Random walk uniformly hitting a compact set Suppose $G$ is a locally compact compactly generated group. Let $\mu$ be a probability measure that is:


*

*Adapted to $G$, i.e. there is no proper subgroup $H$ such that $\mu(H)=1$. 

*Symmetric, i.e. $\mu(A)=\mu(A^{-1})$.

*compactly supported, and $1_G$ is in the support.

*$\mu<<\lambda$, where $\lambda$ is Haar.  
Let $K$ be a compact set with $\lambda(K)>0$. Let $\mu^{*n}$ be the $n$-fold convolution $\mu*...*\mu$. 

Do there exist $c>0$ and $n>0$ such that $\frac{d\mu^{*n}}{d\lambda}\geq c>0$ on $K$? 

In other words, do a $\mu$ random walk, when hitting $K$, after many steps, hits it (in a sense) uniformly?
 A: In the special case of a finitely generated group, you can easily find a counter-example. So, without any other assumption, the answer is no.
Take $G=\mathbb{Z}$ and $\mu=\frac{1}{2}(\delta_{-1}+\delta_1)$. These satisfy your assumptions. Now, take $K=\{0,1\}$. For every $n$, either $\mu^{*n}$ is supported on even numbers or on odd numbers, so either $\mu^{*n}(1)=0$ or $\mu^{*n}(0)=0$.
EDIT: Now that you added aperiodicity, for finitely generated groups, the answer is yes if yo replace adaptedness by irreducibility. More precisely, you only have to assume that there is $n_0\geq 0$ such that $\mu^{*n_0}(e)>0$ and that the symmetric support of $\mu$ generates $G$. Indeed, let $K$ be a compact (i.e. finite set). For every $x\in K$, there exists $n_x$ such that $\mu^{*n_x}(x)>0$ and using aperiodicity, for $n\geq n_x+n_0$, $\mu^{*n}(x)>0$. Since $K$ is finite, you can take $n=\max (n_x+n_0,x\in K)$, so that for every $x\in K$, $\mu^{*n}(x)>0$.
The exact same proof works in the general case if the following holds: for every $x\in G$, there exists an open neighborhood $U_x$ of $x$, some positive $c_x$ and $n_x\geq 0$ such that $\frac{d\mu^{*n_x}}{d\lambda}\geq c_x>0$ on $U_x$.
A: The answer is positive under the assumptions boldfaced below, which hold at least when 
$G$ is a complete connected Riemannian manifold (see the Remark at the end of this answer), so that the case of $G=\mathbb R^d$ is of course included. 
On the other hand, the simple counterexample given by M. Dus shows that without a connectedness condition the desired conclusion will fail to hold in general. 
Since the measure $\mu$ is nonzero, the set $E:=\{x\in G\colon f(x)>t\}$ is of nonzero $\lambda$-measure for some real $t>0$, where $f:=\frac{d\mu}{d\lambda}$. Also, $E=E^{-1}$, since $\mu$ is symmetric. Hence,
\begin{equation*}
 f^{*2}(x)=\int f(y^{-1}x)f(y)dy\ge t\int_E f(y^{-1}x)\,dy\ge t^2|E\cap(xE^{-1})|=t^2|E\cap(xE)|,
\end{equation*}
where $dy:=\lambda(dy)$, and $|\cdot|:=\lambda(\cdot)$. By the regularity of the measure $\lambda$, 
\begin{equation*}
 |C_E|>\tfrac12\,|C| 
\end{equation*}
for some compact set $C$, where $C_E:=C\cap E$. 
Assuming that $G$ has a countable basis of neighborhoods of $1_G$, 
it follows, by continuity, that 
\begin{equation*}
 |(xC)_E\cap C_E|>\tfrac12\,|C| \tag{1}
\end{equation*}
for all $x\in B$, where $B$ is 
some (without loss of generality, symmetric) neighborhood of $1_G$. So, 
\begin{equation}
 f^{*2}(x)\ge t^2|(xE)\cap E|\ge t^2|(xC)_E\cap C_E|>\tfrac{t^2}2\,|C|=:c_1>0 \tag{2}
\end{equation}
for all $x\in B$. 
Suppose next that for some symmetric neighborhood $\hat B$ of $1_G$ such that $\hat B\subseteq B$ and all natural $n\ge2$ we have 
\begin{equation*}
 a_n:=\inf_{x\in \hat B^n}\,|(xB)\cap \hat B^{n-1}|>0.  \tag{3}
\end{equation*}
(If e.g. $G=\mathbb R^d$, $B$ is the ball centered at $0$ of "radius" $r>0$ with respect to (say) the $\ell_\infty^d$ norm, and $\hat B=\frac12\,B$, then $a_n=r^d$.) 
Define recursively $c_n:=c_{n-1}c_1 a_n$ for $n\ge2$, with $c_1$ as above. 
It follows by induction that 
\begin{equation*}
 f^{*2n}\ge c_n>0
\end{equation*}
on $\hat B^n$. Indeed, for $n=1$ this follows by (2) and the condition $\hat B\subseteq B$. For $n\ge2$ and any $x\in \hat B^n$, 
by induction and (2), 
\begin{multline*}
 f^{*2n}(x)=\int f^{*2}(y^{-1}x)f^{*2(n-1)}(y)dy\ge c_{n-1}\int_{\hat B^{n-1}} f^{*2}(y^{-1}x)dy \\ 
 \ge c_{n-1}\int_{(xB)\cap \hat B^{n-1}} f^{*2}(y^{-1}x)dy
  \ge c_{n-1}c_1 a_n=c_n>0.    
\end{multline*} 
(For the penultimate inequality in the above display, we used the fact that for any $y\in xB$ we have $y^{-1}x\in B^{-1}=B$ and hence, by (2), $f^{*2}(y^{-1}x)\ge c_1$.) 
Thus, the desired result follows if we finally assume that for any compact $K$ and any (or, equivalently, any symmetric) neighborhood $U$ of $1_G$ there is a natural $n$ such that 
\begin{equation}
 U^n\supseteq K. \tag{4}
\end{equation}

