[proofplan] The proof is an immediate consequence of Bishop-Gromov monotonicity after normalizing at arbitrarily small radii. The ratio of the volume of a geodesic ball in $M$ to the volume of the corresponding ball in the constant-curvature model is nonincreasing in the radius, while both numerator and denominator have the same Euclidean small-radius asymptotic. Therefore the ratio is bounded above by its limiting value $1$ at radius $0$. When $k>0$, Bonnet-Myers supplies the diameter bound, and the endpoint estimate gives the global volume bound. [/proofplan]

[step:Define the comparison ratio on positive admissible radii] Let $R_k=\infty$ when $k\leq 0$ and let \begin{align*} R_k=\frac{\pi}{\sqrt{k}} \end{align*} when $k>0$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$, and let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $S^{n-1}\subset\mathbb{R}^n$. Define the model radial function $s_k:[0,R_k]\to[0,\infty)$ by \begin{align*} s_k(t)= \begin{cases} \frac{1}{\sqrt{k}}\sin(\sqrt{k}t),& k>0,\\ t,& k=0,\\ \frac{1}{\sqrt{-k}}\sinh(\sqrt{-k}t),& k<0. \end{cases} \end{align*} Define the model ball-volume function $V_k^n:[0,R_k]\to[0,\infty)$ by \begin{align*} V_k^n(r)=\mathcal{H}^{n-1}(S^{n-1})\int_0^r s_k(t)^{n-1}\,d\mathcal{L}^1(t). \end{align*} For $r>0$, let $B_g(p,r)=\{q\in M:\operatorname{dist}_g(p,q)<r\}$ denote the open geodesic ball centered at $p$ with radius $r$, and let $\operatorname{Vol}_g$ denote the Riemannian volume measure on $(M,g)$. For $r\in(0,R_k)$, define the comparison ratio \begin{align*} F_p:(0,R_k)&\to [0,\infty)\\ r&\mapsto \frac{\operatorname{Vol}_g(B_g(p,r))}{V_k^n(r)}. \end{align*} The denominator is positive for every $r\in(0,R_k)$ because $s_k(t)>0$ for $t\in(0,R_k)$ and \begin{align*} V_k^n(r)=\mathcal{H}^{n-1}(S^{n-1})\int_0^r s_k(t)^{n-1}\,d\mathcal{L}^1(t)>0. \end{align*} [/step]

[step:Use Bishop-Gromov monotonicity to compare every radius to smaller radii]By the [Bishop-Gromov Volume Comparison Theorem](/theorems/5371), applied to the complete connected Riemannian manifold $(M,g)$ with Ricci lower bound $\operatorname{Ric}_g\geq (n-1)kg$ at the base point $p$, the function $F_p$ is nonincreasing on $(0,R_k)$. Hence, whenever $0<\varepsilon<R<R_k$, \begin{align*} F_p(R)\leq F_p(\varepsilon). \end{align*}[/step]

[guided]The comparison theorem we need is Bishop-Gromov monotonicity. Its hypotheses are exactly the global hypotheses in the statement: $(M,g)$ is complete, and the Ricci tensor satisfies the lower bound \begin{align*} \operatorname{Ric}_g(v,v)\geq (n-1)k\,g(v,v) \end{align*} for every tangent vector $v$. Under these hypotheses, the Bishop-Gromov Volume Comparison Theorem states that, for each fixed base point $p\in M$, the ratio \begin{align*} F_p(r)=\frac{\operatorname{Vol}_g(B_g(p,r))}{V_k^n(r)} \end{align*} is nonincreasing as $r$ increases through the interval $(0,R_k)$. This monotonicity is the entire mechanism of the proof. If $0<\varepsilon<R<R_k$, then $R$ is the larger radius and $\varepsilon$ is the smaller radius, so nonincreasingness gives \begin{align*} F_p(R)\leq F_p(\varepsilon). \end{align*} Thus the value of the comparison ratio at any fixed positive radius is controlled by its values at arbitrarily small radii. The remaining task is to compute the limiting value of $F_p(\varepsilon)$ as $\varepsilon\to 0$.[/guided]

