

3.2 Preparation for the $L^{1}$ case

The natural framework for the ergodic theorem is $L^1$ . In this section we will firstly prove Theorem 1.4, which is the main result of this paper.

The strategy of the proof involves constructing good approximations by continuous functions, and obtaining the result for $L^1$ functions as a consequence of the theorem for continuous functions (Theorem 1.3 above). So, the first questions that arise are: what does good approximation mean and what should we require of the approximation in order to get the $L^1$ case as a limit of the continuous case? The next two lemmas contain the clues to answer these two questions. The first lemma can be found as [Reference Sturm, Auscher, Coulhon and Grigor’yan30, Theorem 6.3], and it is often called fundamental contraction property. For the sake of completeness we include a simple proof of this fact.

Lemma 3.6. Let $(\Omega ,\mathcal {B}, P)$ be a probability space, and $A, B \in L^1(X, M)$ . If

$$ \begin{align*} {\beta}_{\mbox{A}} & = \arg\!\min_{z \in M} \int_{\Omega} [\delta^2(A(\omega), z) - \delta^2(A(\omega), y)]\,d{\kern-1pt}P(\omega), \\[3pt] {\beta}_{\mbox {B}} & = \arg\!\min_{z \in M} \int_{\Omega} [\delta^2(B(\omega), z) - \delta^2(B(\omega), y)]\,d{\kern-1pt}P(\omega), \end{align*} $$

then

(3.3) $$ \begin{align} \delta({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}}) \leq \int_\Omega \delta(A(\omega), B(\omega))\,d{\kern-1pt}P(\omega). \end{align} $$

Proof. By the variance inequality (2.5) we get

$$ \begin{align*} \delta^2({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}}) & \leq \int_\Omega \delta^2({\beta}_{\mbox{A}}, B(\omega)) - \delta^2({\beta}_{\mbox {B}}, B(\omega))\,d{\kern-1pt}P(\omega), \\[3pt] \delta^2({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}}) & \leq \int_\Omega \delta^2({\beta}_{\mbox {B}}, A(\omega)) - \delta^2({\beta}_{\mbox{A}}, A(\omega))\,d{\kern-1pt}P(\omega), \end{align*} $$

and the combination of these two inequalities leads to

$$ \begin{align*} 2\delta^2({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}}) & \leq \int_\Omega \delta^2({\beta}_{\mbox{A}}, B(\omega)) + \delta^2({\beta}_{\mbox {B}}, A(\omega))\\[3pt] & \quad - \delta^2({\beta}_{\mbox {B}}, B(\omega)) - \delta^2({\beta}_{\mbox{A}}, A(\omega))\,d{\kern-1pt}P(\omega). \end{align*} $$

Finally, using the Reshetnyak quadruple comparison theorem (Theorem 2.2), we obtain

$$ \begin{align*} 2\delta^2({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}}) & \leq 2\delta({\beta}_{\mbox{A}}, {\beta}_{\mbox {B}})\ \int_\Omega \delta(A(\omega), B(\omega))\,d{\kern-1pt}P(\omega), \end{align*} $$

which is, after some algebraic simplification, the desired result.

Lemma 3.8. Let $A, B \in L^1(G, M)$ . Given $\varepsilon>0$ , for almost every $g\in G$ there exists $n_0$ , which may depend on g, such that

(3.4) $$ \begin{align} \delta(S_{n}(a^{\tau}(g)), S_{n}(b^{\tau}(g))) \leq \varepsilon + \int_{G} \delta(A(g), B(g))\,d{\kern-0.6pt}m(g), \end{align} $$

provided $n\geq n_0$ .

Proof. Indeed, by Corollary 2.4,

$$ \begin{align*}\delta(S_n(a^{\tau}(g)), S_n(b^{\tau}(g))) \leq \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta(a_k^{\tau}(g), b_k^{\tau}(g)) = \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta(A(\tau^k(g)), B(\tau^k(g))), \end{align*} $$

and therefore, the lemma follows from the Birkhoff ergodic theorem.

3.3 Good approximation by continuous functions

The previous two lemmas indicate that we need a kind of $L^1$ approximation. More precisely, given $A\in L^1(G,M)$ and $\varepsilon>0$ , we are looking for a continuous function $A_\varepsilon :G\to M$ such that

$$ \begin{align*}\int_G \delta(A(g),A_\varepsilon(g))\,d{\kern-0.6pt}m(g)<\varepsilon. \end{align*} $$

In some cases there exists an underlying finite-dimensional vector space. This is the case, for instance, when M is the set of (strictly) positive matrices, or more generally, when M is a Riemannian manifold with non-positive curvature. In these cases, the function $A_\varepsilon $ can be constructed by using mollifiers. This idea was used by Karcher in [Reference Karcher16]. In the general case, we can use a similar idea.

