Volume doubling implies that the degree is uniformly bounded above?

Question

Let $G=(V,E)$ be a connected graph. Here $V$ is the set of all vertices of $G$, and $E$ is the set of all edges of $G$. Suppose that $G$ is locally finite, i.e., $\sharp\{y\sim x:y \in V \}$ is finite for any $x\in V$. We denote by $m(x)=\sharp\{y\sim x:y \in V \}$. We denote by $d(\cdot,\cdot):V\times V\to [0,\infty)$ the combinatorial metric on $G$, i.e., for any $x,y\in V$, $d(x,y)$ is the smallest number of edges among all paths connecting $x$ and $y$. For any $x\in V$, $B(x,r)$ is the open ball centered at $x$ with respect to the metric $d$, of radius $r$. Suppose that $G$ satisfies the volume doubling condition, i.e., for any $x\in V$ and any $r>0$, $m(B(x,2r))\leq C\cdot m(B(x,r))$, where $C$ is a universal constant, and $m(B(x,r)):=\sum_{y\in B(x,r)}m(x)$.

Question: Can we prove that $\sup_{x\in V} m(x)<\infty$ ?

Answer 1

Yes. More precisely for every $x \in V$, $m(x) \leq (C-1)^2$.

This follows from the volume doubling inequality for $r=1$, which says that for all $x \in V$, $m(x) + \sum_{y \sim x} m(y) \leq C m(x)$, or equivalently $\frac{1}{m(x)} \sum_{y \sim x} m(y) \leq C-1$. In particular, there is $y \sim x$ such that $m(y) \leq C-1$. Applying the volume doubling centered at $y$ and $r=1$ leads to $\sum_{z \sim y} m(z) \leq (C-1)^2$, and (since $x \sim y$) $m(x) \leq (C-1)^2$.

Answer 2

EDIT: as noted by Mikael de la Salle, my answer below is incorrect according to the definitions in the question. Notice, however, that this relies specifically on the case $r=1$. If we restrict our attention to $r>1$, or consider closed instead of open balls, then my answer is valid. I am leaving it here for this reason.

ORIGINAL ANSWER: No. Take $\mathbb{N}$ and connect vertices $x$ and $y$ if $$|x-y|<\log (x+y) .$$ This graph has unbounded degree and satisfies volume doubling.