Let $(\kg,*)$ be a simply connected nilpotent Lie group. Let $\mu$ be  an aperiodic probability measure on $\kg$ with a finite moment  of order strictly greater than $2$ for the weight filtration induced by $\XXab=\E(\mu_{ab})$. Let $\nu$ be the limit measure arising in the central limit theorem (<ref>) for some choice of weight decomposition $\kg=\oplus \km^{(b)}$. Fix a right invariant distance  on $(\kg,*)$ and let

$$\mathscr{L}=\{F: \kg\rightarrow \R \,\,|\,\, F\, \,1\emph{-lipschitz},\, \|F\|_{\infty}\leq 1\}$$
$$\sup_{F\in \mathscr{L}}\, |\mu^{*N}(F)-(\DilN \nu*N\XX)(F)|\underset{N\to +\infty}{\longrightarrow} 0 $$

1) Note that $F$ may not decay at infinity. The idea of the proof is to pixelize $F$, and then apply the uniform local limit theorem to each pixel.

2) If we replace the right-invariant metric by a left invariant metric, then the statement still holds, following the same line of argument. We will need the metric to be right-invariant below in order to regularize bounded harmonic functions for the right-random walk using convolution on the left.

We fix  $0<\gamma_{0}\lll 1$ so that the truncation $\Theta^N$ defined in  <Ref> satisfies the uniform local limit theorem <ref>. As $\mu$ has in particular finite second moment, <Ref> yields that

$$\| \mu^{*N} -\Theta^N\|\to0 $$
in total variation. Hence, we may replace $\mu^{*N}$ by $\Theta^N$ in order to prove <Ref>.

Let $\eps>0$.  We fix a “bump function” $\chi_{\eps} : \kg \rightarrow \R^+$, i.e.  a continuous function supported on the ball  $B_{\kg}(0, \eps)$ with total sum $1$. We introduce the convolution $F_\eps:=\chi_\eps*F$, given by

$$F_{\eps}(x)= \int_{\kg}\chi_{\eps}(u^{-1}) F(u x)du=\int_{\kg}\chi_{\eps}(xu^{-1}) F(u)du$$

where, here and below, we write $xy=x*y$ and $x^{-1}=-x$ to lighten notations.

As $F$ is $1$-Lispchitz we have $\|F-F_{\eps}\|_{\infty}\leq \eps$, so for all $N\geq 1$,
\begin{equation}\label{fee} |\Theta^N(F)-(\DilN \nu*N\XX)(F)|  \leq |\Theta^N(F_{\eps})-(\DilN \nu*N\XX)(F_{\eps})|   +2\eps\end{equation}

On the other hand

\begin{align} \label{muNFeps}
\Theta^N(F_{\eps})&=\int_{\kg}(\Theta^N*\delta_{u^{-1}}) (\chi_{\eps})\,  F(u) \,du
\end{align}

The uniform local limit theorem <ref> with arbitrary deviations by right multiplication implies that uniformly in $u$

\begin{align} \label{TLLchi}
\Theta^N*\delta_{u^{-1}} (\chi_{\eps}) = \DilN \nu *N\XX* \delta_{u^{-1}} (\chi_{\eps}) + o_{\gamma_{0}, \chi_{\eps}}(N^{-\frac{\dd}{2}})
\end{align}

where the $o_{\gamma_{0}, \chi_{\eps}}(.)$ does not depend on $u$. However, we may not replace directly $\Theta^N$ by $\DilN \nu *N\XX$ in the integral because the error term is a priori not integrable. To remedy this, we truncate the domain of integration in a part on which the error integrates to   $o_{\gamma_{0}, \chi_{\eps}}(1)$, and a part that is negligible thanks to the central limit theorem. More precisely, for $C>0$ we set

$$Q_{C, N}=\{u\in \kg \,:\, \forall i,\,\,\|u^{(i)}\|\leq C N^{i/2}\}$$

By the central limit theorem (<ref>), and the observation that $Q_{C,N}= \DilN (Q_{C,1})$, we may choose $C=C(\mu, \eps)$ large enough so that for $N\geq 1$,

