[proofplan] Write $V$ as the sum of the Jacobi field $J$ and an endpoint-vanishing error field $W$. The bilinearity of the index form expands $I_\gamma[V,V]$ into the two diagonal terms and a cross term. The cross term vanishes by [integration by parts](/theorems/2098), using the Jacobi equation for $J$ and the endpoint conditions on $W$. The absence of conjugate points gives nonnegativity of the index form on endpoint-vanishing fields, and positive definiteness gives the equality case. [/proofplan]

[step:Decompose $V$ into the Jacobi field and an endpoint-vanishing error field] Define the piecewise smooth vector field $W$ along $\gamma$ by \begin{align*} W:[a,b]&\to TM,\\ t&\mapsto V(t)-J(t), \end{align*} where $W(t)\in T_{\gamma(t)}M$ for each $t\in[a,b]$. Since $V(a)=0$, $J(a)=0$, and $J(b)=V(b)$, we have \begin{align*} W(a)=V(a)-J(a)=0,\qquad W(b)=V(b)-J(b)=0. \end{align*} Thus $W$ is a continuous piecewise smooth vector field along $\gamma$ vanishing at both endpoints, and \begin{align*} V=J+W. \end{align*} [/step]

[step:Show that the Jacobi field is index-orthogonal to endpoint-vanishing fields]Let $g$ denote the Riemannian metric on $M$, let $R$ denote 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*} and let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. For continuous piecewise smooth vector fields $X,Y:[a,b]\to TM$ along $\gamma$, meaning $X(t),Y(t)\in T_{\gamma(t)}M$, define the index form by \begin{align*} I_\gamma[X,Y] := \int_a^b \left( g_{\gamma(t)}(\nabla_t X(t),\nabla_t Y(t)) - g_{\gamma(t)}(R(X(t),\dot{\gamma}(t))\dot{\gamma}(t),Y(t)) \right) \,d\mathcal{L}^1(t), \end{align*} where $\nabla_t X$ denotes covariant differentiation of $X$ along $\gamma$ and $\dot{\gamma}$ denotes the velocity field of $\gamma$. We prove that \begin{align*} I_\gamma[J,W]=0. \end{align*} Let $a=t_0<t_1<\cdots<t_m=b$ be a partition such that $W$ is smooth on each closed subinterval $[t_{k-1},t_k]$. On each such subinterval, metric compatibility of the Levi-Civita connection gives \begin{align*} \frac{d}{dt}g_{\gamma(t)}(\nabla_t J(t),W(t)) = g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) + g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)). \end{align*} Therefore, \begin{align*} \int_{t_{k-1}}^{t_k} g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) \,d\mathcal{L}^1(t) &= g_{\gamma(t_k)}(\nabla_t J(t_k),W(t_k)) - g_{\gamma(t_{k-1})}(\nabla_t J(t_{k-1}),W(t_{k-1}))\\ &\quad - \int_{t_{k-1}}^{t_k} g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t). \end{align*} Summing over $k=1,\dots,m$, the interior boundary terms telescope because $J$ is smooth and $W$ is continuous at the break points. Since $W$ is continuous at each interior break point and $W(a)=W(b)=0$, the endpoint terms vanish. Hence \begin{align*} \int_a^b g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) \,d\mathcal{L}^1(t) = - \int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t). \end{align*} Because $J$ is a Jacobi field along $\gamma$, it satisfies \begin{align*} \nabla_t\nabla_t J(t)+R(J(t),\dot{\gamma}(t))\dot{\gamma}(t)=0. \end{align*} Substituting this identity into the definition of the index form gives \begin{align*} I_\gamma[J,W] &= \int_a^b \left( g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) - g_{\gamma(t)}(R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \right) \,d\mathcal{L}^1(t)\\ &= -\int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t) - \int_a^b g_{\gamma(t)}(R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \,d\mathcal{L}^1(t)\\ &= -\int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t)+R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \,d\mathcal{L}^1(t)\\ &=0. \end{align*}[/step]

