[proofplan] The proof is an immediate application of Bishop-Gromov volume comparison with model sectional curvature $-K$. The comparison says that the ratio between the volume of a geodesic ball in $M$ and the corresponding model volume is nonincreasing in the radius. Applying this monotonicity at the two radii $R$ and $2R$ gives the displayed doubling estimate. The uniform local doubling constant follows by showing that the model ratio $V_{-K}^n(2R)/V_{-K}^n(R)$ remains bounded for $0<R\leq R_0$, using the small-radius asymptotic $V_{-K}^n(r)\sim c_n r^n$. [/proofplan]

[step:Apply Bishop-Gromov monotonicity at the radii $R$ and $2R$]Fix $p \in M$ and $R>0$. Since $(M^n,g)$ is complete and satisfies $\operatorname{Ric}_g \geq -(n-1)K g$, Bishop-Gromov volume comparison with model curvature $-K$ applies (citing a result not yet in the wiki: [Bishop-Gromov Volume Comparison Theorem](/theorems/5371)). It states that the function \begin{align*} \Theta_p:(0,\infty)&\to (0,\infty)\\ r&\mapsto \frac{\operatorname{vol}_g(B_g(p,r))}{V_{-K}^n(r)} \end{align*} is nonincreasing. Because $R \leq 2R$, monotonicity gives \begin{align*} \frac{\operatorname{vol}_g(B_g(p,2R))}{V_{-K}^n(2R)} \leq \frac{\operatorname{vol}_g(B_g(p,R))}{V_{-K}^n(R)}. \end{align*} Multiplying both sides by $V_{-K}^n(2R)>0$ gives \begin{align*} \operatorname{vol}_g(B_g(p,2R)) \leq \frac{V_{-K}^n(2R)}{V_{-K}^n(R)} \operatorname{vol}_g(B_g(p,R)). \end{align*}[/step]

[guided]Fix $p \in M$ and $R>0$. The theorem is designed to be read directly from Bishop-Gromov comparison: we compare the actual volume growth of $(M,g)$ with the volume growth in the simply [connected space](/page/Connected%20Space) form whose sectional curvature is the constant $-K$. The hypotheses needed for Bishop-Gromov volume comparison are exactly present here. The manifold $(M^n,g)$ is complete, and the Ricci curvature lower bound is \begin{align*} \operatorname{Ric}_g(v,v) \geq -(n-1)K\,g(v,v) \end{align*} for every tangent vector $v$. Therefore Bishop-Gromov volume comparison with model curvature $-K$ applies (citing a result not yet in the wiki: Bishop-Gromov Volume Comparison Theorem). It says that the map \begin{align*} \Theta_p:(0,\infty)&\to (0,\infty)\\ r&\mapsto \frac{\operatorname{vol}_g(B_g(p,r))}{V_{-K}^n(r)} \end{align*} is nonincreasing. Now choose the two radii in the statement. Since $R \leq 2R$, nonincreasingness of $\Theta_p$ gives \begin{align*} \Theta_p(2R) \leq \Theta_p(R). \end{align*} Writing out the definition of $\Theta_p$ at these two radii gives \begin{align*} \frac{\operatorname{vol}_g(B_g(p,2R))}{V_{-K}^n(2R)} \leq \frac{\operatorname{vol}_g(B_g(p,R))}{V_{-K}^n(R)}. \end{align*} The model volume $V_{-K}^n(2R)$ is positive because it is the volume of a nonempty geodesic ball of positive radius in the model space. Multiplying the inequality by $V_{-K}^n(2R)$ therefore preserves the inequality and yields \begin{align*} \operatorname{vol}_g(B_g(p,2R)) \leq \frac{V_{-K}^n(2R)}{V_{-K}^n(R)} \operatorname{vol}_g(B_g(p,R)). \end{align*} This is the first asserted estimate.[/guided]

[step:Bound the model volume ratio on bounded radius intervals] Fix $R_0>0$. Define the model radial function \begin{align*} s_K:[0,\infty)&\to [0,\infty)\\ t&\mapsto \begin{cases} t, & K=0,\\ \frac{\sinh(\sqrt{K}\,t)}{\sqrt{K}}, & K>0. \end{cases} \end{align*} Let $\omega_{n-1}:=\mathcal{H}^{n-1}(S^{n-1})$ denote the Euclidean $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) of the unit sphere. The model volume is \begin{align*} V_{-K}^n(r)=\omega_{n-1}\int_0^r s_K(t)^{n-1}\,d\mathcal{L}^1(t). \end{align*} Define \begin{align*} A:(0,\infty)&\to (0,\infty)\\ r&\mapsto \frac{V_{-K}^n(r)}{r^n}. \end{align*} Since $s_K(t)/t \to 1$ as $t\to 0^+$, we have \begin{align*} \lim_{r\to 0^+} A(r) = \frac{\omega_{n-1}}{n}. \end{align*} Thus $A$ extends continuously to $[0,\infty)$ by setting $A(0):=\omega_{n-1}/n$. For $0<R\leq R_0$, \begin{align*} \frac{V_{-K}^n(2R)}{V_{-K}^n(R)} = 2^n\frac{A(2R)}{A(R)}. \end{align*} The function $A$ is continuous and positive on the compact intervals $[0,2R_0]$ and $[0,R_0]$. Hence \begin{align*} M&:=\max_{0\leq r\leq 2R_0} A(r),\\ m&:=\min_{0\leq r\leq R_0} A(r) \end{align*} are well-defined finite constants, and $m>0$. Therefore, for every $0<R\leq R_0$, \begin{align*} \frac{V_{-K}^n(2R)}{V_{-K}^n(R)} \leq 2^n\frac{M}{m}. \end{align*} Set \begin{align*} C(n,K,R_0):=2^n\frac{M}{m}. \end{align*} Combining this bound with the estimate from the previous step gives \begin{align*} \operatorname{vol}_g(B_g(p,2R)) \leq C(n,K,R_0)\operatorname{vol}_g(B_g(p,R)) \end{align*} for every $p\in M$ and every $0<R\leq R_0$. This proves the local volume doubling conclusion. [/step]