[proofplan] We write the reduced volume as the integral of Perelman's reduced-volume density and use the distributional differential inequality satisfied by that density along a complete backward Ricci flow. After verifying that the completeness and bounded-curvature hypotheses give the required cutoff [integration by parts](/theorems/2098), we integrate the inequality in time to obtain monotonicity. The regularity of the base point identifies the small-time limit of the reduced volume with the Euclidean [Gaussian integral](/theorems/1140), which gives the upper bound, while positivity follows from positivity of the density. [/proofplan]

[step:Define the reduced-volume density and state its differential inequality]Let $n = \dim M$. For each $\tau \in (0,t_0]$, let $d\mu_\tau$ denote the Riemannian measure induced by $g(\tau)$, let $R(\cdot,\tau): M \to \mathbb{R}$ denote the scalar curvature of $g(\tau)$, and let $\ell: M \times (0,t_0] \to \mathbb{R}$ denote Perelman's reduced distance based at $(p,0)$. Define the reduced-volume density $v: M \times (0,t_0] \to (0,\infty)$ by \begin{align*} v(q,\tau)=(4\pi\tau)^{-n/2} e^{-\ell(q,\tau)}. \end{align*} Then \begin{align*} \widetilde V(\tau)=\int_M v(q,\tau)\,d\mu_\tau(q) \end{align*} whenever the integral is finite. Perelman's reduced-distance differential inequality for complete backward Ricci flows with bounded curvature on compact time subintervals gives, in the distributional sense on $M \times (0,t_0]$, \begin{align*} (\partial_\tau - \Delta_{g(\tau)} + R(\cdot,\tau))v \leq 0. \end{align*} The hypotheses of that result are exactly the completeness of $(M,g(\tau))$ for each $\tau$, the backward Ricci flow equation $\partial_\tau g = 2\operatorname{Ric}_{g(\tau)}$, and the bounded-curvature assumption on each compact subinterval of $(0,t_0]$.[/step]

[guided]The first task is to put the reduced volume into a form to which a parabolic inequality can be applied. Let $n = \dim M$. For each time $\tau \in (0,t_0]$, write $d\mu_\tau$ for the Riemannian measure of $g(\tau)$, write $R(\cdot,\tau): M \to \mathbb{R}$ for the scalar curvature of $g(\tau)$, and write $\ell: M \times (0,t_0] \to \mathbb{R}$ for Perelman's reduced distance from the base point $(p,0)$. We define the reduced-volume density as the map $v: M \times (0,t_0] \to (0,\infty)$ given by \begin{align*} v(q,\tau)=(4\pi\tau)^{-n/2} e^{-\ell(q,\tau)}. \end{align*} With this notation the reduced volume is \begin{align*} \widetilde V(\tau)=\int_M v(q,\tau)\,d\mu_\tau(q), \end{align*} provided this integral is finite. The key input is Perelman's reduced-distance differential inequality. It says that for a complete backward Ricci flow satisfying $\partial_\tau g = 2\operatorname{Ric}_{g(\tau)}$ and having bounded curvature on compact time subintervals, the density $v$ satisfies \begin{align*} (\partial_\tau - \Delta_{g(\tau)} + R(\cdot,\tau))v \leq 0 \end{align*} in the distributional sense on $M \times (0,t_0]$. The theorem applies here because completeness is assumed, the family $g(\tau)$ is a backward Ricci flow by hypothesis, and the curvature is bounded on each compact subinterval of $(0,t_0]$. This is the analytic inequality behind reduced-volume monotonicity: the extra scalar-curvature term is exactly the term that compensates for the time derivative of the evolving Riemannian measure.[/guided]

