[proofplan] We prove the comparison along each radial minimizing geodesic before the cut locus. On the smooth distance region, the Hessian of $r_p$ restricted to the orthogonal complement of the radial direction is the shape operator of geodesic spheres, and its trace is $\Delta r_p$. The Jacobi field equation gives a matrix Riccati equation for this shape operator; taking traces and using Cauchy-Schwarz plus the Ricci lower bound gives a scalar Riccati inequality. The model quantity $(n-1)\operatorname{ct}_k$ satisfies the corresponding equality with the same leading singular term near $0$, so a one-dimensional comparison argument yields the desired inequality. [/proofplan]

[step:Reduce the pointwise inequality to a radial geodesic before the cut locus] Fix $x \in \Omega_p$ with $r_p(x) \in I_k$, and set \begin{align*} r_0 := r_p(x). \end{align*} Since $x \in \Omega_p$, there is a unique unit-speed minimizing geodesic \begin{align*} \gamma: [0,r_0] &\to M \\ t &\mapsto \gamma(t) \end{align*} such that $\gamma(0)=p$, $\gamma(r_0)=x$, and $\gamma'(0)=\xi$ for some $\xi \in T_pM$ with $|\xi|_g=1$. Moreover $\gamma(t) \in \Omega_p$ for every $t \in (0,r_0]$, and \begin{align*} r_p(\gamma(t)) = t, \qquad \nabla r_p(\gamma(t)) = \gamma'(t) \end{align*} for every $t \in (0,r_0]$. For $t \in (0,r_0]$, define the orthogonal hyperplane \begin{align*} E_t := \{V \in T_{\gamma(t)}M : g_{\gamma(t)}(V,\gamma'(t)) = 0\}. \end{align*} Let $\operatorname{Hess} r_p$ denote the symmetric covariant $2$-tensor on $\Omega_p$ defined by \begin{align*} (\operatorname{Hess} r_p)_y(V,W) := g_y(\nabla_V \nabla r_p,W) \end{align*} for $y\in\Omega_p$ and $V,W\in T_yM$. Define the shape operator of the geodesic sphere centered at $p$ along $\gamma$ by \begin{align*} S(t): E_t &\to E_t \\ V &\mapsto \nabla_V \nabla r_p. \end{align*} The map $S(t)$ is self-adjoint with respect to $g_{\gamma(t)}$. Define the radial Laplacian function \begin{align*} m:(0,r_0] &\to \mathbb{R} \\ t &\mapsto \operatorname{tr}_{E_t} S(t)=\Delta r_p(\gamma(t)). \end{align*} Thus it suffices to prove \begin{align*} m(r_0) \leq (n-1)\operatorname{ct}_k(r_0). \end{align*} [/step]

[step:Derive the trace Riccati inequality for the radial Laplacian]Parallel transport along $\gamma$ identifies all $E_t$ with the fixed Euclidean space $E_0 := \xi^\perp \subset T_pM$. Under this identification, view $S(t)$ as a self-adjoint endomorphism \begin{align*} S(t): E_0 &\to E_0. \end{align*} Define the radial curvature endomorphism \begin{align*} R_\gamma(t): E_0 &\to E_0 \end{align*} by the identity \begin{align*} g_{\gamma(t)}(R_\gamma(t)V,W) = g_{\gamma(t)}(R(V_t,\gamma'(t))\gamma'(t),W_t), \end{align*} where $V_t,W_t \in E_t$ are the parallel transports of $V,W \in E_0$ along $\gamma$ and $R$ is the Riemann curvature tensor with convention \begin{align*} R(X,Y)Z = \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z. \end{align*} By the cut-locus characterization recorded on the [Cut Locus](/page/Cut%20Locus) page, applied to the complete Riemannian manifold $(M,g)$, the point $p$, and the unit initial vector $\xi$, the condition $x \in \Omega_p$ implies that the segment $\gamma|_{(0,r_0]}$ contains no cut point of $p$ and no conjugate point to $p$ along $\gamma$. Equivalently, the differential of the exponential map in directions orthogonal to $\xi$ is nonsingular for every $t \in (0,r_0]$. Define the Jacobi tensor \begin{align*} J:(0,r_0] &\to \mathcal{L}(E_0,E_0) \end{align*} by identifying, through parallel transport back to $E_0$, the Jacobi field $J_V(t)$ with initial conditions $J_V(0)=0$ and $\nabla_{\gamma'}J_V(0)=V$ for each $V\in E_0$. The nonsingularity of the orthogonal differential of $\exp_p$ implies that $J(t)$ is invertible for every $t\in(0,r_0]$. The shape operator is the logarithmic derivative \begin{align*} S(t)=J'(t)J(t)^{-1}, \end{align*} where $J'(t)$ denotes covariant differentiation along $\gamma$ after the same parallel-transport identification. The Jacobi field equation for radial variations, as recorded on the [Jacobi Field](/page/Jacobi%20Field) page, is \begin{align*} J''(t)+R_\gamma(t)J(t)=0. \end{align*} Differentiating $S(t)=J'(t)J(t)^{-1}$ and using $(J^{-1})'=-J^{-1}J'J^{-1}$ gives \begin{align*} S'(t) &=J''(t)J(t)^{-1}-J'(t)J(t)^{-1}J'(t)J(t)^{-1} \\ &=-R_\gamma(t)-S(t)^2. \end{align*} Thus the matrix Riccati equation for the shape operator of geodesic spheres is \begin{align*} S'(t) + S(t)^2 + R_\gamma(t) = 0. \end{align*} Taking the trace over the $(n-1)$-dimensional space $E_0$ gives \begin{align*} m'(t) + \operatorname{tr}_{E_0}(S(t)^2) + \operatorname{Ric}_{\gamma(t)}(\gamma'(t),\gamma'(t)) = 0. \end{align*} Since $S(t)$ is self-adjoint, the [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) for its eigenvalues gives \begin{align*} \operatorname{tr}_{E_0}(S(t)^2) \geq \frac{m(t)^2}{n-1}. \end{align*} The geodesic has unit speed, so $|\gamma'(t)|_g=1$, and the Ricci lower bound gives \begin{align*} \operatorname{Ric}_{\gamma(t)}(\gamma'(t),\gamma'(t)) \geq (n-1)k. \end{align*} Combining these estimates yields the scalar inequality \begin{align*} m'(t) + \frac{m(t)^2}{n-1} + (n-1)k \leq 0 \end{align*} for every $t \in (0,r_0]$.[/step]

[guided]The geometric quantity we want to compare is $\Delta r_p$, and along the radial geodesic $\gamma$ this becomes a one-variable function \begin{align*} m(t) := \Delta r_p(\gamma(t)). \end{align*} The reason for introducing the shape operator is that $\Delta r_p$ is the trace of the Hessian of $r_p$. Here $\operatorname{Hess} r_p$ denotes the symmetric covariant $2$-tensor on $\Omega_p$ defined by \begin{align*} (\operatorname{Hess} r_p)_y(V,W) := g_y(\nabla_V \nabla r_p,W) \end{align*} for $y\in\Omega_p$ and $V,W\in T_yM$. Since $\nabla r_p=\gamma'$ along $\gamma$ and $|\nabla r_p|_g=1$, the radial component of the Hessian vanishes: \begin{align*} \operatorname{Hess} r_p(\gamma'(t),\gamma'(t)) = \frac{1}{2}\gamma'(t)\bigl(|\nabla r_p|_g^2\bigr)=0. \end{align*} Therefore the full trace of the Hessian of $r_p$ equals the trace over the orthogonal complement $E_t=\gamma'(t)^\perp$: \begin{align*} \Delta r_p(\gamma(t)) = \operatorname{tr}_{E_t}\bigl(V \mapsto \nabla_V\nabla r_p\bigr). \end{align*} To differentiate this operator in one fixed [vector space](/page/Vector%20Space), parallel transport $E_0=\xi^\perp$ along $\gamma$ and use it to identify $E_t$ with $E_0$. Under this identification the