[proofplan] We rewrite the conjugate [heat equation](/page/Heat%20Equation) in terms of the potential $f$ and use the evolving probability measure with density $u$. The core computation differentiates the weighted scalar-curvature-plus-gradient-energy term and shows that its variation is the weighted square of the Ricci tensor plus the Hessian of $f$. We then differentiate Perelman's W-functional, use $\partial_t\tau=-1$, and complete the square to obtain the stated non-negative expression. [/proofplan]

[step:Derive the evolution equation for the potential $f$]Fix $t\in[0,T]$. In this proof all geometric quantities are computed with respect to $g(t)$ unless the time dependence is displayed. Let $S_g:M\to\mathbb{R}$ denote the scalar curvature of $g(t)$, let $\operatorname{Ric}_g$ denote the Ricci curvature, let $\nabla$ denote the Levi-Civita connection of $g(t)$, let $\operatorname{Hess}_{g} f$ denote the Hessian of $f(\cdot,t)$, let $\Delta_g=\operatorname{div}_g\nabla$ denote the Laplace-Beltrami operator, let $\operatorname{tr}_g$ denote trace with respect to the metric $g(t)$, and let $d\mu_g$ denote the Riemannian volume measure. If $\alpha$ is a one-form, write $\alpha^\sharp$ for the vector field obtained by raising its index with $g(t)$. The flow equation $\partial_tg=-2\operatorname{Ric}_g$ is the Ricci flow, and the equation $\partial_tu=-\Delta_gu+S_gu$ is the conjugate heat equation. Since the theorem assumes a smooth Ricci flow on the closed manifold $M$ and a smooth positive conjugate heat density $u$, the standard evolution formulas for scalar curvature, volume measure, and gradient norms under Ricci flow apply at every interior time, with one-sided interpretation at the endpoints. Since \begin{align*} u=(4\pi\tau)^{-n/2}e^{-f}, \end{align*} taking the time derivative of $\log u$ gives \begin{align*} \partial_t\log u = \frac{n}{2\tau}-\partial_t f. \end{align*} On the other hand, the conjugate heat equation and the identity \begin{align*} \frac{\Delta_g u}{u} = -\Delta_g f+|\nabla f|_g^2 \end{align*} give \begin{align*} \partial_t\log u = \frac{\partial_tu}{u} = -\frac{\Delta_gu}{u}+S_g = \Delta_g f-|\nabla f|_g^2+S_g. \end{align*} Equating these two formulas yields \begin{align*} \partial_t f = -\Delta_g f+|\nabla f|_g^2-S_g+\frac{n}{2\tau}. \end{align*} The volume-form evolution under Ricci flow gives \begin{align*} \partial_t(d\mu_g)=-S_g\,d\mu_g. \end{align*} Let $\mathcal B(M)$ denote the Borel $\sigma$-algebra of the closed manifold $M$. Therefore the weighted measure $\nu_t:\mathcal B(M)\to[0,1]$ defined by \begin{align*} \nu_t(A)=\int_A u(\cdot,t)\,d\mu_{g(t)} \end{align*} for each $A\in\mathcal B(M)$ satisfies, as a density identity, \begin{align*} \partial_t(u\,d\mu_g) = (\partial_tu-S_gu)\,d\mu_g = -\Delta_gu\,d\mu_g. \end{align*}[/step]

[guided]We first translate the equation for $u$ into an equation for $f$, because $\mathcal W$ is written using $f$ rather than $u$. The relation between them is \begin{align*} u=(4\pi\tau)^{-n/2}e^{-f}. \end{align*} Taking logarithms gives \begin{align*} \log u=-\frac n2\log(4\pi\tau)-f. \end{align*} Since $\partial_t\tau=-1$, differentiating in time gives \begin{align*} \partial_t\log u = -\frac n2\frac{\partial_t\tau}{\tau}-\partial_t f = \frac{n}{2\tau}-\partial_t f. \end{align*} Now use the conjugate heat equation. Because $u>0$, we may divide by $u$: \begin{align*} \partial_t\log u = \frac{\partial_tu}{u} = -\frac{\Delta_gu}{u}+S_g. \end{align*} The spatial