[proofplan] The proof compares the second variation of radial geodesics in $M$ with the corresponding one-dimensional model Jacobi field in the space form of curvature $k$. Since $x$ is in the regular domain $\Omega_p$, the distance function is smooth at $x$, and the radial Jacobi field with endpoint value $X$ computes $\operatorname{Hess} r_p(X,X)$ through the index form. The hypothesis $\operatorname{sn}_k(s)>0$ on $(0,t]$ supplies the no-conjugate condition for the model. Under $K \ge k$, the [index lemma](/theorems/5349) and the explicit model variation give the upper bound; under $K \le k$, Rauch comparison for radial Jacobi tensors gives the lower bound for the radial shape operator. [/proofplan] [step:Reduce the Hessian to the index form of the radial Jacobi field] Fix $X \in T_xM$ with $g(X,\nabla r_p(x))=0$. If $X=0$, both desired inequalities read $0 \le 0$ or $0 \ge 0$, so assume $X \ne 0$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,t]$. The completeness of $(M,g)$ ensures, by the [Hopf-Rinow geodesic existence theorem](/page/Hopf-Rinow%20geodesic%20existence%20theorem), that the global distance function $r_p:M\to[0,\infty)$ is realized by minimizing geodesics from $p$. In this proof, the regular domain $\Omega_p$ denotes the complement of $p$ and its cut locus. Thus $x\in\Omega_p$ means that $x$ is reached by a unique minimizing radial geodesic and that no conjugate point to $p$ occurs before or at $x$. Define this unit-speed minimizing geodesic by \begin{align*} \gamma: [0,t] &\to M \\ s &\mapsto \gamma(s), \end{align*} with $\gamma(0)=p$, $\gamma(t)=x$, and $|\dot\gamma(s)|=1$ for every $s\in[0,t]$. We use the standard model functions $\operatorname{sn}_k:[0,\infty)\to\mathbb{R}$ and $\operatorname{ct}_k:(0,\infty)\to\mathbb{R}$ defined by \begin{align*} \operatorname{sn}_k''+k\operatorname{sn}_k=0,\qquad \operatorname{sn}_k(0)=0,\qquad \operatorname{sn}_k'(0)=1, \end{align*} and, where $\operatorname{sn}_k(s)>0$, by \begin{align*} \operatorname{ct}_k(s):=\frac{\operatorname{sn}_k'(s)}{\operatorname{sn}_k(s)}. \end{align*} Hence the radial Jacobi boundary problem has a unique solution \begin{align*} J: [0,t] &\to TM \end{align*} along $\gamma$ satisfying \begin{align*} J(0)=0, \qquad J(t)=X, \qquad J(s) \perp \dot{\gamma}(s) \quad \text{for all } s \in [0,t]. \end{align*} Let $R$ denote the Riemann curvature tensor of the Riemannian manifold $(M,g)$, using the convention \begin{align*} R(U_1,U_2)U_3 := \nabla_{U_1}\nabla_{U_2}U_3 - \nabla_{U_2}\nabla_{U_1}U_3 - \nabla_{[U_1,U_2]}U_3 \end{align*} for smooth vector fields $U_1,U_2,U_3 \in \mathfrak{X}(M)$. All comparison results cited below are used with this same curvature convention; in particular, for orthonormal vectors $U,V\in T_qM$, the sectional curvature of $\operatorname{span}\{U,V\}$ is $g(R(V,U)U,V)$. Define the index form along $\gamma$ on piecewise smooth vector fields $V,W$ along $\gamma$ with $V(s),W(s)\perp \dot{\gamma}(s)$ by \begin{align*} I(V,W) := \int_0^t \left( g(\nabla_{\dot{\gamma}}V,\nabla_{\dot{\gamma}}W) - g(R(V,\dot{\gamma})\dot{\gamma},W) \right) \, d\mathcal{L}^1(s). \end{align*} By the [Hessian-index identity for the distance function](/page/Hessian-index%20identity%20for%20the%20distance%20function), applied on the cut-locus-free interval $[0,t]$, \begin{align*} \operatorname{Hess} r_p(X,X)=I(J,J). \end{align*} [/step] [step:Compare with the model endpoint field under the lower curvature bound] Assume first that every radial sectional curvature along $\gamma$ is at