Given $\eta>0$ , let $U_\eta $ be a neighborhood of the identity of G so that $m(U_\eta )<\eta $ , whose diameter is also less than $\eta $ . Fix any $y\in M$ , and define

(3.5) $$ \begin{align} A_\eta(g_0)=\arg\!\min_{z\in M} \int_{U_\eta} [\delta^2(z, A(g+g_0)) - \delta^2(y, A(g+g_0)) ]\,d{\kern-0.6pt}m(g). \end{align} $$

Equivalently, $A_\eta (g_0)$ is the barycenter of the pushforward by A of the Haar measure restricted to $g_0+U_\eta $ . This definition follows the idea of mollifiers, replacing the arithmetic mean by the average induced by barycenters. We will prove that, as in the case of usual mollifiers, these continuous functions provide good approximation in $L^1$ (Theorem 3.12 below). With this aim, we first prove the following lemma.

Lemma 3.9. Let $A \in L^p(G, M)$ where $1 \leq p < \infty $ . The function defined by $\varphi : G \rightarrow [0,+\infty )$ by

$$ \begin{align*}\varphi(h) = \int_{G} \delta^p(A(g), A(g + h))\,d{\kern-0.6pt}m(g) \end{align*} $$

is a continuous function.

Proof. Fix $z_0 \in M$ , and define the measure

$$ \begin{align*}\nu(B) := \int_{B} \delta^p(A(g), z_0)\,d{\kern-0.6pt}m(g) \end{align*} $$

on the Borel sets of G. By definition, $\nu $ is absolutely continuous with respect to the Haar measure m. In consequence, given $\varepsilon> 0$ , there exists $\eta> 0$ , such that, whenever a Borel set B satisfies

$$ \begin{align*}\int_{B} \,d{\kern-0.6pt}m(g) < \eta, \end{align*} $$

the inequality

(3.6) $$ \begin{align} \nu(B) = \int_{B} \delta^p(A(g), z_0)\,d{\kern-0.6pt}m(g) < \frac{\varepsilon}{2^{p+1}} \end{align} $$

holds. By the Lusin theorem [Reference Dudley11, Theorem 7.5.2], there is a compact set $C_{\eta } \subset G$ such that $m(C_{\eta }) \geq 1 - \eta /2$ and the restriction of A to $C_{\eta }$ is (uniformly) continuous.

Since m is a Haar measure, it is enough to prove the continuity of $\varphi $ at the identity. With this aim in mind, take a neighborhood of the identity U so that whenever $g_1,g_2\in C_\eta $ satisfy that $g_1-g_2\in U$ , we have

$$ \begin{align*}\delta^p(A(g_1), A(g_2)) \leq \frac{\varepsilon}{2}. \end{align*} $$

Given $h\in U$ , define $\Omega := C_{\eta } \cap (C_{\eta } + h)$ , and $\Omega ^c :=G \setminus \Omega $ . Then

$$ \begin{align*} \int_G \delta^p(A(g), A(g+ h))\,d{\kern-0.6pt}m(g) & \leq \int_{\Omega}\frac{\varepsilon}{2}\,d{\kern-0.6pt}m(g) +\int_{\Omega^c} \delta^p(A(g), A(g + h))\,d{\kern-0.6pt}m(g) \\[3pt] & \leq \frac{\varepsilon}{2}+ \int_{\Omega^c} \delta^p(A(g), A(g+h))\,d{\kern-0.6pt}m(g) \nonumber\\& \leq \frac{\varepsilon}{2} + \int_{\Omega^c} [\delta(A(g), z_0) + \delta(A(g + h), z_0)]^p\, d{\kern-0.6pt}m(g) \\[3pt] & \leq \frac{\varepsilon}{2} + 2^{p} \int_{\Omega^c} \delta^p(A(g), z_0)\,d{\kern-0.6pt}m(g), \end{align*} $$

where in the last identity we have used that m is shift invariant. Since $|\Omega ^c|<\eta $ we obtain that

$$ \begin{align*} \int_G \delta^p(A(g), A(g+ h))\,d{\kern-0.6pt}m(g) & \leq \frac{\varepsilon}{2}+ \frac{\varepsilon}{2}=\varepsilon.\\[-3.4pc] \end{align*} $$