identity for the displayed definition of $u$ is \begin{align*} \nabla u=-u\nabla f, \qquad \Delta_gu = -u\Delta_g f+u|\nabla f|_g^2. \end{align*} Thus \begin{align*} -\frac{\Delta_gu}{u}+S_g = \Delta_g f-|\nabla f|_g^2+S_g. \end{align*} Comparing the two formulas for $\partial_t\log u$ gives \begin{align*} \partial_t f = -\Delta_g f+|\nabla f|_g^2-S_g+\frac{n}{2\tau}. \end{align*} We also record how the weighted measure evolves. Under Ricci flow, \begin{align*} \partial_t(d\mu_g)=-S_g\,d\mu_g. \end{align*} Combining this with $\partial_tu=-\Delta_gu+S_gu$ gives \begin{align*} \partial_t(u\,d\mu_g) = (\partial_tu)\,d\mu_g+u\,\partial_t(d\mu_g) = (-\Delta_gu+S_gu)\,d\mu_g-S_gu\,d\mu_g = -\Delta_gu\,d\mu_g. \end{align*} This identity is the reason closedness of $M$ will remove boundary terms: every time derivative of an integral against $u\,d\mu_g$ produces a Laplacian term, and on a closed manifold Laplacians can be integrated by parts without boundary contributions.[/guided]

[step:Differentiate Perelman's $\mathcal F$-integrand against the conjugate heat measure]Define $Q:M\times[0,T]\to\mathbb{R}$ by \begin{align*} Q(x,t)=S_{g(t)}(x)+|\nabla f(x,t)|_{g(t)}^2. \end{align*} Define the functional $\mathcal F:[0,T]\to\mathbb{R}$ by \begin{align*} \mathcal F(t)=\int_M Q(\cdot,t)\,u(\cdot,t)\,d\mu_{g(t)}. \end{align*} Because $g$, $f$, and $u$ are smooth in space and time on the compact manifold $M$, the scalar curvature, Ricci curvature, Hessian, Laplacian, and all time derivatives used below are smooth and bounded on each compact subinterval of $[0,T]$. This regularity justifies differentiating this integral under the integral sign and applying the standard Ricci-flow evolution formulas pointwise at every interior time. We claim that \begin{align*} \frac{d}{dt}\mathcal F(t) = 2\int_M |\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2 u\,d\mu_g. \end{align*} Indeed, since $M$ is closed, [integration by parts](/theorems/2098) has no boundary term. Using \begin{align*} \partial_t(u\,d\mu_g)=-\Delta_gu\,d\mu_g \end{align*} and the self-adjointness of $\Delta_g$ on $M$, we obtain \begin{align*} \frac{d}{dt}\mathcal F(t)=\int_M \partial_t Q\,u\,d\mu_g-\int_M Q\,\Delta_g u\,d\mu_g. \end{align*} By self-adjointness of $\Delta_g$ on the closed manifold $M$, this becomes \begin{align*} \frac{d}{dt}\mathcal F(t)=\int_M \left(\partial_t Q-\Delta_g Q\right)u\,d\mu_g. \end{align*} The scalar-curvature evolution under Ricci flow applies because $g(t)$ is a smooth Ricci flow on the closed manifold $M$, and gives \begin{align*} \partial_tS_g=\Delta_gS_g+2|\operatorname{Ric}_g|_g^2. \end{align*} The metric variation $\partial_tg=-2\operatorname{Ric}_g$ gives \begin{align*} \partial_t|\nabla f|_g^2=2\operatorname{Ric}_g(\nabla f,\nabla f)+2\langle\nabla f,\nabla\partial_t f\rangle_g. \end{align*} The Bochner identity gives \begin{align*} \Delta_g|\nabla f|_g^2=2\langle\nabla f,\nabla\Delta_g f\rangle_g+2|\operatorname{Hess}_{g} f|_g^2+2\operatorname{Ric}_g(\nabla f,\nabla f). \end{align*} Combining these identities gives \begin{align*} \partial_tQ-\Delta_gQ = 2|\operatorname{Ric}_g|_g^2 -2|\operatorname{Hess}_{g} f|_g^2 + 2\left\langle \nabla f, \nabla(\partial_t f-\Delta_g f) \right\rangle_g. \end{align*} Substituting \begin{align*} \partial_t f-\Delta_g f=-2\Delta_g f+|\nabla f|_g^2-S_g+\frac{n}{2\tau} \end{align*} reduces the remaining term to a weighted integration-by-parts calculation. Here $\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g$ denotes the metric trace [inner product](/page/Inner%20Product) of symmetric $2$-tensors. [claim:Evaluate the weighted gradient term] One has \begin{align*} \int_M\left\langle\nabla f,\nabla(\partial_t f-\Delta_g f)\right\rangle_gu\,d\mu_g=2\int_M\left(|\operatorname{Hess}_{g} f|_g^2+\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g\right)u\,d\mu_g. \end{align*} [/claim] [proof] Since $u$ has the displayed form, its spatial gradient is \begin{align*} \nabla u=-u\nabla f. \end{align*} The term \begin{align*} \frac{n}{2\tau} \end{align*} is spatially constant, so it disappears after applying $\nabla$. Therefore the left-hand side equals \begin{align*} -2\int_M\langle\nabla f,\nabla\Delta_g f\rangle_gu\,d\mu_g+2\int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g-\int_M\langle\nabla f,\nabla S_g\rangle_gu\,d\mu_g. \end{align*} For the first term, integrate the Bochner identity against $u$. Since $M$ is closed, the self-adjointness of the Laplacian gives \begin{align*} \int_M\frac12\Delta_g|\nabla f|_g^2u\,d\mu_g=\int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g. \end{align*} Thus Bochner's identity implies \begin{align*} \int_M\langle\nabla f,\nabla\Delta_g f\rangle_gu\,d\mu_g=\int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g-\int_M|\operatorname{Hess}_{g} f|_g^2u\,d\mu_g-\int_M\operatorname{Ric}_g(\nabla f,\nabla f)u\,d\mu_g. \end{align*} For the scalar-curvature term, the contracted Bianchi identity gives $\operatorname{div}_g\operatorname{Ric}_g=\frac12\nabla S_g$. Applying the [divergence theorem](/theorems/2754) on the closed manifold $M$ to the vector field $u\operatorname{Ric}_g(\nabla f,\cdot)^{\sharp}$ gives \begin{align*} \int_M\langle\nabla S_g,\nabla f\rangle_gu\,d\mu_g=-2\int_M\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_gu\,d\mu_g+2\int_M\operatorname{Ric}_g(\nabla f,\nabla f)u\,d\mu_g. \end{align*} Substituting these two identities into the expanded left-hand side cancels the two terms containing $\operatorname{Hess}_{g} f(\nabla f,\nabla f)$ and also cancels the two terms containing $\operatorname{Ric}_g(\nabla f,\nabla f)$. The result is exactly \begin{align*} 2\int_M\left(|\operatorname{Hess}_{g} f|_g^2+\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g\right)u\,d\mu_g. \end{align*} [/proof] Consequently \begin{align*} \frac{d}{dt}\mathcal F(t)=2\int_M|\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2u\,d\mu_g. \end{align*}[/step]

[guided]The goal of this step is to isolate the exact square that later appears in Perelman's monotonicity formula. For each $t\in[0,T]$, let $S_g:M\to\mathbb{R}$ denote the scalar curvature of $g(t)$, let $\operatorname{Ric}_g$ denote the Ricci curvature of $g(t)$, let $\nabla$ denote the Levi-Civita connection of $g(t)$, let $\operatorname{Hess}_{g} f$ denote the Hessian of $f(\cdot,t)$, let $\Delta_g=\operatorname{div}_g\nabla$ denote the Laplace-Beltrami operator, and let $d\mu_g$ denote the Riemannian volume measure. Define $Q:M\times[0,T]\to\mathbb{R}$ by \begin{align*} Q(x,t)=S_{g(t)}(x)+|\nabla f(x,t)|_{g(t)}^2. \end{align*} Define the functional $\mathcal F:[0,T]\to\mathbb{R}$ by \begin{align*} \mathcal F(t)=\int_MQ(\cdot,t)u(\cdot,t)\,d\mu_{g(t)}. \end{align*} Because $g$, $f$, and $u$ are smooth in space and time on compact $M$, all curvature quantities and time derivatives below are smooth and bounded. This justifies differentiating under the integral sign. Since $M$ is closed, the Laplace-Beltrami