least $k$. Let \begin{align*} E: [0,t] &\to TM \end{align*} be the parallel vector field along $\gamma$ satisfying \begin{align*} E(t)=\frac{X}{|X|}. \end{align*} Then $E(s)\perp \dot{\gamma}(s)$ and $|E(s)|=1$ for every $s \in [0,t]$, because parallel transport preserves inner products and $E(t)\perp \dot{\gamma}(t)=\nabla r_p(x)$. Define the scalar comparison function \begin{align*} a: [0,t] &\to \mathbb{R} \\ s &\mapsto \frac{\operatorname{sn}_k(s)}{\operatorname{sn}_k(t)} \end{align*} and define the model endpoint field \begin{align*} V: [0,t] &\to TM \\ s &\mapsto |X|\,a(s)\,E(s). \end{align*} Then $V(0)=0$ and $V(t)=X$. By the [Index Lemma for Jacobi Fields](/page/Index%20Lemma%20for%20Jacobi%20Fields), applied on an interval with no conjugate point to $p$ before or at $t$, \begin{align*} I(J,J) \le I(V,V). \end{align*} Since $\nabla_{\dot{\gamma}}E=0$, we have \begin{align*} \nabla_{\dot{\gamma}}V(s)=|X|\,a'(s)\,E(s). \end{align*} The radial curvature lower bound gives \begin{align*} g(R(E(s),\dot{\gamma}(s))\dot{\gamma}(s),E(s)) = K(\operatorname{span}\{\dot{\gamma}(s),E(s)\}) \ge k. \end{align*} Therefore \begin{align*} I(V,V) &= |X|^2 \int_0^t \left( (a'(s))^2 - a(s)^2 g(R(E(s),\dot{\gamma}(s))\dot{\gamma}(s),E(s)) \right) \, d\mathcal{L}^1(s) \\ &\le |X|^2 \int_0^t \left( (a'(s))^2-k a(s)^2 \right) \, d\mathcal{L}^1(s). \end{align*} [guided] The comparison field should have the same endpoints as the true Jacobi field, because the index lemma compares fields with fixed endpoint values. Since $X$ is orthogonal to the radial direction at $x$, we parallel transport the unit vector $X/|X|$ backward along $\gamma$. This gives a vector field \begin{align*} E: [0,t] &\to TM \end{align*} with $E(t)=X/|X|$, $\nabla_{\dot{\gamma}}E=0$, $|E(s)|=1$, and $E(s)\perp \dot{\gamma}(s)$ for every $s$. Orthogonality is preserved because \begin{align*} \frac{d}{ds}g(E(s),\dot{\gamma}(s)) = g(\nabla_{\dot{\gamma}}E,\dot{\gamma}) + g(E,\nabla_{\dot{\gamma}}\dot{\gamma}) = 0, \end{align*} where the first term vanishes by parallelness and the second because $\gamma$ is a geodesic. The model scalar factor is chosen so that it vanishes at $s=0$ and equals $1$ at $s=t$: \begin{align*} a: [0,t] &\to \mathbb{R} \\ s &\mapsto \frac{\operatorname{sn}_k(s)}{\operatorname{sn}_k(t)}. \end{align*} This is well-defined because the theorem assumes $\operatorname{sn}_k(s)>0$ for every $s\in(0,t]$, and in particular $\operatorname{sn}_k(t)>0$. This same hypothesis records that the model radial Jacobi field has no zero before or at time $t$. The corresponding comparison field is \begin{align*} V: [0,t] &\to TM \\ s &\mapsto |X|\,a(s)\,E(s). \end{align*} It satisfies $V(0)=0$ because $\operatorname{sn}_k(0)=0$, and $V(t)=X$ because $a(t)=1$ and $E(t)=X/|X|$. The actual field $J$ is the unique Jacobi field with the same endpoint values. Since $x$ is outside the cut locus of $p$, there is no conjugate point to $p$ along $\gamma|_{(0,t]}$. The index lemma for Jacobi fields applies to piecewise smooth perpendicular fields along a geodesic segment with fixed endpoint values when the initial endpoint has no conjugate point in the open terminal interval. Those hypotheses are satisfied here: $J$ is the Jacobi field with $J(0)=V(0)=0$, $J(t)=V(t)=X$, both fields are perpendicular to $\dot\gamma$, and the preceding no-conjugate condition holds. Hence \begin{align*} I(J,J) \le I(V,V). \end{align*} We now compute $I(V,V)$. Since $E$ is parallel, \begin{align*} \nabla_{\dot{\gamma}}V(s)=|X|\,a'(s)\,E(s), \end{align*} and