Proof. Indeed, by Lemma 3.6

$$ \begin{align*} \delta(A_\eta(h_1),A_\eta(h_2))&\leq\frac{1}{m(U_\eta)}\int_{U_\eta} \delta(A(g+h_1),A(g+h_2))\,d{\kern-0.6pt}m(g)\\[2pt] &\leq\frac{1}{m(U_\eta)}\int_G \delta(A(g+h_1),A(g+h_2))\,d{\kern-0.6pt}m(g)\\[2pt] &\leq\frac{1}{m(U_\eta)}\int_G \delta[\,A(g),A(g+(h_2-h_1))\,]\,d{\kern-0.6pt}m(g). \end{align*} $$

So the continuity of $A_\eta $ is a consequence of the continuity of $\varphi $ at the identity.

The map $A\mapsto A_\varepsilon $ has the following useful continuity property.

Lemma 3.11. Let $A, B \in L^1(G, M)$ , and $\eta> 0$ . For every $\varepsilon>0$ , there exists $\rho>0$ such that if

$$ \begin{align*}\int_G \delta(A(g), B(g))\,d{\kern-0.6pt}m(g) \leq \rho, \end{align*} $$

then the corresponding continuous functions $A_\eta $ and $B_\eta $ satisfy that

$$ \begin{align*}\max_{g \in G} \, \delta(A_{\eta}(g), B_{\eta}(g)) \leq \varepsilon. \end{align*} $$

Proof. Indeed, given $\varepsilon>0$ , take $\rho =m(U_{\eta })\varepsilon $ . Then, by Lemma 3.6,

$$ \begin{align*} \delta(A_{\eta}(g), B_{\eta}(g)) & \leq \frac{1}{| U_{\eta} | }\int_{U_{\eta}} \delta(A(g + h), B(g + h))\,d{\kern-0.6pt}m(h)\\[2pt] & \leq \frac{1}{| U_{\eta} | }\int_{G} \delta(A(h), B(h))\,d{\kern-0.6pt}m(h) \leq \varepsilon, \end{align*} $$

for all $g \in G$ .

We arrive at the main result on approximation.

Proposition 3.12. Given a function $A\in L^1(G,M)$ , if $A_\eta $ are the continuous functions defined by (3.5) then

$$ \begin{align*}\lim_{\eta\to0^+}\int_G\delta(A(g),A_\eta(g))\,d{\kern-0.6pt}m(g) = 0. \end{align*} $$

Proof. First, assume that $A\in L^2(G,M)$ . In this case, by the variance inequality, the inequality

$$ \begin{align*}\delta^2(A(g), A_{\eta}(g)) \leq \frac{1}{| U_{\eta} | }\int_{U_{\eta}} \delta^2(A(g), A(g + h))\,d{\kern-0.6pt}m(h) \end{align*} $$

holds. So, using Fubini’s theorem, we obtain

$$ \begin{align*} \int_G\delta^2(A(g),A_\eta(g))\,d{\kern-0.6pt}m(g) &\leq \frac{1}{| U_{\eta} | }\int_{U_{\eta}} \int_G \delta^2(A(g), A(g + h))\,d{\kern-0.6pt}m(g)\,d{\kern-0.6pt}m(h)\\[3pt] &= \frac{1}{| U_{\eta} | }\int_{U_{\eta}} \varphi (h) \,d{\kern-0.6pt}m(h). \end{align*} $$

By Lemma 3.9, the function $\varphi $ is continuous. In consequence, if e denotes the identity of G, then

$$ \begin{align*}\lim_{\eta\to 0^+} \frac{1}{| U_{\eta} | }\int_{U_{\eta}} \varphi (h) \,d{\kern-0.6pt}m(h) = \varphi (e)=0. \end{align*} $$

This proves the result for functions in $L^2(G,M)$ since, by Jensen’s inequality,

$$ \begin{align*}\ \ \int_G\delta(A(g),A_\eta(g))\,d{\kern-0.6pt}m(g)\leq \bigg( \int_G\delta^2(A(g),A_\eta(g))\,d{\kern-0.6pt}m(g) \bigg)^{1/2}. \end{align*} $$

