Let $M$ be a smooth manifold, $g\_M$ a Riemannian metric, and consider for $x\in M$ the *volume growth function*, $gr\_x$ that maps $r>0$ to the volume $vol\_{g\_M}(B(x,r))$. My interest is to see whether or not any differentiable manifold can be endowed with a metric structure such that the family of growth functions $\{gr\_x, x\in M\}$ has bounded variation. The answers posted in [Example of a manifold which is not a homogeneous space of any Lie group](https://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group) make clear that some kind of variation must be allowed. 


A Riemannian manifold is said to be of *bounded geometry* if the injectivity radius is bounded away from 0 and the norms of the covariant derivatives of the curvature tensor $|\nabla^iR|$ are bounded with bounds independent of the point $x\in M$ (but they may depend on $i$ of course). It is well known that every differentiable manifold admits a metric of bounded geometry. This was proved by R.E. Greene in his paper *Complete metrics of bounded curvature on noncompact manifolds*.


We say that the growth of $M$ is *uniformly superlinear* or *uniformly subexponential* if the limits $$\lim \inf\_{s\to \infty} \frac{vol(B(x,s))}{s}=\infty,\quad \lim \sup\_{s\to \infty} \frac{\ln[ vol(B(x,s))]}{s}=0$$
converge uniformly on $x\in M$ (that is, in the first case for example, given $\epsilon >0$, we can choose $\delta(\epsilon)$ independent of $x\in M$).


**Question:** Can we endow every differentiable manifold $M$ with a metric tensor such that $M$ is of bounded geometry and exhibits either uniform superlinear or uniform subexponential growth? 



Does every smooth manifold admit a metric with bounded geometry and uniform growth?
The answer is "yes".


Read page 96 in ["Volume and bounded cohomology"](http://www.numdam.org/item?id=PMIHES_1982__56__5_0) by Gromov or "Manifolds with quadratic curvature decay and slow volume growth" by Lott and Shen.