[step:Evaluate the limiting ratio at the base point scale]Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let $B_{\mathbb{R}^n}(0,1)=\{x\in\mathbb{R}^n:|x|<1\}$ denote the Euclidean unit ball. Define \begin{align*} \omega_n:=\mathcal{L}^n(B_{\mathbb{R}^n}(0,1)). \end{align*} The Riemannian small-ball volume asymptotic in [normal coordinates](/theorems/2713) at $p$ gives \begin{align*} \operatorname{Vol}_g(B_g(p,\varepsilon))=\omega_n\varepsilon^n+o(\varepsilon^n) \end{align*} as $\varepsilon\to 0$. The model volume has the same first-order asymptotic. Since $s_k(t)/t\to 1$ as $t\to 0^+$, we have \begin{align*} s_k(t)^{n-1}=t^{n-1}+o(t^{n-1}) \end{align*} uniformly for $0<t<\varepsilon$ as $\varepsilon\to 0^+$. Hence \begin{align*} \int_0^\varepsilon s_k(t)^{n-1}\,d\mathcal{L}^1(t) &=\int_0^\varepsilon t^{n-1}\,d\mathcal{L}^1(t)+o(\varepsilon^n)\\ &=\frac{\varepsilon^n}{n}+o(\varepsilon^n). \end{align*} Using $\mathcal{H}^{n-1}(S^{n-1})=n\omega_n$, we obtain \begin{align*} V_k^n(\varepsilon) &=\mathcal{H}^{n-1}(S^{n-1})\int_0^\varepsilon s_k(t)^{n-1}\,d\mathcal{L}^1(t)\\ &=\omega_n\varepsilon^n+o(\varepsilon^n). \end{align*} Therefore \begin{align*} \lim_{\varepsilon\to 0}F_p(\varepsilon)=1. \end{align*}[/step]

[guided]The small-radius limit compares two quantities that both look Euclidean at first order. Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let \begin{align*} \omega_n:=\mathcal{L}^n(B_{\mathbb{R}^n}(0,1)). \end{align*} The Riemannian small-ball volume asymptotic in normal coordinates at $p$ gives \begin{align*} \operatorname{Vol}_g(B_g(p,\varepsilon))=\omega_n\varepsilon^n+o(\varepsilon^n) \end{align*} as $\varepsilon\to 0^+$. For the model volume, the defining property of $s_k$ near the origin is $s_k(t)/t\to 1$. Raising to the power $n-1$ preserves this first-order comparison, so \begin{align*} s_k(t)^{n-1}=t^{n-1}+o(t^{n-1}) \end{align*} uniformly for $0<t<\varepsilon$ as $\varepsilon\to 0^+$. Integrating this estimate with respect to $\mathcal{L}^1$ gives \begin{align*} \int_0^\varepsilon s_k(t)^{n-1}\,d\mathcal{L}^1(t) &=\int_0^\varepsilon t^{n-1}\,d\mathcal{L}^1(t)+o(\varepsilon^n)\\ &=\frac{\varepsilon^n}{n}+o(\varepsilon^n). \end{align*} Since $\mathcal{H}^{n-1}(S^{n-1})=n\omega_n$, this yields \begin{align*} V_k^n(\varepsilon)=\omega_n\varepsilon^n+o(\varepsilon^n). \end{align*} Dividing the Riemannian asymptotic by the model asymptotic gives \begin{align*} \lim_{\varepsilon\to 0}F_p(\varepsilon)=1. \end{align*}[/guided]

[step:Pass from monotonicity to the ball volume bound] Fix $R$ with $0<R<R_k$. For every $\varepsilon\in(0,R)$, the previous monotonicity step gives \begin{align*} \frac{\operatorname{Vol}_g(B_g(p,R))}{V_k^n(R)} \leq \frac{\operatorname{Vol}_g(B_g(p,\varepsilon))}{V_k^n(\varepsilon)}. \end{align*} Taking $\liminf_{\varepsilon\to 0}$ on the right-hand side and using the limiting ratio computed above yields \begin{align*} \frac{\operatorname{Vol}_g(B_g(p,R))}{V_k^n(R)}\leq 1. \end{align*} Since $V_k^n(R)>0$, this is exactly \begin{align*} \operatorname{Vol}_g(B_g(p,R))\leq V_k^n(R). \end{align*} For $R=0$, both sides are $0$, so the same inequality holds. [/step]

[step:Obtain the positive curvature endpoint by continuity from below]Assume $k>0$ and let $R_k=\pi/\sqrt{k}$. Choose any increasing sequence $(R_j)_{j=1}^\infty$ in $(0,R_k)$ with $R_j\to R_k$. Since the geodesic balls $B_g(p,R_j)$ increase to $B_g(p,R_k)$, continuity from below for the Riemannian volume measure gives \begin{align*} \operatorname{Vol}_g(B_g(p,R_k)) = \lim_{j\to\infty}\operatorname{Vol}_g(B_g(p,R_j)). \end{align*} The function $V_k^n$ is continuous on $[0,R_k]$ by its integral definition, so \begin{align*} \lim_{j\to\infty}V_k^n(R_j)=V_k^n(R_k). \end{align*} Applying the already proved estimate at each $R_j$ and passing to the limit gives \begin{align*} \operatorname{Vol}_g(B_g(p,R_k))\leq V_k^n(R_k). \end{align*}[/step]