Now consider a general $A \in L^1(G, M)$ . Fix $z_0\in M$ , and for each natural number N define the truncations

$$ \begin{align*}A^{(N)}(g) := \begin{cases} A(g) & \mbox{if }\delta(A(g), z_0) < N, \\ z_0 & \mbox{if }\delta(A(g), z_0) \geq N. \end{cases} \end{align*} $$

For each N we have that $A^{(N)}\in L^1(G,M)\cap L^\infty (G,M)$ , and therefore it also belongs to $L^2(G,M)$ . On the other hand, since the function defined on G by $g\mapsto \delta (A(g),z_0)$ is integrable, we have that

(3.7) $$ \begin{align} \int_G \delta(A(g),A^{(N)}(g))\,d{\kern-0.6pt}m(g)= \int_{\{g:\,\delta(A(g), z_0) \geq N\}} \delta(A(g),z_0)\,d{\kern-0.6pt}m(g) \xrightarrow[N\to \infty]{}0. \end{align} $$

So, if $A_\eta $ and $A^{(N)}_\eta $ are the continuous functions associated to A and $A^{(N)}$ respectively, then

$$ \begin{align*} \int_G \delta(A(g),A_\eta(g))\,d{\kern-0.6pt}m(g) &\leq \int_G \delta(A(g),A^{(N)}(g))\,d{\kern-0.6pt}m(g) \\[3pt] &\quad + \int_G \delta(A^{(N)}(g),A^{(N)}_\eta(g))\,d{\kern-0.6pt}m(g) \\[3pt] &\quad + \int_G \delta(A^{(N)}_\eta(g),A_\eta(g))\,d{\kern-0.6pt}m(g). \end{align*} $$

Note that each term of the right-hand side tends to zero: the first by (3.7), the second by the $L^2$ case done in the first part, and the last by Lemma 3.11.

3.4 The $L^{1}$ case and almost everywhere convergence

Let $\varepsilon>0$ . For each $k\in \mathbb {N}$ , let $A_k$ be a continuous function such that

$$ \begin{align*}\int_G \delta(A(g),A_k(g))\,d{\kern-0.6pt}m(g)\leq\frac{1}{k}. \end{align*} $$

By Lemma 3.8, we can take a set of measure zero $N\subseteq G$ such that if we take $g\in G\setminus N$ and $k\in \mathbb {N}$ , there exists $n_0$ , which may depend on g and k, such that

$$ \begin{align*}\delta(S_{n}(a^{\tau}(g)), S_{n}(a^{\tau}_{(k)}(g))) \leq \frac{\varepsilon}{4} + \int_{G} \delta(A(g), A_k(g))\,d{\kern-0.6pt}m(g), \end{align*} $$

provided $n\geq n_0$ . In this expression, $a_{(k)}^{\tau }$ is the sequence defined in terms of $A_k$ and $\tau $ as in (1.6). Fix $g\in G\setminus N$ . Taking k so that $1/k<\varepsilon /4$ , we get that

$$ \begin{align*}\delta(S_{n}(a^{\tau}(g)), S_{n}(a^{\tau}_{(k)}(g))) \leq \frac{\varepsilon}{2}, \end{align*} $$

for every $n\geq n_0$ . By Lemma 3.6, we also have that $ \delta ({\beta }_{\mbox{A}}, {\beta }_{\mbox {A_k}}) \leq {\varepsilon }/{4}, $ where

$$ \begin{align*} {\beta}_{\mbox{A}} & = \arg\!\min_{z \in M} \int_G [\delta^2(A(g), z) - \delta^2(A(g), y)]\,d{\kern-0.6pt}m(g), \\ {\beta}_{\mbox{A_k}} & = \arg\!\min_{z \in M} \int_G [\delta^2(A_k(g), z) - \delta^2(A_k(g), y)]\,d{\kern-0.6pt}m(g). \end{align*} $$

Finally, by Theorem 1.3, there exists $n_1\geq 1$ such that, for every $n\geq n_1$ ,

$$ \begin{align*}\delta(S_{n}(a_{(k)}^{\tau}(g),{\beta}_{\mbox{A_k}})\leq\frac{\varepsilon}{4}. \end{align*} $$

Combining all these inequalities, we obtain that

$$ \begin{align*} \delta(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})&\leq \delta(S_{n}(a^{\tau}(g)), S_{n}(a^{\tau}_{(k)}(g)))\\[3pt] &\quad +\delta(S_{n}(a^{\tau}_{(k)}(g)),{\beta}_{\mbox{A_k}})+\delta({\beta}_{\mbox{A_k}}, {\beta}_{\mbox{A}})\leq\varepsilon, \end{align*} $$

which concludes the proof.