operator is self-adjoint with no boundary contribution. Using $\partial_t(u\,d\mu_g)=-\Delta_gu\,d\mu_g$, we get \begin{align*} \frac{d}{dt}\mathcal F(t)=\int_M(\partial_tQ-\Delta_gQ)u\,d\mu_g. \end{align*} The scalar-curvature evolution under Ricci flow gives \begin{align*} \partial_tS_g=\Delta_gS_g+2|\operatorname{Ric}_g|_g^2. \end{align*} The metric variation $\partial_tg=-2\operatorname{Ric}_g$ gives \begin{align*} \partial_t|\nabla f|_g^2=2\operatorname{Ric}_g(\nabla f,\nabla f)+2\langle\nabla f,\nabla\partial_tf\rangle_g. \end{align*} Bochner's identity gives \begin{align*} \Delta_g|\nabla f|_g^2=2\langle\nabla f,\nabla\Delta_gf\rangle_g+2|\operatorname{Hess}_{g} f|_g^2+2\operatorname{Ric}_g(\nabla f,\nabla f). \end{align*} Combining these three formulas cancels the two Ricci-gradient terms and yields \begin{align*} \partial_tQ-\Delta_gQ=2|\operatorname{Ric}_g|_g^2-2|\operatorname{Hess}_{g} f|_g^2+2\langle\nabla f,\nabla(\partial_tf-\Delta_gf)\rangle_g. \end{align*} Now substitute the equation for $f$, \begin{align*} \partial_tf-\Delta_gf=-2\Delta_gf+|\nabla f|_g^2-S_g+\frac{n}{2\tau}. \end{align*} The final summand in the displayed equation is spatially constant, so its gradient is zero. Therefore \begin{align*} \int_M\langle\nabla f,\nabla(\partial_tf-\Delta_gf)\rangle_gu\,d\mu_g =-2\int_M\langle\nabla f,\nabla\Delta_gf\rangle_gu\,d\mu_g +2\int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g -\int_M\langle\nabla f,\nabla S_g\rangle_gu\,d\mu_g. \end{align*} We next compute the first and third terms. Since $u$ has the displayed form, its spatial gradient is \begin{align*} \nabla u=-u\nabla f. \end{align*} Integrating Bochner's identity against $u$ and using self-adjointness of $\Delta_g$ on closed $M$ gives \begin{align*} \int_M\frac12\Delta_g|\nabla f|_g^2u\,d\mu_g =-\int_M\frac12\langle\nabla|\nabla f|_g^2,\nabla u\rangle_g\,d\mu_g. \end{align*} Since $\nabla u=-u\nabla f$, this equals \begin{align*} \int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g. \end{align*} Thus Bochner's identity implies \begin{align*} \int_M\langle\nabla f,\nabla\Delta_gf\rangle_gu\,d\mu_g =\int_M\operatorname{Hess}_{g} f(\nabla f,\nabla f)u\,d\mu_g -\int_M|\operatorname{Hess}_{g} f|_g^2u\,d\mu_g -\int_M\operatorname{Ric}_g(\nabla f,\nabla f)u\,d\mu_g. \end{align*} For the scalar-curvature term, use the contracted Bianchi identity \begin{align*} \operatorname{div}_g\operatorname{Ric}_g=\frac12\nabla S_g. \end{align*} Applying the [divergence theorem](/theorems/3614) on the closed manifold $M$ to the vector field $u\operatorname{Ric}_g(\nabla f,\cdot)^{\sharp}$ gives \begin{align*} 0 =\int_M\operatorname{div}_g\left(u\operatorname{Ric}_g(\nabla f,\cdot)^{\sharp}\right)\,d\mu_g. \end{align*} Expanding the divergence gives \begin{align*} 0 =\int_M\langle\nabla u,\operatorname{Ric}_g(\nabla f,\cdot)^{\sharp}\rangle_g\,d\mu_g +\int_Mu\langle\operatorname{div}_g\operatorname{Ric}_g,\nabla f\rangle_g\,d\mu_g +\int_Mu\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g\,d\mu_g. \end{align*} Using $\nabla u=-u\nabla f$ and the contracted Bianchi identity, this becomes \begin{align*} 0 =-\int_M\operatorname{Ric}_g(\nabla f,\nabla f)u\,d\mu_g +\frac12\int_M\langle\nabla S_g,\nabla f\rangle_gu\,d\mu_g +\int_M\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_gu\,d\mu_g. \end{align*} Solving this identity for the scalar-curvature integral yields \begin{align*} \int_M\langle\nabla S_g,\nabla f\rangle_gu\,d\mu_g =2\int_M\operatorname{Ric}_g(\nabla