[guided]The inequality has already been proved for radii strictly smaller than $R_k$. To reach the endpoint, choose an increasing sequence $(R_j)_{j=1}^\infty$ in $(0,R_k)$ such that $R_j\to R_k$. The open balls are nested and satisfy \begin{align*} \bigcup_{j=1}^\infty B_g(p,R_j)=B_g(p,R_k). \end{align*} Continuity from below for the Riemannian volume measure therefore gives \begin{align*} \operatorname{Vol}_g(B_g(p,R_k)) = \lim_{j\to\infty}\operatorname{Vol}_g(B_g(p,R_j)). \end{align*} For each $j$, the strict-radius estimate gives \begin{align*} \operatorname{Vol}_g(B_g(p,R_j))\leq V_k^n(R_j). \end{align*} The model volume function $V_k^n$ is continuous on $[0,R_k]$ because it is defined as the integral of the [continuous function](/page/Continuous%20Function) $s_k(t)^{n-1}$ over $[0,R]$ with respect to $\mathcal{L}^1$. Hence \begin{align*} \lim_{j\to\infty}V_k^n(R_j)=V_k^n(R_k). \end{align*} Passing to the limit in the strict-radius inequalities yields \begin{align*} \operatorname{Vol}_g(B_g(p,R_k))\leq V_k^n(R_k). \end{align*}[/guided]

[step:Use Bonnet-Myers to convert the endpoint ball estimate into the total volume estimate]Assume $k>0$. Since the theorem assumes $n\geq 2$, the constant $(n-1)k$ is positive. By the Bonnet-Myers Theorem, applied to the complete connected Riemannian manifold $(M,g)$ with Ricci lower bound $\operatorname{Ric}_g\geq (n-1)kg$, the diameter of $M$ satisfies \begin{align*} \operatorname{diam}_g(M)\leq \frac{\pi}{\sqrt{k}}. \end{align*} Let $\overline{B}_g(p,\pi/\sqrt{k})=\{q\in M:\operatorname{dist}_g(p,q)\leq \pi/\sqrt{k}\}$ denote the closed geodesic ball. The diameter bound gives $M\subseteq \overline{B}_g(p,\pi/\sqrt{k})$. The distance sphere \begin{align*} S_g\left(p,\frac{\pi}{\sqrt{k}}\right) := \left\{q\in M:\operatorname{dist}_g(p,q)=\frac{\pi}{\sqrt{k}}\right\} \end{align*} has zero $n$-dimensional Riemannian volume: outside the cut locus it is covered by a smooth geodesic-polar hypersurface, and the cut locus has zero Riemannian volume. Therefore the open and closed balls at this radius have the same Riemannian volume, and \begin{align*} \operatorname{Vol}_g(M) = \operatorname{Vol}_g\left(B_g\left(p,\frac{\pi}{\sqrt{k}}\right)\right). \end{align*} Using the endpoint estimate, \begin{align*} \operatorname{Vol}_g(M) \leq V_k^n\left(\frac{\pi}{\sqrt{k}}\right). \end{align*} Finally, the simply connected $n$-sphere $S_k^n$ of constant sectional curvature $k$ has geodesic polar volume density $s_k(t)^{n-1}$ up to radius $\pi/\sqrt{k}$, so \begin{align*} \operatorname{Vol}(S_k^n) = V_k^n\left(\frac{\pi}{\sqrt{k}}\right). \end{align*} Therefore \begin{align*} \operatorname{Vol}_g(M)\leq \operatorname{Vol}(S_k^n), \end{align*} which completes the proof.[/step]

[guided]The total-volume estimate uses the positive Ricci lower bound to turn a ball estimate into a global estimate. Since $k>0$ and $n\geq 2$, the lower bound $\operatorname{Ric}_g\geq (n-1)kg$ has positive constant $(n-1)k$. The Bonnet-Myers Theorem applies to complete connected Riemannian manifolds with such a positive Ricci lower bound, and gives \begin{align*} \operatorname{diam}_g(M)\leq \frac{\pi}{\sqrt{k}}. \end{align*} Thus every point of $M$ lies in the closed geodesic ball \begin{align*} \overline{B}_g\left(p,\frac{\pi}{\sqrt{k}}\right) = \left\{q\in M:\operatorname{dist}_g(p,q)\leq \frac{\pi}{\sqrt{k}}\right\}. \end{align*} The boundary sphere at this radius has zero $n$-dimensional Riemannian volume: away from the cut locus it is a smooth geodesic-polar hypersurface, and the cut locus has zero Riemannian volume. Therefore the open and closed balls at radius $\pi/\sqrt{k}$ have the same Riemannian volume, and \begin{align*} \operatorname{Vol}_g(M) = \operatorname{Vol}_g\left(B_g\left(p,\frac{\pi}{\sqrt{k}}\right)\right). \end{align*} The endpoint estimate gives \begin{align*} \operatorname{Vol}_g(M) \leq V_k^n\left(\frac{\pi}{\sqrt{k}}\right). \end{align*} By the definition of the model sphere $S_k^n$, its geodesic polar volume density is $s_k(t)^{n-1}$ up to radius $\pi/\sqrt{k}$, hence \begin{align*} \operatorname{Vol}(S_k^n) = V_k^n\left(\frac{\pi}{\sqrt{k}}\right). \end{align*} Combining these identities and inequalities gives \begin{align*} \operatorname{Vol}_g(M)\leq \operatorname{Vol}(S_k^n). \end{align*}[/guided]