[guided]The point of this step is to show that the Jacobi part $J$ and the error part $W$ do not interact inside the index form. We first fix the notation used in the computation. Let $g$ denote the Riemannian metric on $M$, let $R$ denote 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*} and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For continuous piecewise smooth vector fields $X,Y:[a,b]\to TM$ along $\gamma$, meaning $X(t),Y(t)\in T_{\gamma(t)}M$, the index form is \begin{align*} I_\gamma[X,Y] := \int_a^b \left( g_{\gamma(t)}(\nabla_t X(t),\nabla_t Y(t)) - g_{\gamma(t)}(R(X(t),\dot{\gamma}(t))\dot{\gamma}(t),Y(t)) \right) \,d\mathcal{L}^1(t), \end{align*} where $\nabla_t X$ denotes covariant differentiation of $X$ along $\gamma$ and $\dot{\gamma}$ denotes the velocity field of $\gamma$. Since $W$ is only assumed piecewise smooth, choose a partition $a=t_0<t_1<\cdots<t_m=b$ such that $W$ is smooth on each $[t_{k-1},t_k]$. On such an interval, the product rule for the Levi-Civita connection and metric compatibility give \begin{align*} \frac{d}{dt}g_{\gamma(t)}(\nabla_t J(t),W(t)) = g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) + g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)). \end{align*} Integrating this identity with respect to $\mathcal{L}^1$ on $[t_{k-1},t_k]$ gives \begin{align*} \int_{t_{k-1}}^{t_k} g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) \,d\mathcal{L}^1(t) &= g_{\gamma(t_k)}(\nabla_t J(t_k),W(t_k)) - g_{\gamma(t_{k-1})}(\nabla_t J(t_{k-1}),W(t_{k-1}))\\ &\quad - \int_{t_{k-1}}^{t_k} g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t). \end{align*} Now sum over all subintervals. The boundary terms at interior partition points cancel because $W$ is continuous there and the same value of $g(\nabla_tJ,W)$ appears once with a plus sign and once with a minus sign. The remaining endpoint terms vanish because $W(a)=W(b)=0$. Thus \begin{align*} \int_a^b g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) \,d\mathcal{L}^1(t) = - \int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t). \end{align*} The Jacobi equation is exactly the equation needed to cancel the curvature term in the index form: \begin{align*} \nabla_t\nabla_t J(t)+R(J(t),\dot{\gamma}(t))\dot{\gamma}(t)=0. \end{align*} Substituting the integration-by-parts identity into $I_\gamma[J,W]$ yields \begin{align*} I_\gamma[J,W] &= \int_a^b \left( g_{\gamma(t)}(\nabla_t J(t),\nabla_t W(t)) - g_{\gamma(t)}(R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \right) \,d\mathcal{L}^1(t)\\ &= -\int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t),W(t)) \,d\mathcal{L}^1(t) - \int_a^b g_{\gamma(t)}(R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \,d\mathcal{L}^1(t)\\ &= -\int_a^b g_{\gamma(t)}(\nabla_t\nabla_t J(t)+R(J(t),\dot{\gamma}(t))\dot{\gamma}(t),W(t)) \,d\mathcal{L}^1(t)\\ &=0. \end{align*} Thus the Jacobi field $J$ is orthogonal, with respect to the index form, to every endpoint-vanishing variation field $W$.[/guided]

[step:Apply index form positivity to the endpoint-vanishing error field]The index form is symmetric and bilinear in its two vector-field arguments. Since $V=J+W$ and $I_\gamma[J,W]=0$, we obtain \begin{align*} I_\gamma[V,V] &= I_\gamma[J+W,J+W]\\ &= I_\gamma[J,J]+2I_\gamma[J,W]+I_\gamma[W,W]\\ &= I_\gamma[J,J]+I_\gamma[W,W]. \end{align*} We now invoke the [Index Form Positivity for Geodesic Segments without Conjugate Points](/theorems/1006) in its continuous piecewise smooth endpoint-vanishing form. Its hypotheses require that $\gamma:[a,b]\to M$ is a geodesic, that there is no conjugate point to $\gamma(a)$ in $(a,b]$, and that the tested vector field is continuous piecewise smooth and vanishes at both endpoints. These conditions hold by the theorem statement and by the construction of $W$. Therefore \begin{align*} I_\gamma[W,W]\geq 0. \end{align*} Consequently, \begin{align*} I_\gamma[V,V]\geq I_\gamma[J,J]. \end{align*}[/step]

[guided]The algebraic part of this step uses only bilinearity. Since $V=J+W$, symmetry and bilinearity of the index form give \begin{align*} I_\gamma[V,V] &= I_\gamma[J+W,J+W]\\ &= I_\gamma[J,J]+I_\gamma[J,W]+I_\gamma[W,J]+I_\gamma[W,W]\\ &= I_\gamma[J,J]+2I_\gamma[J,W]+I_\gamma[W,W]. \end{align*} The previous step proved $I_\gamma[J,W]=0$, so this reduces to \begin{align*} I_\gamma[V,V]=I_\gamma[J,J]+I_\gamma[W,W]. \end{align*} Thus the desired inequality is exactly the statement that the endpoint-vanishing error field $W$ has nonnegative index. We apply the [Index Form Positivity for Geodesic Segments without Conjugate Points](/theorems/1006) in its continuous piecewise smooth endpoint-vanishing form. The theorem requires three hypotheses. First, $\gamma:[a,b]\to M$ is a geodesic, which is part of the present theorem statement. Second, there is no conjugate point to $\gamma(a)$ in $(a,b]$, again by hypothesis. Third, the vector field being tested must be continuous piecewise smooth along $\gamma$ and vanish at both endpoints. The field $W$ satisfies this because $W=V-J$, the field $V$ is continuous piecewise smooth, the Jacobi field $J$ is smooth, and the endpoint identities give \begin{align*} W(a)=V(a)-J(a)=0,\qquad W(b)=V(b)-J(b)=0. \end{align*} The positivity theorem therefore yields \begin{align*} I_\gamma[W,W]\geq 0. \end{align*} Substituting this into the decomposition of $I_\gamma[V,V]$ gives \begin{align*} I_\gamma[V,V]=I_\gamma[J,J]+I_\gamma[W,W]\geq I_\gamma[J,J]. \end{align*} This is the asserted index comparison.[/guided]

[step:Identify the equality case under positive definiteness] Assume now that $I_\gamma$ is positive definite on endpoint-vanishing piecewise smooth vector fields along $\gamma$. From the identity \begin{align*} I_\gamma[V,V]-I_\gamma[J,J]=I_\gamma[W,W], \end{align*} equality $I_\gamma[V,V]=I_\gamma[J,J]$ holds if and only if \begin{align*} I_\gamma[W,W]=0. \end{align*} By positive definiteness, this is equivalent to $W=0$ on $[a,b]$. Since $W=V-J$, this is equivalent to \begin{align*} V=J \end{align*} on $[a,b]$. This proves both the inequality and the stated equality characterization. [/step]