f,\nabla f)u\,d\mu_g -2\int_M\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_gu\,d\mu_g. \end{align*} Substituting the Bochner and Bianchi evaluations into the expanded weighted gradient term, the terms involving $\operatorname{Hess}_{g} f(\nabla f,\nabla f)$ cancel and the terms involving $\operatorname{Ric}_g(\nabla f,\nabla f)$ cancel. The remaining terms are \begin{align*} \int_M\langle\nabla f,\nabla(\partial_tf-\Delta_gf)\rangle_gu\,d\mu_g =2\int_M\left(|\operatorname{Hess}_{g} f|_g^2+\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g\right)u\,d\mu_g. \end{align*} Substituting this identity into the derivative of $\mathcal F$ gives \begin{align*} \frac{d}{dt}\mathcal F(t) =2\int_M|\operatorname{Ric}_g|_g^2u\,d\mu_g -2\int_M|\operatorname{Hess}_{g} f|_g^2u\,d\mu_g +4\int_M\left(|\operatorname{Hess}_{g} f|_g^2+\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g\right)u\,d\mu_g. \end{align*} Collecting the three terms gives \begin{align*} \frac{d}{dt}\mathcal F(t) =2\int_M\left(|\operatorname{Ric}_g|_g^2+2\langle\operatorname{Ric}_g,\operatorname{Hess}_{g} f\rangle_g+|\operatorname{Hess}_{g} f|_g^2\right)u\,d\mu_g. \end{align*} The last integrand is the metric square of the symmetric $2$-tensor $\operatorname{Ric}_g+\operatorname{Hess}_{g} f$, so \begin{align*} \frac{d}{dt}\mathcal F(t)=2\int_M|\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2u\,d\mu_g. \end{align*}[/guided]

[step:Compute the derivative of the entropy term]Define the entropy functional $E:[0,T]\to\mathbb{R}$ by \begin{align*} E(t)=\int_M f(\cdot,t)\,u(\cdot,t)\,d\mu_{g(t)}. \end{align*} The smoothness of $f$, $u$, and $g$ on compact $M$ justifies differentiating this integral under the integral sign. Using the evolution of $u\,d\mu_g$ and [integration by parts](/theorems/210) on the closed manifold $M$, \begin{align*} \frac{d}{dt}E(t)=\int_M \partial_t f\,u\,d\mu_g-\int_M f\,\Delta_g u\,d\mu_g. \end{align*} By self-adjointness of $\Delta_g$ on the closed manifold $M$, this is \begin{align*} \frac{d}{dt}E(t)=\int_M \partial_t f\,u\,d\mu_g-\int_M u\,\Delta_g f\,d\mu_g. \end{align*} Substituting the equation for $\partial_t f$ gives \begin{align*} \frac{d}{dt}E(t) &= \int_M \left( -2\Delta_g f+|\nabla f|_g^2-S_g+\frac{n}{2\tau} \right) u\,d\mu_g. \end{align*} Since $\nabla u=-u\nabla f$, integration by parts gives \begin{align*} \int_M \Delta_g f\,u\,d\mu_g = -\int_M \langle\nabla f,\nabla u\rangle_g\,d\mu_g = \int_M |\nabla f|_g^2u\,d\mu_g. \end{align*} Therefore, using the normalization $\int_Mu\,d\mu_g=1$, \begin{align*} \frac{d}{dt}E(t)=-\int_M\left(S_g+|\nabla f|_g^2\right)u\,d\mu_g+\frac{n}{2\tau}. \end{align*}[/step]

[guided]Define the entropy functional $E:[0,T]\to\mathbb{R}$ by $E(t)=\int_Mf(\cdot,t)u(\cdot,t)\,d\mu_{g(t)}$. The smoothness of $f$, $u$, and $g$ on compact $M$ justifies differentiating under the integral sign. Differentiating an integral against the evolving measure $u\,d\mu_g$ gives two contributions: the explicit derivative of $f$ and the derivative of the measure. Since $\partial_t(u\,d\mu_g)=-\Delta_gu\,d\mu_g$, we obtain \begin{align*} \frac{d}{dt}E(t)=\int_M\partial_tfu\,d\mu_g-\int_Mf\Delta_gu\,d\mu_g. \end{align*} Self-adjointness of $\Delta_g$ on the closed manifold $M$ changes the second integral to $-\int_Mu\Delta_gf\,d\mu_g$. Substituting the displayed formula for $\partial_tf$ gives \begin{align*} \frac{d}{dt}E(t)=\int_M\left(-2\Delta_gf+|\nabla f|_g^2-S_g+\frac{n}{2\tau}\right)u\,d\mu_g. \end{align*} Finally, $\nabla u=-u\nabla f$ and the absence of boundary terms imply \begin{align*} \int_M\Delta_gfu\,d\mu_g=\int_M|\nabla f|_g^2u\,d\mu_g. \end{align*} Using the mass normalization $\int_Mu\,d\mu_g=1$, we conclude \begin{align*} \frac{d}{dt}E(t)=-\int_M\left(S_g+|\nabla f|_g^2\right)u\,d\mu_g+\frac{n}{2\tau}. \end{align*}[/guided]