[step:Test the distributional inequality and pass to the reduced-volume integral]Fix $0<a<b\leq t_0$ such that $\widetilde V(a)$ and $\widetilde V(b)$ are finite. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. We use the integrated form of Perelman's reduced-volume inequality: for every nonnegative time [test function](/page/Test%20Function) $\eta \in C_c^\infty((a,b))$, the distribution \begin{align*} \tau \mapsto \int_M v(q,\tau)\,d\mu_\tau(q) \end{align*} has nonpositive [distributional derivative](/page/Distributional%20Derivative) on $(a,b)$. This integrated inequality is obtained by testing \begin{align*} (\partial_\tau-\Delta_{g(\tau)}+R(\cdot,\tau))v\leq 0 \end{align*} against spacetime test functions $\phi_j(q)\eta(\tau)$, where $\phi_j\in C_c^\infty(M)$ is a complete-manifold exhaustion cutoff with $0\leq \phi_j\leq 1$ and $\phi_j\uparrow 1$. The bounded-curvature hypothesis on $[a,b]$, completeness, and Perelman's Gaussian lower bound for the reduced distance give the required tail estimate for the reduced-density measure, so the Laplacian cutoff term tends to $0$ after integration over $M\times(a,b)$ with respect to $d\mu_\tau(q)\,d\mathcal{L}^1(\tau)$. This is precisely the point at which global reduced-distance tail control is used; endpoint finiteness alone is not being used to justify the spacetime cutoff error. The distributional inequality means that for every nonnegative $\eta\in C_c^\infty((a,b))$, \begin{align*} -\int_a^b \eta'(\tau)\widetilde V(\tau)\,d\mathcal{L}^1(\tau)\leq 0. \end{align*} Hence the distributional derivative of $\widetilde V$ on $(a,b)$ is a nonpositive Radon measure. Therefore $\widetilde V$ has a representative that is nonincreasing on $(a,b)$. Since the endpoint values are finite, applying this representative and taking one-sided limits gives \begin{align*} \widetilde V(b)\leq \widetilde V(a). \end{align*} Thus $\widetilde V$ is nonincreasing in $\tau$ on the set of times where the reduced-volume integral is finite.[/step]

[guided]We cannot differentiate $\widetilde V(\tau)$ as though $v$ were smooth, because the reduced distance is generally nonsmooth and the inequality for $v$ is only distributional. The correct way to extract monotonicity is to test the spacetime inequality. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$, fix $0<a<b\leq t_0$, and choose a nonnegative time test function $\eta\in C_c^\infty((a,b))$. For spatial localization, choose complete-manifold exhaustion cutoffs $\phi_j\in C_c^\infty(M)$ with $0\leq \phi_j\leq 1$ and $\phi_j\uparrow 1$ pointwise. We test \begin{align*} (\partial_\tau-\Delta_{g(\tau)}+R(\cdot,\tau))v\leq 0 \end{align*} against $\phi_j(q)\eta(\tau)$. The derivative of the evolving measure is \begin{align*} \partial_\tau d\mu_\tau=R(\cdot,\tau)d\mu_\tau, \end{align*} so the scalar-curvature term in the distributional inequality cancels exactly with the derivative of $d\mu_\tau$. After integrating by parts in the spatial variable on the compact support of $\phi_j$, the only spatial error is the Laplacian cutoff term involving $v\Delta_{g(\tau)}\phi_j$. The vanishing of this cutoff error is not a consequence of endpoint finiteness alone. It uses the global Gaussian tail estimate for the reduced-volume density coming from Perelman's reduced-distance estimates, together with completeness and bounded curvature on the compact time interval $[a,b]$. These hypotheses justify passing to the limit $j\to\infty$ in the tested inequality over $M\times(a,b)$ with respect to $d\mu_\tau(q)\,d\mathcal{L}^1(\tau)$. The limiting inequality is \begin{align*} -\int_a^b \eta'(\tau)\widetilde V(\tau)\,d\mathcal{L}^1(\tau)\leq 0 \end{align*} for every nonnegative $\eta\in C_c^\infty((a,b))$. This says exactly that the distributional derivative of $\widetilde V$ on $(a,b)$ is a nonpositive Radon measure. A locally integrable function with nonpositive distributional derivative has a nonincreasing representative. Since $\widetilde V(a)$ and $\widetilde V(b)$ are finite, taking the corresponding one-sided endpoint limits gives \begin{align*} \widetilde V(b)\leq \widetilde V(a). \end{align*} Thus the reduced volume is nonincreasing at all finite times.[/guided]

[step:Identify the small-time limit from regularity of the base point]Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$. The regularity assumption at $(p,0)$ says that the pointed parabolic rescalings converge locally smoothly to Euclidean space as $\tau\downarrow 0$. Choose rescaled coordinate maps $E_\tau: U_\tau\subset\mathbb{R}^n\to M$ based at $p$ for the pointed parabolic rescalings, defined on exhausting open sets $U_\tau$ with $E_\tau(0)=p$. In these coordinates the small-time asymptotic theorem for reduced distance at a regular base point gives, locally uniformly for $z\in\mathbb{R}^n$, \begin{align*} \ell(E_\tau(z),\tau)\to \frac{|z|^2}{4}. \end{align*} The rescaled Riemannian measures also converge locally smoothly to Lebesgue measure, so \begin{align*} \tau^{-n/2}(E_\tau)^*d\mu_\tau\to d\mathcal{L}^n \end{align*} locally smoothly as measures on $\mathbb{R}^n$. Perelman's regular-basepoint asymptotic theorem also includes a global Gaussian upper bound for the reduced-volume density outside compact rescaled balls, and this tail estimate is the domination needed for the global integral. Applying dominated convergence in the coordinates $q=E_\tau(z)$ gives \begin{align*} \lim_{\tau\downarrow 0}\widetilde V(\tau)=(4\pi)^{-n/2}\int_{\mathbb{R}^n}e^{-|z|^2/4}\,d\mathcal{L}^n(z). \end{align*} The standard $n$-dimensional Gaussian integral with covariance matrix $2I_n$ is \begin{align*} (4\pi)^{-n/2}\int_{\mathbb{R}^n}e^{-|z|^2/4}\,d\mathcal{L}^n(z)=1. \end{align*} Therefore \begin{align*} \lim_{\tau\downarrow 0}\widetilde V(\tau)=1. \end{align*}[/step]

