The question is Lemma 5.3 in [1] (with-out detailed proof). But I don't know how to prove. 


Let $M$ be a (finite dim) manifold satisfying the following two assumptions:


(1) for any $x\in M$, and any $r>0$, we have $\mu(B\_x(2r))\leq C\_1\cdot \mu(B\_x(r))$. Here $\mu$ is the volume measure on $M$;


(2) for any $p\geq1$, any $f\in C^\infty(M)$, any $x\in M$, and $r>0$, we have, 
$\int\_{B\_x(r)}|f-f\_{x,r}|^pd\mu \leq C\_2\cdot r^p\cdot\int\_{B\_x(2r)} |\nabla f|^pd\mu$. Here $f\_{x,r}$ is the mean value of $f$ on $B\_x(r)$.


We want to prove that: Given $\epsilon>0$ and any $x,y\in M$ with $d(x,y)\leq2 \epsilon$, we have $$|{f\_{x,\epsilon}}-{f\_{y,\epsilon}}|\leq\frac{C(\epsilon,C\_1,C\_2)}{\mu(B\_x(r))}\int\_{B\_x(6r)}|\nabla f|d\mu.$$


Reference:
[1] T. Coulhon & L. Saloff-Coste, Variétés riemanniennes isométriques à l´ ınfini, Rev. Mat. Iberoamericana 11 (1995), no. 3, 687726.



Prove a consequence of Poincare inequality and volume doubling
I proved this by myself just now. Given any $u\in C^\infty$. 
Let $z$ be a point such that $d(x,z) \leq \epsilon$ and $d(y,z)\leq\epsilon$. 
Then, $|u\_{x,\epsilon}-u\_{z,2\epsilon}| \leq \int\_{B\_x(\epsilon)} |u-u\_{z,2\epsilon}|/\mu(B\_{x}(\epsilon)) \leq \int\_{B\_z(2\epsilon)} |u-u\_{z,2\epsilon}|/\mu(B\_{x}(\epsilon)) \leq C\epsilon $$\int\_{B\_z(4\epsilon)}|\nabla u | /\mu(B\_x(\epsilon))$. 


We can prove an inequality for $|u\_{y,\epsilon}-u\_{z,2\epsilon}|$. Combining the two inequalities and using the Doubling condition, we can get the answer.