shape operator is a family of self-adjoint maps \begin{align*} S(t): E_0 &\to E_0. \end{align*} Because $x \in \Omega_p$, the segment from $p$ to $x$ stays before the cut locus. By the cut-locus characterization recorded on the [Cut Locus](/page/Cut%20Locus) page, applied to the complete Riemannian manifold $(M,g)$, the point $p$, and the unit initial vector $\xi$, for every $t \in (0,r_0]$ the radial geodesic is still minimizing and has no conjugate point to $p$ along $\gamma$; equivalently, the differential of the exponential map in directions orthogonal to $\xi$ is nonsingular. This nonsingularity is the hypothesis needed to pass from Jacobi fields to a logarithmic derivative: the Jacobi tensor along $\gamma$ is invertible on $E_0$, and the shape operator $S(t)$ is that logarithmic derivative. The Jacobi equation for radial variations says that this tensor satisfies a second-order equation involving the radial curvature endomorphism. Written in logarithmic derivative form, this is exactly the matrix Riccati equation \begin{align*} S'(t)+S(t)^2+R_\gamma(t)=0, \end{align*} where \begin{align*} g_{\gamma(t)}(R_\gamma(t)V,W) = g_{\gamma(t)}(R(V_t,\gamma'(t))\gamma'(t),W_t) \end{align*} for parallel fields $V_t,W_t$ along $\gamma$. Now take traces. The trace of $S'(t)$ is $m'(t)$, the trace of $S(t)^2$ remains $\operatorname{tr}_{E_0}(S(t)^2)$, and the trace of $R_\gamma(t)$ over the hyperplane perpendicular to $\gamma'(t)$ is the Ricci curvature in the radial direction: \begin{align*} m'(t)+\operatorname{tr}_{E_0}(S(t)^2) +\operatorname{Ric}_{\gamma(t)}(\gamma'(t),\gamma'(t))=0. \end{align*} This is the exact point where the hypothesis on Ricci curvature enters. Since $\gamma$ is unit speed, $|\gamma'(t)|_g=1$, and hence \begin{align*} \operatorname{Ric}_{\gamma(t)}(\gamma'(t),\gamma'(t))\geq (n-1)k. \end{align*} It remains to replace the full operator $S(t)$ by its trace $m(t)$. Because $S(t)$ is self-adjoint on the $(n-1)$-dimensional [inner product space](/page/Inner%20Product%20Space) $E_0$, choose an orthonormal eigenbasis with eigenvalues $\lambda_1(t),\dots,\lambda_{n-1}(t)$. Then \begin{align*} m(t)=\sum_{i=1}^{n-1}\lambda_i(t), \qquad \operatorname{tr}_{E_0}(S(t)^2)=\sum_{i=1}^{n-1}\lambda_i(t)^2. \end{align*} The [Cauchy-Schwarz Inequality](/page/Cauchy-Schwarz%20Inequality) applied to the vectors $(\lambda_1(t),\dots,\lambda_{n-1}(t))$ and $(1,\dots,1)$ gives \begin{align*} m(t)^2 = \left(\sum_{i=1}^{n-1}\lambda_i(t)\right)^2 \leq (n-1)\sum_{i=1}^{n-1}\lambda_i(t)^2. \end{align*} Thus \begin{align*} \operatorname{tr}_{E_0}(S(t)^2)\geq \frac{m(t)^2}{n-1}. \end{align*} Substituting this and the Ricci lower bound into the trace Riccati identity gives \begin{align*} m'(t)+\frac{m(t)^2}{n-1}+(n-1)k\leq 0. \end{align*}[/guided]