$$(\Theta^N*-N\XX)(Q^c_{C,N}) <\eps\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\DilN \nu(Q_{C,N}) < \eps $$

where $Q^c_{C,N}:=\kg \smallsetminus Q^c_{C,N}$. Then we split the integral in (<ref>) into two parts corresponding to the integration domains $Q_{2C,N}*N\XX$ and $Q^c_{2C,N}*N\XX$. Since $\|F\|_\infty \leq 1$, the second part is bounded above by $(\Theta^N*-N\XX) (B_{\kg}(0,\eps)*Q^c_{2C,N})$, which is itself at most $(\Theta^N*-N\XX)(Q^c_{C,N}  )<\eps$. The same applies to $\DilN \nu$ in place of $\Theta^N*-N\XX$, so in the end in view of (<ref>), we get:

\begin{align*}
|\Theta^N(F_{\eps}) - \DilN \nu *N\XX(F_{\eps}) | \leq 2\eps + o_{\gamma_{0}, \chi_{\eps}}(1)
\end{align*}

Finally, for large $N$  (depending on $\mu$, $\gamma_0$, $\eps$, $\chi_{\eps}$ but not on $F\in \mathscr{L}$), the left-hand side in (<ref>) is at most $5\eps$, and the final result follows letting  $\eps$ go to $0$.

Finally, we show  <Ref>.

We  identify $G$ with its Lie algebra $\kg$ via the exponential map, and denote by $*$ the induced group structure. Up to replacing $\mu$ by $\frac{1}{2}\mu+\frac{1}{2}\delta_0$, which has the same harmonic functions, we may assume that $\mu$ is aperiodic.

We endow $\kg$ with a $*$-right invariant Riemannian metric. Without loss of generality we may further restrict attention to harmonic functions $F$ that are Lipschitz. Indeed, we may replace $F$ by the convolution $\chi *F$, where $\chi$ is any smooth bump function with integral $1$. This new function remains harmonic and bounded. It is furthermore Lipschitz.  Finally if $\chi *F$ is a.e. constant for all such $\chi$, then $F$ too is a.e. constant.

Arguing by induction on the dimension of $\kg$ (applying the result to $\kg/\mathfrak{z}(\kg)$), it is enough to prove that $F$ is invariant under translation by the center $\mathfrak{z}(\kg)$ of $\kg$.

So, assume that $F$ is Lipschitz, pick $x_0 \in \kg$ and $z \in \mathfrak{z}(\kg)$. The rest of the proof is very similar to Nelson's proof of Liouville's theorem [68].  The left translate $F_{x_{0}}(x)=F(x_0x)$ is again bounded, harmonic and it remains Lipschitz[The Lipschitz constant of $F_{x_{0}}$ depends on $x_{0}$ but this is not an issue.]. Thus by <Ref> we have

\begin{equation}\label{Fx0F}F(x_{0}) =\mu^{*N}(F_{x_{0}}) = (\DilN \nu*N\XX)(F_{x_{0}})+o_{F_{x_{0}}}(1)
\end{equation}

and the same holds with $x_{0}z$ in place of $x_{0}$.
But since $z$ lies in the center $\mathfrak{z}(\kg)$ we have:

$$(\DilN \nu*N\XX)(F_{x_{0}z})= \int_{\kg} F(x_{0} z x *N\XX) \,d\DilN \nu(x) = \int_{\kg} F(x_{0}x *N\XX) \,d(\DilN \nu*\delta_z)(x)$$

In view of (<ref>), we can conclude that $F(x_0)=F(x_0z)$ provided we show that

$$\int_{\kg} F(x_{0}x *N\XX) \,d(\DilN \nu*\delta_z)- \DilN \nu )(x)\longrightarrow 0.$$

But this is indeed the case, since setting $\DilN \nu= u_{N}(x) dx$ where $v_{N} =N^{-\frac{\dd}{2}} v \circ \DilsN $, the above integral is bounded above by:

\begin{align*}
\|F\|_{\infty}\int_{\kg} |v_{N}(x+z)-v_{N}(x)| dx =\|F\|_{\infty}\int_{\kg} |v(x+\DilsN z)-v(x)| dx
\end{align*}

which tends to $0$ as $N\to+\infty$.