therefore \begin{align*} g(\nabla_{\dot{\gamma}}V,\nabla_{\dot{\gamma}}V)=|X|^2(a'(s))^2. \end{align*} Also, \begin{align*} g(R(V,\dot{\gamma})\dot{\gamma},V) = |X|^2 a(s)^2 g(R(E,\dot{\gamma})\dot{\gamma},E). \end{align*} Because $E(s)$ is unit and orthogonal to $\dot{\gamma}(s)$, the last curvature term is exactly the radial sectional curvature of the plane $\operatorname{span}\{\dot{\gamma}(s),E(s)\}$. The hypothesis $K \ge k$ gives \begin{align*} g(R(E(s),\dot{\gamma}(s))\dot{\gamma}(s),E(s)) \ge k. \end{align*} Substituting this into the index form yields \begin{align*} I(V,V) &= |X|^2 \int_0^t \left( (a'(s))^2 - a(s)^2 g(R(E(s),\dot{\gamma}(s))\dot{\gamma}(s),E(s)) \right) \, d\mathcal{L}^1(s) \\ &\le |X|^2 \int_0^t \left( (a'(s))^2-k a(s)^2 \right) \, d\mathcal{L}^1(s). \end{align*} This is the precise point where the curvature lower bound is converted into an upper bound for the Hessian. [/guided] [/step] [step:Evaluate the model index integral] Since $\operatorname{sn}_k$ satisfies $\operatorname{sn}_k''+k\operatorname{sn}_k=0$, the function $a$ satisfies \begin{align*} a''+ka=0, \qquad a(0)=0, \qquad a(t)=1. \end{align*} Hence \begin{align*} \frac{d}{ds}\bigl(a(s)a'(s)\bigr) = (a'(s))^2+a(s)a''(s) = (a'(s))^2-k a(s)^2. \end{align*} Integrating with respect to one-dimensional Lebesgue measure gives \begin{align*} \int_0^t \left( (a'(s))^2-k a(s)^2 \right) \, d\mathcal{L}^1(s) = a(t)a'(t)-a(0)a'(0) = a'(t). \end{align*} Since \begin{align*} a'(t)=\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)} = \operatorname{ct}_k(t), \end{align*} we obtain \begin{align*} \operatorname{Hess} r_p(X,X) = I(J,J) \le I(V,V) \le \operatorname{ct}_k(t)|X|^2. \end{align*} This proves the comparison estimate under the radial curvature lower bound. [/step] [step:Apply Rauch comparison in the opposite curvature direction] Assume now that every radial sectional curvature along $\gamma$ is at most $k$. Let $P_s:T_xM\to T_{\gamma(s)}M$ denote parallel transport backward along $\gamma$ from time $t$ to time $s$, and let \begin{align*} A(s): \{Y\in T_xM:g(Y,\dot{\gamma}(t))=0\} &\to \{\dot{\gamma}(s)\}^{\perp}\subset T_{\gamma(s)}M \end{align*} be the radial Jacobi tensor determined by \begin{align*} A(0)=0, \qquad \nabla_{\dot{\gamma}}A(0)=P_0. \end{align*} Here $\nabla_{\dot{\gamma}}A(0)=P_0$ means that for every $Y\in T_xM$ with $g(Y,\dot{\gamma}(t))=0$, the vector field $s\mapsto A(s)Y$ satisfies \begin{align*} \left(\nabla_{\dot{\gamma}}(A(\cdot)Y)\right)(0)=P_0Y. \end{align*} Because $x\in \Omega_p$, the operator $A(t)$ is invertible on the orthogonal complement of $\dot{\gamma}(t)$. Define the radial shape operator \begin{align*} S(t): \{\dot{\gamma}(t)\}^{\perp} &\to \{\dot{\gamma}(t)\}^{\perp} \end{align*} by \begin{align*} S(t)Y := \nabla_Y \nabla r_p. \end{align*} For $Y\perp \dot{\gamma}(t)$, the Jacobi tensor identity gives \begin{align*} g(S(t)Y,Y)=\operatorname{Hess}r_p(Y,Y). \end{align*} The [Rauch Comparison Theorem for Jacobi Tensors](/page/Rauch%20Comparison%20Theorem%20for%20Jacobi%20Tensors) in its radial Hessian form applies on $[0,t]$: with the curvature convention fixed above, if the manifold radial Jacobi tensor is nonsingular on $(0,t]$, the model radial Jacobi tensor of constant curvature $k$ is nonsingular on $(0,t]$, and the radial sectional curvatures satisfy $K \le k$, then the radial shape operator is bounded below by the model shape operator as a quadratic form. The nonsingularity hypotheses hold because $x\in\Omega_p$ for the manifold