[step:Identify the model solution and its initial singular behaviour] Define \begin{align*} m_k: I_k &\to \mathbb{R} \\ t &\mapsto (n-1)\operatorname{ct}_k(t). \end{align*} Since $s_k$ satisfies \begin{align*} s_k''(t)+k s_k(t)=0, \qquad s_k(0)=0, \qquad s_k'(0)=1, \end{align*} and $\operatorname{ct}_k=s_k'/s_k$, differentiation gives \begin{align*} \operatorname{ct}_k'(t)+\operatorname{ct}_k(t)^2+k=0. \end{align*} Multiplying by $n-1$ gives \begin{align*} m_k'(t)+\frac{m_k(t)^2}{n-1}+(n-1)k=0. \end{align*} We use the following local meaning of big-$O$ notation: for functions defined near $0$, the assertion $f(t)=O(t^q)$ as $t\downarrow 0$ means that there exist constants $C>0$ and $\delta>0$ such that $|f(t)|\leq C t^q$ for every $t\in(0,\delta)$. We justify the singular expansion of $S(t)$ from the Jacobi tensor rather than using it as an unexplained normal-coordinate fact. For each $V\in E_0$, the Jacobi field $J_V$ satisfies $J_V(0)=0$, $\nabla_{\gamma'}J_V(0)=V$, and the Jacobi field equation recorded on the [Jacobi Field](/page/Jacobi%20Field) page. Let \begin{align*} I_{E_0}:E_0 &\to E_0 \\ V &\mapsto V \end{align*} denote the identity operator on $E_0$. Since the curvature endomorphism $R_\gamma(t)$ is smooth along $\gamma$, Taylor expansion of this linear ordinary differential equation gives, as endomorphisms of $E_0$, \begin{align*} J(t)=tI_{E_0}+O(t^3), \qquad J'(t)=I_{E_0}+O(t^2) \end{align*} as $t\downarrow 0$. Inverting the first expansion gives \begin{align*} J(t)^{-1}=\frac{1}{t}I_{E_0}+O(t). \end{align*} Using the logarithmic derivative formula $S(t)=J'(t)J(t)^{-1}$, we obtain \begin{align*} S(t)=\frac{1}{t}I_{E_0}+O(t) \end{align*} as $t \downarrow 0$. Taking the trace over the $(n-1)$-dimensional space $E_0$ gives \begin{align*} m(t)=\frac{n-1}{t}+O(t). \end{align*} Also $s_k(t)=t+O(t^3)$ and $s_k'(t)=1+O(t^2)$, so \begin{align*} m_k(t)=\frac{n-1}{t}+O(t). \end{align*} Thus, for \begin{align*} w:(0,r_0] &\to \mathbb{R} \\ t &\mapsto m(t)-m_k(t), \end{align*} we have \begin{align*} w(t)=O(t) \end{align*} as $t \downarrow 0$. [/step]

[step:Compare the scalar Riccati inequalities from the common singular initial value]Subtracting the equality for $m_k$ from the inequality for $m$ gives \begin{align*} w'(t)+a(t)w(t)\leq 0, \end{align*} where \begin{align*} a:(0,r_0] &\to \mathbb{R} \\ t &\mapsto \frac{m(t)+m_k(t)}{n-1}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. For $0<\varepsilon<t\leq r_0$, define the integrating factor \begin{align*} A_{\varepsilon,t}:=\int_\varepsilon^t a(\tau)\,d\mathcal{L}^1(\tau). \end{align*} Multiplying the differential inequality by \begin{align*} \exp\left(\int_\varepsilon^t a(\tau)\,d\mathcal{L}^1(\tau)\right) \end{align*} and integrating from $\varepsilon$ to $t$ yields \begin{align*} w(t)\leq w(\varepsilon)\exp(-A_{\varepsilon,t}). \end{align*} The asymptotics of $m$ and $m_k$ give \begin{align*} a(\tau)=\frac{2}{\tau}+O(\tau) \end{align*} as $\tau\downarrow 0$. Hence, for each fixed $t\in(0,r_0]$, \begin{align*} \exp(-A_{\varepsilon,t}) = O(\varepsilon^2) \end{align*} as $\varepsilon\downarrow 0$. Since $w(\varepsilon)=O(\varepsilon)$, we obtain \begin{align*} w(\varepsilon)\exp(-A_{\varepsilon,t})\to 0. \end{align*} Letting $\varepsilon\downarrow 0$ gives \begin{align*} w(t)\leq 0. \end{align*} Therefore \begin{align*} m(t)\leq m_k(t) \end{align*} for every $t\in(0,r_0]$.[/step]