§.§ Equidistribution in homogeneous spaces

In this section we prove Theorem <ref> regarding equidistribution of unipotent random walks on homogeneous spaces of finite volume. In a nutshell, we will use our LLT (twice!) in the guise of Lemma <ref> above in order to reduce to a version of Ratner's equidistribution theorem proved by Shah, Theorem <ref> below. A similar strategy was employed in [21].

We first reduce to the case of aperiodic walks thanks to the following general lemma.

[Reduction to aperiodic case]
Let $\mu$ be a probability measure on a group $G$ and  $\mu'=\frac{1}{2}\mu + \frac{1}{2}\delta_{1}$. Then for all $N>1$,

$$\|\frac{1}{2N}\sum_{k=1}^{2N}\mu'^k- \frac{1}{N}\sum_{k=1}^N\mu^k\| \leq C \frac{\log N}{\sqrt{N}},$$

where $\|.\|$ is the total variation norm and $C$ an absolute constant.

Expanding the power $\mu'^k$, and setting $\binom{k}{i}=0$ if $i>k$,  we may write

\begin{align*}
\sum_{k=1}^{2N}\mu'^k = \sum_{i=1}^{2N}  \sum_{k=1}^{2N} \frac{1}{2^k} \binom{k}{i}  \mu^i = \sum_{i=1}^{2N} \E( N_{2N}(i)) \mu^i
\end{align*}

where $N_p(i)$ is the occupation time $N_p(i):=|\{k\leq p, S_{k}=i\}|$ and $(S_{k})_{k\geq0}$ is the Markov chain on $\N_{\ge 0}$ starting at $0$ with transition probabilities $p(i,i)=p(i,i+1)=1/2$, for $i \ge 0$. We can then write:

\begin{equation}\label{bbmar}
\|\frac{1}{2N}\sum_{k=1}^{2N}\mu'^k- \frac{1}{N}\sum_{k=1}^N\mu^k\| \leq \frac{1}{N}\sum_{i=1}^{2N} |\frac{1}{2}\E( N_{2N}(i)) -\1_{i\leq N}|
\end{equation}

We split the right hand side in three terms $A_1+A_2+A_3$ according to the position of $i$ with respect to $p_{N}:=\lfloor N- \sqrt{N}\log N \rfloor$ and $q_{N}=\lfloor N+ \sqrt{N}\log N \rfloor$.

Recall, either from the general theory of discrete Markov chains or in this case by a direct calculation, that $\E(N_\infty(i))=2$ for all $i$. In particular, if $i \in [p_N,q_N]$, each term in the sum $A_2$ is bounded by $2$, so $A_2 \leq 4\log N/\sqrt{N}$.  Next:
\begin{align*}A_3 & = \frac{1}{N} \sum_{i \ge q_N} \E(N_{2N}(i)) = \frac{1}{N} \sum_{k\leq 2N}\sum_{i \ge q_N} \PP(S_k=i) = \frac{1}{N} \sum_{k\leq 2N}  \PP(S_k\ge q_N)\\ & \leq  2 \PP(S_{2N} \ge q_N). \end{align*}
However $S_{2N}$ is a  binomial  random variable and by classical tail bounds $\PP(S_{2N} \ge q_N)$ decays faster than any inverse power of $N$. So $A_3 \leq \log N/\sqrt{N}$ for $N$ large enough.

Finally if $i\le p_N$,

\begin{align*}
|\frac{1}{2}\E( N_{2N}(i)) -1| = \frac{1}{2} |\E( N_{2N}(i)) -\E( N_{\infty}(i)) | \leq \frac{1}{2} \sum_{k>2N} \PP(S_k=i)
\end{align*}

Summing over $i\le p_N$, we get:

$$A_1\leq \frac{1}{2}\sum_{k>2N} \PP(S_k \leq p_N) \leq \frac{1}{2}\sum_{k>2N} \PP(S_k \leq k/2- \sqrt{k/2}\log(k/2)).$$

Again each term of the latter sum decays faster than any inverse power of $k$, hence so does the sum as a function of $N$. This ends the proof.

