[proofplan] We prove the weak inequality by separating the smooth radial region from the cut locus. Away from $p$ and $\operatorname{Cut}(p)$, the distance function is smooth and the classical smooth [Laplacian comparison theorem](/theorems/5360) gives $\Delta_g r_p \leq (n-1)\operatorname{ct}_k(r_p)$ before the cut time. The cut locus is handled by the distributional structure theorem for semiconcave distance functions: $\Delta_g r_p$ is the sum of its classical absolutely continuous part on the smooth radial region and a nonpositive Radon measure supported on $\operatorname{Cut}(p)$. Testing this measure inequality against a nonnegative compactly supported function gives the desired weak inequality. [/proofplan]

[step:Fix a nonnegative test function and record the regularity of the distance function] Let $d_g:M\times M\to[0,\infty)$ denote the Riemannian distance induced by $g$, and let $d\operatorname{vol}_g$ denote the Riemannian volume measure on $M$. Define the model interval $I_k$ by $I_k=(0,\pi/\sqrt{k})$ if $k>0$ and $I_k=(0,\infty)$ if $k\leq0$, and define the model cotangent function \begin{align*} \operatorname{ct}_k:I_k&\to\mathbb{R}\\ s&\mapsto \begin{cases} \sqrt{k}\cot(\sqrt{k}s), & k>0,\\ 1/s, & k=0,\\ \sqrt{-k}\coth(\sqrt{-k}s), & k<0. \end{cases} \end{align*} Let \begin{align*} \Omega_k:=\{x\in M\setminus\{p\}: r_p(x)\in I_k\} \end{align*} denote the domain on which the model expression is defined away from $p$. Let $\varphi \in C_c^\infty(\Omega_k)$ be nonnegative, and define its compact support by \begin{align*} K := \operatorname{supp}\varphi \subset \Omega_k. \end{align*} The distance function \begin{align*} r_p: M &\to [0,\infty)\\ x &\mapsto d_g(p,x) \end{align*} is $1$-Lipschitz with respect to the Riemannian distance $d_g$. Let $W^{1,\infty}_{\mathrm{loc}}$ denote the local Sobolev space of functions whose local weak derivatives are locally essentially bounded with respect to the Riemannian volume measure $d\operatorname{vol}_g$. Since $K$ is compact and $K \subset M\setminus\{p\}$, the restriction of $r_p$ to a neighbourhood of $K$ belongs to $W^{1,\infty}_{\mathrm{loc}}$, and its weak gradient agrees almost everywhere with the metric gradient $\nabla r_p$. Moreover, \begin{align*} |\nabla r_p|_g = 1 \end{align*} for $d\operatorname{vol}_g$-almost every point of $M\setminus\{p\}$. Let $\operatorname{Cut}(p) \subset M$ denote the [cut locus](/page/Cut%20Locus) of $p$, namely the set of endpoints $x \in M\setminus\{p\}$ where a minimizing geodesic segment starting at $p$ ceases to minimize beyond $x$, equivalently the complement in $M\setminus\{p\}$ of the largest [open set](/page/Open%20Set) on which the distance function from $p$ is smooth along a unique minimizing radial geodesic. Define \begin{align*} \mathcal{R} := M\setminus\bigl(\{p\}\cup \operatorname{Cut}(p)\bigr). \end{align*} Then $\mathcal{R}$ is the smooth radial domain. On $\mathcal{R}$, the function $r_p$ is smooth and $\nabla r_p$ is the unit radial vector field. [/step]

[step:Apply the classical pointwise comparison on the smooth radial domain] Let $x\in \mathcal{R}\cap\Omega_k$. By definition of $\mathcal{R}$, there is a unique minimizing unit-speed geodesic \begin{align*} \gamma_x:[0,r_p(x)]&\to M\\ t&\mapsto \gamma_x(t) \end{align*} from $p$ to $x$, and $r_p$ is smooth in a neighbourhood of $x$. Since $(M,g)$ is complete, the radial geodesic exists up to $x$. Since $x$ lies before the cut time, the radial Jacobi fields along $\gamma_x$ have no degeneracy on $(0,r_p(x))$. Since $\operatorname{Ric}_g\geq(n-1)k\,g$ and $r_p(x)$ lies in the interval on which the model function $\operatorname{ct}_k$ is defined, the [Laplacian Comparison Theorem](/theorems/2726) applies to $r_p$ along $\gamma_x$. Therefore \begin{align*} \Delta_g r_p(x) \leq (n-1)\operatorname{ct}_k(r_p(x)) \end{align*} for every $x \in \mathcal{R}\cap \Omega_k$. [/step]

