# Does every smooth manifold admit a metric with bounded geometry and uniform growth?

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 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?