We keep the notation of Theorem <ref> and consider the Alexandrov compactification $\Omega\cup \{\infty\}$ of the homogeneous space $\Omega=H/\Lambda$. The next proposition ensures that the sequence of measures has no escape of mass and that any limit measure is absolutely continuous with respect to the homogeneous measure $m_\omega$ given by Ratner's theorem:

There exists $C>0$ such that every weak limit $m$ of the sequence of probability measures $(\frac{1}{N}\sum_{k=1}^N\mu^k*\delta_{\w})_{n\geq 1}$ on $\Omega \cup \{\infty\}$ satisfies $m\leq C m_{\w}$.
The result holds without  Cesàro if $\mu$ is centered.

Before giving the proof, we recall the Ratner-Shah equidistribution theorem. Let $\ku$ be the Lie algebra of the connected $Ad$-unipotent subgroup $U \leq H$, so $(\ku,*)$ is the simply connected covering of $U$.  We say that a vector space  basis  $(e_{1}, \dots, e_{d})$ of $\ku$ is  triangular if it satisfies $[e_{i}, e_{j}]\subseteq \text{Vect}\{e_{k} , \, k>\max(i,j)\}$ for all $i,j$. A basis is regular if it can be obtained by permuting  the elements in a triangular basis.

Let $(e_{1}, \dots, e_{d})$ be a regular basis  of $\ku$.  For $\ut=(t_{1}, \dots, t_{d})\in \R^d$, set $R_{\ut}:=[0, t_{1}] e_{1}*  \dots *[0, t_{d}]e_{d}$. Then for  $\w\in \Omega$,

$$\frac{\leb_{|R_{\ut}}}{\leb(R_{\ut})}*\delta_{\w} \underset{\inf t_{i}\to +\infty}{\longrightarrow}  m_{\w}$$

where $\leb$ stands for the Lebesgue measure on $\ku$.

This theorem is due to Ratner when $U$ is one-dimensional. It was subsequently generalized by Shah <cit.> in the present form. Note that Shah states it for simply connected $U$, but his proof does not require this assumption.

Applying this result to a finite union of rectangles, it clearly holds with $R_{\ut}$ replaced by $B_{\ut}:=[-t_{1}, t_{1}]e_{1}* \dots * [-t_{d}, t_{d}]e_{d}$, as $\inf t_{i} \to +\infty$.

We now assume that  $e_{1}, \dots, e_{d}$ enumerates the basis $(e^{(i)}_{j})$ introduced  in <Ref>.
It is indeed a regular basis (thanks to the relation $[\kg^{(i_{1})}, \kg^{(i_{2})}]\subseteq \kg^{(i_{1}+i_{2})}$). In the non-centred case, we also assume $e_{d}=\XX$.

First we observe that <Ref> allows us to replace $\mu^k$ by the measure $D_{\sqrt{k}}\nu*k\XX$ without loss of generality. Indeed, endow $H$ with a right invariant Riemannian metric, and $\Omega$  with the quotient metric. For every Lipschitz function $f$ on $\Omega$, we set $F:\ku \to \R$, $F(u)=f(\exp(u)\omega)$. Then $F$ is Lipschitz on $\ku$ for say a right-invariant Riemannian metric on $(\ku,*)$, and <Ref> guarantees that
$$(\mu^k-D_{\sqrt{k}}\nu*k\XX)(F) \underset{k\to+\infty}{\longrightarrow} 0.$$

Hence, the set of weak limits of  $\frac{1}{N}\sum_{k=1}^N\mu^k*\delta_\omega$ (or $\mu^N*\delta_{\w}$ in the centered case)  does not change if we consider $D_{\sqrt{k}}\nu*k\XX$ in place of  $\mu^k$.

<Ref> now follows from the following claim.