[step:Differentiate $\mathcal W$ and complete the square]By definition, \begin{align*} \mathcal W[g(t),f(t),\tau(t)] = \tau(t)\mathcal F(t)+E(t)-n. \end{align*} Hence, since $\partial_t\tau=-1$, \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=-\mathcal F(t)+\tau(t)\frac{d}{dt}\mathcal F(t)+\frac{d}{dt}E(t). \end{align*} Substituting the formulas for $d\mathcal F/dt$ and $dE/dt$ gives \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=2\tau\int_M|\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2u\,d\mu_g-2\int_M(S_g+|\nabla f|_g^2)u\,d\mu_g+\frac{n}{2\tau}. \end{align*} Because \begin{align*} \int_M \Delta_g f\,u\,d\mu_g = \int_M|\nabla f|_g^2u\,d\mu_g, \end{align*} we may rewrite the middle term as \begin{align*} -2\int_M(S_g+|\nabla f|_g^2)u\,d\mu_g = -2\int_M \operatorname{tr}_g(\operatorname{Ric}_g+\operatorname{Hess}_{g} f) u\,d\mu_g. \end{align*} Also, \begin{align*} \left|\frac{1}{2\tau}g\right|_g^2 = \frac{n}{4\tau^2}, \end{align*} and the normalization of $u\,d\mu_g$ gives \begin{align*} 2\tau\int_M\left|\frac{1}{2\tau}g\right|_g^2u\,d\mu_g = \frac{n}{2\tau}. \end{align*} Therefore \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=2\tau\int_M\left|\operatorname{Ric}_g+\operatorname{Hess}_{g} f-\frac{1}{2\tau}g\right|_g^2u\,d\mu_g. \end{align*} The integrand is pointwise non-negative and $\tau>0$, so \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]\geq 0. \end{align*} This is the claimed monotonicity formula.[/step]

[guided]Perelman's W-functional is written here as \begin{align*} \mathcal W[g(t),f(t),\tau(t)]=\tau(t)\mathcal F(t)+E(t)-n. \end{align*} Differentiating and using $\partial_t\tau=-1$ gives \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=-\mathcal F(t)+\tau(t)\frac{d}{dt}\mathcal F(t)+\frac{d}{dt}E(t). \end{align*} The previous two steps give \begin{align*} \frac{d}{dt}\mathcal F(t)=2\int_M|\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2u\,d\mu_g \end{align*} and \begin{align*} \frac{d}{dt}E(t)=-\mathcal F(t)+\frac{n}{2\tau}. \end{align*} Therefore \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=2\tau\int_M|\operatorname{Ric}_g+\operatorname{Hess}_{g} f|_g^2u\,d\mu_g-2\int_M(S_g+|\nabla f|_g^2)u\,d\mu_g+\frac{n}{2\tau}. \end{align*} The identity $\int_M\Delta_gfu\,d\mu_g=\int_M|\nabla f|_g^2u\,d\mu_g$ rewrites the middle term as the trace of the symmetric $2$-tensor $\operatorname{Ric}_g+\operatorname{Hess}_{g} f$: \begin{align*} \int_M(S_g+|\nabla f|_g^2)u\,d\mu_g=\int_M\operatorname{tr}_g(\operatorname{Ric}_g+\operatorname{Hess}_{g} f)u\,d\mu_g. \end{align*} Since $|g|_g^2=n$ and $\int_Mu\,d\mu_g=1$, the remaining scalar term satisfies \begin{align*} \frac{n}{2\tau}=2\tau\int_M\left|\frac{1}{2\tau}g\right|_g^2u\,d\mu_g. \end{align*} Completing the square for symmetric $2$-tensors gives \begin{align*} \frac{d}{dt}\mathcal W[g(t),f(t),\tau(t)]=2\tau\int_M\left|\operatorname{Ric}_g+\operatorname{Hess}_{g} f-\frac{1}{2\tau}g\right|_g^2u\,d\mu_g. \end{align*} The squared norm is non-negative at every point of $M$, and $\tau>0$, hence $d\mathcal W/dt\geq0$. This proves Perelman's W entropy monotonicity.[/guided]