[guided]Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$. The local smooth convergence of the pointed parabolic rescalings means that, after choosing rescaled coordinate maps $E_\tau: U_\tau\subset\mathbb{R}^n\to M$ with $E_\tau(0)=p$ and with $U_\tau$ exhausting $\mathbb{R}^n$, the pulled-back metrics converge locally smoothly to the Euclidean metric. The regular-basepoint asymptotic theorem for reduced distance then gives \begin{align*} \ell(E_\tau(z),\tau)\to \frac{|z|^2}{4} \end{align*} locally uniformly for $z\in\mathbb{R}^n$, and the pulled-back rescaled measures satisfy \begin{align*} \tau^{-n/2}(E_\tau)^*d\mu_\tau\to d\mathcal{L}^n \end{align*} locally smoothly as measures. Local convergence is not enough by itself, because the reduced volume is an integral over all of $M$. The additional input from Perelman's regular-basepoint asymptotics is a Gaussian tail bound for the reduced-volume density outside compact rescaled balls. That bound gives an integrable dominating function on $\mathbb{R}^n$ after writing $q=E_\tau(z)$. Therefore dominated convergence applies to the global reduced-volume integral and yields \begin{align*} \lim_{\tau\downarrow 0}\widetilde V(\tau)=(4\pi)^{-n/2}\int_{\mathbb{R}^n}e^{-|z|^2/4}\,d\mathcal{L}^n(z). \end{align*} Finally, the standard Gaussian integral in dimension $n$ gives \begin{align*} (4\pi)^{-n/2}\int_{\mathbb{R}^n}e^{-|z|^2/4}\,d\mathcal{L}^n(z)=1. \end{align*} Thus \begin{align*} \lim_{\tau\downarrow 0}\widetilde V(\tau)=1. \end{align*}[/guided]

[step:Combine monotonicity with the Euclidean initial limit]Let $\tau\in(0,t_0]$ be a time for which $\widetilde V(\tau)$ is finite. For every $\sigma\in(0,\tau)$ such that $\widetilde V(\sigma)$ is finite, monotonicity gives \begin{align*} \widetilde V(\tau)\leq \widetilde V(\sigma). \end{align*} Taking $\sigma\downarrow 0$ and using $\lim_{\sigma\downarrow 0}\widetilde V(\sigma)=1$ gives \begin{align*} \widetilde V(\tau)\leq 1. \end{align*} Since $v(q,\tau)>0$ for every $q\in M$ and $d\mu_\tau$ is a positive Riemannian measure on the nonempty manifold $M$, finiteness of the integral implies \begin{align*} \widetilde V(\tau)=\int_M v(q,\tau)\,d\mu_\tau(q)>0. \end{align*} Therefore \begin{align*} 0<\widetilde V(\tau)\leq 1, \end{align*} and the monotonicity and bounds asserted in the theorem follow.[/step]

[guided]Fix a time $\tau\in(0,t_0]$ for which $\widetilde V(\tau)$ is finite. For every finite time $\sigma\in(0,\tau)$, the monotonicity already proved gives \begin{align*} \widetilde V(\tau)\leq \widetilde V(\sigma). \end{align*} The small-time limit established above is \begin{align*} \lim_{\sigma\downarrow 0}\widetilde V(\sigma)=1. \end{align*} Taking $\sigma\downarrow 0$ in the monotonicity inequality gives \begin{align*} \widetilde V(\tau)\leq 1. \end{align*} For the lower bound, the density satisfies $v(q,\tau)>0$ for every $q\in M$, and $d\mu_\tau$ is a positive Riemannian measure on the nonempty manifold $M$. Since the integral is finite by hypothesis, it is a finite positive integral: \begin{align*} \widetilde V(\tau)=\int_M v(q,\tau)\,d\mu_\tau(q)>0. \end{align*} Combining the lower and upper bounds gives \begin{align*} 0<\widetilde V(\tau)\leq 1. \end{align*} This proves both the monotonicity and the asserted bounds.[/guided]