There is $C>0$, such that for all   $N,\eps >0$ there are positive measures $m_{N}$ and $\sigma_{N}$ on $\Omega$, with  $\|m_{N}\|=1$ and $\|\sigma_{N}\|\leq \eps$, such that

$$\frac{1}{N}\sum_{k=1}^N D_{\sqrt{k}}\nu*k\XX \leq C m_{N} + \sigma_{N}$$

and  $m_{N}\to_{N \to +\infty} m_{\w}$. The statement also holds without  Cesàro average if $\XX=0$.

Let $K_1$ be a compact neighbourhood of the origin in $\ku$ and let $K_r=D_{\sqrt{r}}K_1$. Since $\nu$ has a continuous density, there is $C_1>0$ such that $\nu\leq C_{1}\leb_{|K_r}+ \nu_{\ku\smallsetminus K_r}$ for all $r\ge 1$.

For all $N\geq1$,

\begin{align*}
\DilN \nu &  \leq \DilN  C_{1}\leb_{K_r}+\DilN \nu_{\ku\smallsetminus K_r}= C_{1}N^{-\dd/2}\leb_{D_{\sqrt{rN}}K_1}+\DilN \nu_{\ku\smallsetminus K_r}\\
&\leq C_{1}M\frac{\leb_{B_{\ut_{Nr}}}}{\leb(B_{\ut_{Nr}})}+\DilN \nu_{\ku\smallsetminus K_r} = Cm_N + \sigma_N
\end{align*}

setting $C=C_1M$, where the constant $M$ and the box $B_{\ut_{rN}} \supset D_{\sqrt{rN}}K_1$ are given by   <Ref> below applied to $K_1$. Choosing $r$ such that $\|\nu_{\ku\smallsetminus K_r}\|<\eps$, the result follows in case $\mu$ is centered.
In the general case when $\XX \neq 0$, we translate by $k\XX$ on the right and average over $k$:

$$\frac{1}{N}\sum_{k=1}^N D_{\sqrt{k}}\nu*k\XX \leq C\frac{1}{N}\sum_{k=1}^N \frac{\leb_{B_{\ut_{rN}} *k\XX}}{\leb(B_{\ut_{rN}})} + \sigma_{N}$$

where $\sigma_{N}=\frac{1}{N}\sum_{k=1}^N(D_{\sqrt{k}}\nu_{\ku\smallsetminus K_r})*k\XX$ has mass less than $\eps$.
As we took care of choosing $e_{d}=\XX$,  the diffeomorphism $t_{1}e_{1}*\dots *t_{d}e_{d}\mapsto (t_{1}, \dots, t_{d})$ sends the Lebesgue measure $\leb_{B_{\ut_{rN}} *k\XX}$ to the product measure

$$\leb_{[-\ut_{rN,1}, \ut_{rN,1}]}\otimes \dots \otimes \leb_{[-\ut_{rN,d-1}, \ut_{rN,d-1}]}\otimes \leb_{[-\ut_{rN,d}, \ut_{rN,d}]+k}.$$

By construction of $\ut_{rN}$, we have  $\ut_{rN,d}=M \sqrt{rN}$, so for $N$ large enough

$$ \frac{1}{N}\sum_{k=1}^N\leb_{[-\ut_{rN,d}, \ut_{rN,d}]+k}\leq  \frac{2M\sqrt{Nr}+1}{N}\leb_{[-2N,2N]}.$$

$$\frac{1}{N}\sum_{k=1}^N D_{\sqrt{k}}\nu*k\XX \leq C(2M+1)\frac{\sqrt{r}\leb_{B_{\ut'_{rN}}}}{\sqrt{N}\leb(B_{\ut_{rN}})} + \sigma_{N}$$

where $\ut'_{rN}$ is obtained from $\ut_{rN}$ by replacing the coordinate $\ut_{rN,d}$ par $2N$. As  $\leb(B_{\ut'_{rN}})$  is comparable to $\frac{2N}{\sqrt{rN}}\leb(B_{\ut_{rN}})$ the claim follows by <Ref>.