tensor and because the theorem assumes $\operatorname{sn}_k(s)>0$ for every $s\in(0,t]$ for the model tensor. Thus Rauch comparison gives \begin{align*} g(S(t)Y,Y) \ge \operatorname{ct}_k(t)|Y|^2 \end{align*} for every $Y\in \{\dot{\gamma}(t)\}^{\perp}$. Taking $Y=X$ yields \begin{align*} \operatorname{Hess}r_p(X,X)\ge \operatorname{ct}_k(t)|X|^2. \end{align*} [guided] For the reverse inequality, the index-minimizing argument alone is not enough: the index lemma always says that the true Jacobi field has no larger index than any competitor with the same endpoints. To obtain a lower bound for the Hessian, we use the stronger Rauch comparison theorem for radial Jacobi tensors. The geometric object being compared is the radial shape operator. Let $P_s:T_xM\to T_{\gamma(s)}M$ be parallel transport backward along $\gamma$ from time $t$ to time $s$. Define the Jacobi tensor \begin{align*} A(s): \{Y\in T_xM:g(Y,\dot{\gamma}(t))=0\} &\to \{\dot{\gamma}(s)\}^{\perp}\subset T_{\gamma(s)}M \end{align*} by the initial conditions \begin{align*} A(0)=0, \qquad \nabla_{\dot{\gamma}}A(0)=P_0. \end{align*} For each endpoint vector $Y\perp \dot{\gamma}(t)$, the field $s\mapsto A(s)A(t)^{-1}Y$ is the unique perpendicular Jacobi field along $\gamma$ that vanishes at $s=0$ and equals $Y$ at $s=t$. The assumption $x\in\Omega_p$ ensures that no conjugate point occurs along $\gamma|_{(0,t]}$, so $A(t)$ is invertible on the orthogonal complement of $\dot{\gamma}(t)$. The radial shape operator is the map \begin{align*} S(t): \{\dot{\gamma}(t)\}^{\perp} &\to \{\dot{\gamma}(t)\}^{\perp} \end{align*} defined by \begin{align*} S(t)Y := \nabla_Y\nabla r_p. \end{align*} Since $r_p$ is smooth at $x=\gamma(t)$, this operator is well-defined, and for every $Y\perp \dot{\gamma}(t)$ we have \begin{align*} g(S(t)Y,Y)=\operatorname{Hess}r_p(Y,Y). \end{align*} Rauch comparison for Jacobi tensors compares this operator with the corresponding model operator in the simply [connected space](/page/Connected%20Space) form of constant curvature $k$. The model radial Jacobi factor is $\operatorname{sn}_k(s)$, so the model shape operator at radius $t$ is multiplication by \begin{align*} \frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)} = \operatorname{ct}_k(t). \end{align*} The hypotheses required for the comparison are exactly the ones in force: the manifold interval contains no conjugate point before or at $t$ because $x\in\Omega_p$, the model tensor is nonsingular throughout $(0,t]$ because the theorem assumes $\operatorname{sn}_k(s)>0$ for every $s\in(0,t]$, and the radial sectional curvatures satisfy $K\le k$. Therefore the [Rauch Comparison Theorem for Jacobi Tensors](/page/Rauch%20Comparison%20Theorem%20for%20Jacobi%20Tensors) gives \begin{align*} g(S(t)Y,Y)\ge \operatorname{ct}_k(t)|Y|^2 \end{align*} for every $Y\perp \dot{\gamma}(t)$. Since $\dot{\gamma}(t)=\nabla r_p(x)$ and $X\perp\nabla r_p(x)$, we may take $Y=X$ and obtain \begin{align*} \operatorname{Hess}r_p(X,X) = g(S(t)X,X) \ge \operatorname{ct}_k(t)|X|^2. \end{align*} [/guided] [/step] [step:Combine the two curvature directions] The first comparison proves the stated upper Hessian bound when all radial sectional curvatures along $\gamma$ are bounded below by $k$. The Rauch comparison step proves the stated lower Hessian bound when all radial sectional curvatures along $\gamma$ are bounded above by $k$. Together with the already handled case $X=0$, this proves both assertions for every $X\in T_xM$ satisfying $g(X,\nabla r_p(x))=0$. [/step]