3.5 The $L^p$ results

We conclude this section by proving the $L^p$ ergodic theorems.

Theorem 3.13. Let $1 \leq p < \infty $ and $A\in L^p(G, M)$ . Then

(3.8) $$ \begin{align} \lim_{n\to\infty } \int_{G} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) = 0. \end{align} $$

Proof. Let us define the following measure on the Borel sets of G:

$$ \begin{align*}\nu(B) := \int_{B} \delta^p(A(g), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g). \end{align*} $$

By definition, $\nu $ is absolutely continuous with respect to the Haar measure m. In consequence, given $\varepsilon> 0$ , there exists $\eta> 0$ such that, whenever a Borel set B satisfies

$$ \begin{align*}\int_{B} \,d{\kern-0.6pt}m(g) < \eta, \end{align*} $$

we have that

(3.9) $$ \begin{align} \nu(B) = \int_{B} \delta^p(A(g), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) < \frac{\varepsilon}{2}. \end{align} $$

By Egoroff’s theorem, as

$$ \begin{align*}\delta^p(S_n(a^{\tau}(g)), {\beta}_{\mbox{A}}) \xrightarrow[n \rightarrow \infty]{} 0\end{align*} $$

converge almost everywhere on a finite measure space, there exists a set $C_{\eta } \subset G$ with $m(C_{\eta }) < \eta $ such that

(3.10) $$ \begin{align} \delta^p(S_n(a^{\tau}(g)), {\beta}_{\mbox{A}}) \xrightarrow[n \rightarrow \infty]{} 0 \end{align} $$

uniformly on $G \setminus C_\eta .$

On the other hand,

$$ \begin{align*}\delta(S_n(a^{\tau}(g)), {\beta}_{\mbox{A}}) \leq \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta(a_k^{\tau}(g), {\beta}_{\mbox{A}}) = \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta(A(\tau^k(g)), {\beta}_{\mbox{A}}), \end{align*} $$

Therefore, by Jensen’s inequality,

(3.11) $$ \begin{align} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}}) & \leq \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta^p(A(\tau^k(g)), {\beta}_{\mbox{A}}). \end{align} $$

Now, there exists $n_0 \in \mathbb {N}$ such that, for all $n \geq n_0$ ,

$$ \begin{align*}\int_{G \setminus C_\eta} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) < \frac{\varepsilon}{2}, \end{align*} $$

as a consequence of (3.10). Therefore

$$ \begin{align*} \int_{G} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) & = \int_{G \setminus C_\eta} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g)\\[3pt] &\quad + \int_{C_\eta} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) \\[3pt] &= \frac{\varepsilon}{2} + \int_{C_\eta} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g). \end{align*} $$

On the other hand, taking integral over $C_\eta $ in (3.11), we obtain

$$ \begin{align*} \int_{C_\eta} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) &\leq \int_{C_\eta} \displaystyle\frac{1}{n} \displaystyle\sum_{k = 0}^{n-1} \delta^p(A(\tau^k(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) \\[3pt] & = \sum_{k = 0}^{n-1}\displaystyle\frac{1}{n} \int_{C_\eta} \delta^p(A(\tau^k(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g). \end{align*} $$

Since $\displaystyle \int _{C_{\eta }} \,d{\kern-0.6pt}m(g) < \eta , $ by (3.9),

$$ \begin{align*}\nu(C_{\eta}) = \int_{C_{\eta}} \delta^p(A(g), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) < \frac{\varepsilon}{2}. \end{align*} $$

So, for all $n \in \mathbb {N}$ ,

$$ \begin{align*}\displaystyle\sum_{k = 0}^{n-1}\displaystyle\frac{1}{n} \int_{C_\eta} \delta^p(A(\tau^k(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) < \frac{\varepsilon}{2}. \end{align*} $$

Finally, combining this two bounds, given $\varepsilon $ , there exists $n_0 \in \mathbb {N}$ such that, for all $n \geq n_0$ ,

$$ \begin{align*}\int_{G} \delta^p(S_{n}(a^{\tau}(g)), {\beta}_{\mbox{A}})\,d{\kern-0.6pt}m(g) < \varepsilon, \end{align*} $$

which concludes the proof.