[step:Use the distributional Hessian structure of the distance function]Choose a relatively compact open set $V \subset \Omega_k$ with smooth boundary such that $K \subset V$. The function $r_p$ is locally [semiconcave](/page/Semiconcavity) on $V$ because $V\subset M\setminus\{p\}$ and the Riemannian distance from a fixed point is locally semiconcave away from that point. Hence the [distributional Laplacian](/page/Distributional%20Laplacian) $\Delta_g r_p$ on $V$ is a signed Radon measure. We apply the [distributional Hessian structure theorem for semiconcave distance functions](/page/Semiconcavity) to the locally semiconcave function $r_p|_V$. Its hypotheses are satisfied because $V$ is relatively compact, $V\subset M\setminus\{p\}$, and $r_p$ is smooth on $\mathcal{R}\cap V$ with singular set contained in $\operatorname{Cut}(p)\cap V$. In the traced form needed here, the theorem states that the distributional Laplacian decomposes into the classical absolutely continuous density on the smooth radial region, plus a jump measure on the codimension-one part of the cut locus with nonpositive density, plus any remaining singular measure; semiconcavity forces the full singular contribution to be nonpositive as a Radon measure. Therefore it gives the decomposition \begin{align*} \Delta_g r_p = (\Delta_g r_p)_{\mathrm{reg}}\,d\operatorname{vol}_g+\mu_{\operatorname{Cut}} \end{align*} on $V$, where $(\Delta_g r_p)_{\mathrm{reg}}$ agrees with the classical smooth Laplacian on $\mathcal{R}\cap V$ for $d\operatorname{vol}_g$-almost every point, and where $\mu_{\operatorname{Cut}}$ is the total singular contribution, a nonpositive Radon measure supported on $\operatorname{Cut}(p)\cap V$. Equivalently, for every nonnegative $\psi\in C_c^\infty(V)$, \begin{align*} (\Delta_g r_p)(\psi) = \int_{\mathcal{R}\cap V}\psi\,\Delta_g r_p\,d\operatorname{vol}_g+ \int_V \psi\,d\mu_{\operatorname{Cut}} \leq \int_{\mathcal{R}\cap V}\psi\,\Delta_g r_p\,d\operatorname{vol}_g. \end{align*} The distributional action is defined by \begin{align*} (\Delta_g r_p)(\psi) := -\int_V g(\nabla r_p,\nabla\psi)\,d\operatorname{vol}_g, \end{align*} which is well-defined because $r_p\in W^{1,\infty}_{\mathrm{loc}}(V)$.[/step]