[guided]The scalar comparison step turns the Riccati inequality into a first-order linear inequality for the difference between the actual radial Laplacian and the model radial Laplacian. Define \begin{align*} w:(0,r_0] &\to \mathbb{R} \\ t &\mapsto m(t)-m_k(t). \end{align*} The functions $m$ and $m_k$ satisfy \begin{align*} m'(t)+\frac{m(t)^2}{n-1}+(n-1)k\leq 0 \end{align*} and \begin{align*} m_k'(t)+\frac{m_k(t)^2}{n-1}+(n-1)k=0. \end{align*} Subtracting the second identity from the [first inequality](/theorems/2897) gives \begin{align*} w'(t)+\frac{m(t)^2-m_k(t)^2}{n-1}\leq 0. \end{align*} Factor the difference of squares: \begin{align*} m(t)^2-m_k(t)^2=(m(t)+m_k(t))w(t). \end{align*} Thus, with \begin{align*} a:(0,r_0] &\to \mathbb{R} \\ t &\mapsto \frac{m(t)+m_k(t)}{n-1}, \end{align*} we obtain the linear differential inequality \begin{align*} w'(t)+a(t)w(t)\leq 0. \end{align*} Fix $0<\varepsilon<t\leq r_0$. Since $a$ is continuous on the compact interval $[\varepsilon,t]$, the integrating factor is well-defined. Let \begin{align*} A_{\varepsilon,s}:=\int_\varepsilon^s a(\tau)\,d\mathcal{L}^1(\tau) \end{align*} for $s\in[\varepsilon,t]$. Multiplying by $e^{A_{\varepsilon,s}}$ gives \begin{align*} \frac{d}{ds}\left(e^{A_{\varepsilon,s}}w(s)\right) =e^{A_{\varepsilon,s}}(w'(s)+a(s)w(s))\leq 0. \end{align*} Integrating this inequality over $[\varepsilon,t]$ with respect to $\mathcal{L}^1$ yields \begin{align*} e^{A_{\varepsilon,t}}w(t)\leq w(\varepsilon), \end{align*} and therefore \begin{align*} w(t)\leq w(\varepsilon)e^{-A_{\varepsilon,t}}. \end{align*} Now we use the common singular initial behaviour at $0$, and we restate its derivation here so that the guided comparison is self-contained. Let \begin{align*} I_{E_0}:E_0 &\to E_0 \\ V &\mapsto V \end{align*} denote the identity operator on $E_0$. The Jacobi tensor expansion along $\gamma$ gives \begin{align*} J(s)=sI_{E_0}+O(s^3), \qquad J'(s)=I_{E_0}+O(s^2), \end{align*} so $S(s)=J'(s)J(s)^{-1}=s^{-1}I_{E_0}+O(s)$ and hence \begin{align*} m(s)=\frac{n-1}{s}+O(s) \end{align*} as $s\downarrow 0$. The model function satisfies $s_k(s)=s+O(s^3)$ and $s_k'(s)=1+O(s^2)$, so \begin{align*} m_k(s)=\frac{n-1}{s}+O(s). \end{align*} Therefore \begin{align*} w(\varepsilon)=O(\varepsilon) \end{align*} as $\varepsilon\downarrow 0$, and, using the definition $a=(m+m_k)/(n-1)$, \begin{align*} a(\tau)=\frac{2}{\tau}+O(\tau) \end{align*} as $\tau\downarrow 0$. Hence, for fixed $t\in(0,r_0]$, there are constants $C>0$ and $\delta\in(0,t)$ such that for $0<\varepsilon<\delta$, \begin{align*} e^{-A_{\varepsilon,t}}\leq C\varepsilon^2. \end{align*} Combining the two estimates gives \begin{align*} w(\varepsilon)e^{-A_{\varepsilon,t}}\to 0 \end{align*} as $\varepsilon\downarrow 0$. Passing to the limit in \begin{align*} w(t)\leq w(\varepsilon)e^{-A_{\varepsilon,t}} \end{align*} gives \begin{align*} w(t)\leq 0. \end{align*} Since $w(t)=m(t)-m_k(t)$, this proves \begin{align*} m(t)\leq m_k(t) \end{align*} for every $t\in(0,r_0]$.[/guided]

[step:Evaluate the comparison at the chosen point] Taking $t=r_0$ in the scalar comparison gives \begin{align*} \Delta r_p(x) = m(r_0) \leq m_k(r_0) = (n-1)\operatorname{ct}_k(r_0). \end{align*} Since $r_0=r_p(x)$, this is \begin{align*} \Delta r_p(x)\leq (n-1)\operatorname{ct}_k(r_p(x)). \end{align*} The point $x\in\Omega_p$ with $r_p(x)\in I_k$ was arbitrary, so the asserted inequality holds throughout the stated smooth region. [/step]