In the proof we used the following well-known comparison principle between Malcev coordinates of the first and second kind <cit.>.

[comparison principle]
Let $K\subseteq \ku$ be a compact set. There exists $M>0$ such that for all $r>1$, there exists $\ut_{r}\in \R^d$ satisfying

$$D_{\sqrt{r}}K \subseteq B_{\ut_{r}}\,\,\,\,\,\,\,\,\,\,et \,\,\,\,\,\,\,\,\,\,\leb(B_{\ut_{r}}) \leq  M r^{\frac{\dd}{2}}.$$

It is sufficient to check that for all $C>0$, there exists $M>0$ such that for all $r>1$,

\begin{align} \label{eq-incl}
\left\{\sum_{i,j} s^{(i)}_{j}e^{(i)}_{j}\,: \,\forall i,j, \,|s^{(i)}_{j}|\leq C \right\} \subseteq D_{\frac{1}{\sqrt{r}}} \left\{ \prod_{i,j} t^{(i)}_{j}e^{(i)}_{j}\,: \,\forall i,j, \,|t^{(i)}_{j}|\leq Mr^{i/2}  \right\}
\end{align}

where the product follows the order prescribed the enumeration $e_{1}, \dots, e_{d}$.  But we can write  $D_{\frac{1}{\sqrt{r}}} \prod_{i,j} t^{(i)}_{j}e^{(i)}_{j}= \prod^{*'}_{i,j} r^{-i/2}t^{(i)}_{j}e^{(i)}_{j} + P_{r}((r^{-i/2}t^{(i)}_{j})_{i,j})$ where $*'$ is the graded product coming from $*$ for the weight filtration, and $P_{r}$ is obtained by dividing by a  (non zero)  power of $\sqrt{r}$ each coefficent of a polynomial function $P$ with $d$ variables. Then, we choose $r_{0}>0$ such that for $r>r_{0}$,
$$\|P_{r}((r^{-i/2}t^{(i)}_{j})_{i,j})\|\leq 1$$

and $M>0$ large enough so that

$$\left\{\sum_{i,j} s^{(i)}_{j}e^{(i)}_{j}\,: \,\forall i,j, \,|s^{(i)}_{j}|\leq C+1 \right\} \subseteq  \left\{ {\prod_{i,j}}^{*'} t^{(i)}_{j}e^{(i)}_{j}\,: \,\forall i,j, \,|t^{(i)}_{j}|\leq M \right\}$$

This proves (<ref>) for $r>r_{0}$. Up to increasing the value of $M$ even more, we get the result for all $r>1$.

We can finally conclude the proof of <Ref>.

<Ref> guarantees that every weak limit of the sequence
$(\frac{1}{N}\sum_{k=1}^N \mu^N*\delta_{\w})_{N\geq1}$
is a probability measure on $\Omega$ that is absolutely continuous with respect to   $m_{\w}$. It is also $\mu$-stationary, so it must be $U$-invariant by <Ref> and hence $m_\w$ itself, because $m_\w$ is $U$-ergodic by Ratner's theorem [71].

When $\mu$ is centered, we may apply the same argument to the sequence $(\mu^N*\delta_{\w})_{N\geq1}$, but we need to justify why limit measures are $\mu$-stationary. It is sufficient to check that for every compactly supported smooth function  $f$  on $\Omega$, we have $|\mu^{N+1}*\delta_\w(f)- \mu^{N}*\delta_\w(f)|\rightarrow 0$ as $N\to +\infty$.  Setting $F(u)=f(\exp(u)\w)$ we obtain a bounded Lipschitz function on $\ku$ and by <Ref>, it  boils down  to checking that  $|D_{\sqrt{N+1}}\nu(F)- \DilN \nu(F)|\rightarrow 0$. To see this, we observe that for the  norm in total variation,

$$\|D_{\sqrt{N+1}}\,\nu-\DilN \nu\|= \|\nu -D_{\sqrt{\frac{N}{N+1}}}\,\nu\|\longrightarrow 0$$

whence the result.

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