Remark. Conditions (1), (3), and (4), used above, will hold when $G$ is metrizable with a metric $d$ satisfying the following "linear" connectedness condition: 
\begin{equation}
 \forall x,y\in G\ \forall t\in[0,1]\ \exists z\in G\quad d(x,z)=(1-t)d(x,y)\quad \& \quad d(z,y)=td(x,y). \tag{5}
\end{equation}
In particular, condition (5) holds when $G$ is a complete connected Riemannian manifold; see e.g. wiki/Hopf--Rinow theorem and wiki/Riemannian_manifold. 
Let us now verify that (5) implies (1), (3), and (4). This is obvious concerning (1), since $G$ is now a metric space and therefore has a countable basis of neighborhoods of $1_G$. Also, (5) implies 
\begin{equation}
 B_r^n=B_{nr} \tag{6}
\end{equation}
for all natural $n$ and all real $r>0$, where $B_r:=\{x\in G\colon d(x,1_G)\le r\}$, the $d$-ball of radius $r$ centered at $1_G$. 
Take now any real $r>0$ and any compact $K$. Then $K\subseteq FB_r$ for some finite $F\subseteq G$. By (6), for each $x\in F$ there is some natural $n_x$ such that $x\in B_r^{n_x}$ and hence $xB_r\subseteq B_r^{n_x+1}$. So, $K\subseteq FB_r\subseteq B_r^n$ for $n:=1+\max_{x\in F}n_x$, which verifies (4). 
It remains to verify (3). Without loss of generality, there we have $B=B_r$ for some real $r>0$. Take any natural $n\ge2$, then any $a\in(1/n,1)$, and let $\hat B:=B_{ar}$. It is enough to show that 
\begin{equation}
 a_n\ge|B_b|>0, \tag{7}
\end{equation}
where $b:=\frac r2\,(1-a)>0$. 
First here, if $|B_b|=0$ then $|K|=0$ for any compact $K$, which would contradict the condition that $G$ is compactly generated. So, we have the second inequality in (7). To verify the first inequality there, take any $x\in\hat B^n=B_{nar}$, so that $d(x,1_G)=nar_1$ for some $r_1\in[0,r]$. Let 
\begin{equation*}
 s:=nar_1-\tfrac{r_1}2\,(1+a)=(n-1)ar_1-\tfrac{r_1}2\,(1-a)\in[0,(n-1)ar_1]. 
\end{equation*}
By (5), there is some $z\in G$ such that 
\begin{equation*}
 d(z,1_G)=s,\quad d(x,z)=nar_1-s=\tfrac{r_1}2\,(1+a). 
\end{equation*}
For any $y\in zB_b$, we have $d(y,z)\le b$ and hence 
\begin{equation*}
 d(y,x)\le d(y,z)+d(z,x)\le b+\tfrac{r_1}2\,(1+a)\le b+\tfrac r2\,(1+a)=r, 
\end{equation*}
\begin{equation*}
 d(y,1_G)\le d(y,z)+d(z,1_G)\le b+s\le b+(n-1)ar-\tfrac r2\,(1-a)=(n-1)ar, 
\end{equation*}
so that $y\in (xB_r)\cap B_{(n-1)ar}=(xB)\cap \hat B^{n-1}$. 
So, $zB_b\subseteq(xB)\cap \hat B^{n-1}$, whence $|(xB)\cap \hat B^{n-1}|\ge|B_b|$, for all $x\in\hat B^n$, 
which implies the first inequality in (7). 
Thus, all the conditions (1), (3), and (4) are verified. 