[guided]In this step, the [test function](/page/Test%20Function) already fixed in the proof is a nonnegative function $\varphi\in C_c^\infty(\Omega_k)$, and its compact support is \begin{align*} K:=\operatorname{supp}\varphi\subset\Omega_k. \end{align*} We choose a relatively compact open set $V\subset\Omega_k$ with smooth boundary such that $K\subset V$. The obstacle is not [integration by parts](/theorems/2098) on a smooth domain; it is the fact that $r_p$ fails to be smooth on the [cut locus](/page/Cut%20Locus) $\operatorname{Cut}(p)$. The correct object is therefore the [distributional Laplacian](/page/Distributional%20Laplacian). Since $V$ stays away from $p$, the distance function $r_p$ is locally [semiconcave](/page/Semiconcavity) on $V$. Semiconcavity is the regularity input that replaces an ad hoc exhaustion of the cut locus: it implies that the distributional second derivatives, and hence $\Delta_g r_p$, are Radon measures on $V$. We now apply the [distributional Hessian structure theorem for semiconcave distance functions](/page/Semiconcavity). The theorem requires a locally semiconcave distance function on a relatively compact open set away from the base point, together with the smooth radial region on which the distance function is classically smooth. These requirements hold here: $V$ is relatively compact, $V\subset M\setminus\{p\}$, the function $r_p|_V$ is locally semiconcave, and $r_p$ is smooth on $\mathcal{R}\cap V$ by the definition of $\mathcal{R}$. In the traced form needed here, the theorem states that the distributional Laplacian is the sum of the classical absolutely continuous density on the smooth radial region, a jump measure on the codimension-one part of the cut locus whose density is nonpositive, and any remaining singular measure; semiconcavity makes the total singular contribution nonpositive in the Radon-measure order. Hence the theorem gives \begin{align*} \Delta_g r_p = (\Delta_g r_p)_{\mathrm{reg}}\,d\operatorname{vol}_g+\mu_{\operatorname{Cut}} \end{align*} on $V$. Here $(\Delta_g r_p)_{\mathrm{reg}}$ is the density of the absolutely continuous part with respect to $d\operatorname{vol}_g$, and it agrees with the classical smooth Laplacian of $r_p$ on $\mathcal{R}\cap V$ for $d\operatorname{vol}_g$-almost every point. The measure $\mu_{\operatorname{Cut}}$ is the total singular contribution, is supported on $\operatorname{Cut}(p)\cap V$, and is nonpositive. Why is the sign nonpositive? Let $\mathcal{H}_{g,n-1}$ denote the codimension-one [Hausdorff measure](/page/Hausdorff%20Measure) induced by the Riemannian distance $d_g$. At $\mathcal{H}_{g,n-1}$-almost every point of the cut locus where exactly two smooth sheets of the distance function meet, there are two one-sided metric gradients $\nabla r_p^+$ and $\nabla r_p^-$. If $\nu$ is the unit normal pointing from the minus side to the plus side, the jump contribution to the distributional Laplacian has density \begin{align*} g(\nabla r_p^- - \nabla r_p^+,\nu) \end{align*} with respect to $\mathcal{H}_{g,n-1}$ on that sheet. The semiconcavity of $r_p$ gives the monotonicity inequality \begin{align*} g(\nabla r_p^- - \nabla r_p^+,\nu)\leq 0, \end{align*} and the remaining singular part is also nonpositive in the Radon-measure sense. This is the precise form of the informal statement that the cut locus contributes no positive mass to $\Delta_g r_p$. Therefore, for every nonnegative test function $\psi\in C_c^\infty(V)$, \begin{align*} (\Delta_g r_p)(\psi) = \int_{\mathcal{R}\cap V}\psi\,\Delta_g r_p\,d\operatorname{vol}_g+ \int_V \psi\,d\mu_{\operatorname{Cut}} \leq \int_{\mathcal{R}\cap V}\psi\,\Delta_g r_p\,d\operatorname{vol}_g. \end{align*} The distributional action on the left is \begin{align*} (\Delta_g r_p)(\psi) := -\int_V g(\nabla r_p,\nabla\psi)\,d\operatorname{vol}_g, \end{align*} which is legitimate because $r_p\in W^{1,\infty}_{\mathrm{loc}}(V)$ and $\psi$ is smooth with compact support.[/guided]

[step:Test the measure inequality and conclude] Apply the distributional decomposition from the previous step with $\psi=\varphi$. Since $\varphi\geq0$, the nonpositive measure $\mu_{\operatorname{Cut}}$ gives \begin{align*} -\int_V g(\nabla r_p,\nabla\varphi)\,d\operatorname{vol}_g =(\Delta_g r_p)(\varphi) \leq \int_{\mathcal{R}\cap V}\varphi\,\Delta_g r_p\,d\operatorname{vol}_g. \end{align*} Using the pointwise smooth comparison on $\mathcal{R}\cap V$ and the nonnegativity of $\varphi$ gives \begin{align*} \int_{\mathcal{R}\cap V}\varphi\,\Delta_g r_p\,d\operatorname{vol}_g \leq \int_{\mathcal{R}\cap V}(n-1)\operatorname{ct}_k(r_p)\varphi\,d\operatorname{vol}_g. \end{align*} The cut locus has $d\operatorname{vol}_g$-measure zero. Also, because $K=\operatorname{supp}\varphi$ is compactly contained in $\Omega_k$, the function $r_p$ is bounded away from $0$ on $K$. When $k>0$, it is also bounded away from the first positive pole \begin{align*} \frac{\pi}{\sqrt{k}} \end{align*} of $\operatorname{ct}_k$. Hence $\operatorname{ct}_k(r_p)$ is bounded on $K$, so the final integral is finite and changing the domain from $\mathcal{R}\cap V$ to $\Omega_k$ does not change its value. Therefore \begin{align*} -\int_{\Omega_k} g(\nabla r_p,\nabla\varphi)\,d\operatorname{vol}_g \leq \int_{\Omega_k}(n-1)\operatorname{ct}_k(r_p)\varphi\,d\operatorname{vol}_g. \end{align*} Since this holds for every nonnegative $\varphi\in C_c^\infty(\Omega_k)$, the distributional inequality \begin{align*} \Delta_g r_p \leq (n-1)\operatorname{ct}_k(r_p) \end{align*} holds on $\Omega_k$. [/step]