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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.

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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.

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