[proofplan] We prove that the distance from $s(x,t)$ to the fiberwise convex set $\mathcal K_x$ cannot become positive. If a first exit occurred, convexity gives a nearest boundary point $v\in\mathcal K_x$ and a supporting unit covector $\ell\in E_x^*$ separating $s(x,t)$ from $\mathcal K_x$. Parallel-transport invariance extends $v$ and $\ell$ locally by parallel fields, so the scalar separating function has a spatial maximum and the connection Laplacian has the correct sign there. The ODE invariance gives an inward-pointing inequality for $\Phi$ at $v$, while local Lipschitz continuity controls the difference between $\Phi(s,t)$ and $\Phi(v,t)$; after an exponential barrier this contradicts the parabolic maximum principle at the first positive maximum. [/proofplan] [step:Define the global support quantity and fix a uniform Lipschitz neighbourhood] Throughout the proof, write $g(t)$ for the Ricci flow background Riemannian metric on $M$ at time $t$, write $h(t)$ for the bundle metric on $E$ at time $t$, and write $|\cdot|_{h(t)}$ for the induced norm on each fiber $E_x$. For each $t\in[0,T]$, write $\nabla^t$ for the compatible connection on $E$, write $\nabla^{t,*}$ for its induced dual connection on $E^*$, and write $\Delta_t$ for the connection Laplacian on sections of $E$ defined by taking the $g(t)$-trace of the second covariant derivative associated to $\nabla^t$. For a scalar function $u:M\to\mathbb R$, write $\Delta_{g(t)}u$ for the Laplace-Beltrami operator of $g(t)$, with the sign convention that $\Delta_{g(t)}u\le0$ at a spatial maximum. For each $x\in M$ and $t\in[0,T]$, define the distance map $d_x^t:E_x\to[0,\infty)$ by \begin{align*} d_x^t(w):=\inf_{z\in\mathcal K_x}|w-z|_{h(t)}. \end{align*} Define the global distance function $D:[0,T]\to[0,\infty)$ by \begin{align*} D(t):=\sup_{x\in M}d_x^t(s(x,t)). \end{align*} The function $D$ is continuous. Indeed, fix $\tau\in[0,T]$ and $x_0\in M$. Choose a coordinate ball $U\subset M$ around $x_0$ such that each $y\in U$ is joined to $x_0$ by a unique radial curve in $U$, and trivialize $E|_U$ by $\nabla^\tau$-parallel transport along these radial curves. Parallel-transport invariance identifies each $\mathcal K_y$ with the same closed convex subset $K_0\subset E_{x_0}$. In this trivialization the metrics $h(t)_y$ become a smoothly varying family of positive definite inner products on the fixed finite-dimensional [vector space](/page/Vector%20Space) $E_{x_0}$, uniformly equivalent on a smaller compact neighbourhood of $(x_0,\tau)$. If $w,w'\in E_{x_0}$ and $\|\cdot\|_{y,t}$ denotes the transported $h(t)_y$-norm, then the [reverse triangle inequality](/theorems/2300) gives \begin{align*} \left|\operatorname{dist}_{\|\cdot\|_{y,t}}(w,K_0)-\operatorname{dist}_{\|\cdot\|_{y,t}}(w',K_0)\right|\le \|w-w'\|_{y,t}, \end{align*} and uniform equivalence plus smooth dependence of $\|\cdot\|_{y,t}$ gives continuity in $(y,t,w)$. Since $s:M\times[0,T]\to E$ is continuous, $(x,t)\mapsto d_x^t(s(x,t))$ is continuous locally, hence globally by a finite trivializing cover of compact $M$. The supremum over compact $M$ of this [continuous function](/page/Continuous%20Function) is continuous in $t$. If $D(t)=0$ for every $t\in[0,T]$, then the result follows from closedness of the fibers $\mathcal K_x$. Assume for contradiction that $D(t_1)>0$ for some $t_1\in(0,T]$. Let $R>0$ satisfy $|s(x,t)|_{h(t)}\le R$ for all $(x,t)\in M\times[0,T]$, where the norm is computed in the fiber $E_x$. We first construct a bounded reference set in $\mathcal K$. For each $x\in M$, choose a fiber point $a_x\in\mathcal K_x$. Parallel-transport invariance of $\mathcal K$ gives a neighbourhood $U_x\subset M$ of $x$ and a smooth local section $z_x:U_x\to E|_{U_x}$ with $z_x(y)\in\mathcal K_y$, obtained by parallel transport of $a_x$ along radial curves in a coordinate ball. By compactness of $M$, choose finitely many such neighbourhoods $U_{x_1},\dots,U_{x_N}$ covering $M$. Since the metrics $h(t)$ vary smoothly on the compact set $M\times[0,T]$, there is a constant $A>0$ such that for every $y\in M$ and $t\in[0,T]$ one can choose an index $j$ with $y\in U_{x_j}$ and $|z_{x_j}(y)|_{h(t)}\le A$. If $v\in\mathcal K_y$ is a nearest point to $s(y,t)$, then using the admissible comparison point $z_{x_j}(y)\in\mathcal K_y$ gives \begin{align*} |s(y,t)-v|_{h(t)}\le |s(y,t)-z_{x_j}(y)|_{h(t)}\le R+A. \end{align*} Hence $|v|_{h(t)}\le 2R+A$. Fix any smooth reference bundle metric $h_0$ on $E$. Since $M\times[0,T]$ is compact and $h(t)$ varies smoothly, there is a constant $B>0$ such that $|w|_{h_0}\le B|w|_{h(t)}$ for every $w\in E_x$, $x\in M$, and $t\in[0,T]$. Thus every such nearest point satisfies $|v|_{h_0}\le B(2R+A)$, so the set of all such nearest points lies in the compact subset $\{w\in E: |w|_{h_0}\le B(2R+A)\}$. Choose a compact fiberwise neighbourhood $\mathcal U\subset E$ containing $s(M\times[0,T])$ and this compact set of nearest points. By local Lipschitz continuity of $\Phi$ in the fiber variable and compactness of $\mathcal U\times[0,T]$, there exists $L>0$ such that for all $x\in M$, all $t\in[0,T]$, and all $u,v\in\mathcal U\cap E_x$, \begin{align*} |\Phi(u,t)-\Phi(v,t)|_{h(t)}\le L|u-v|_{h(t)}. \end{align*} [/step] [step:Use a maximizing support pair to obtain the Laplacian sign] Fix a time $t_*\in(0,T)$ with $D(t_*)>0$, and choose $x_*\in M$ such that $d_{x_*}^{t_*}(s(x_*,t_*))=D(t_*)$. Since $\mathcal K_{x_*}$ is closed and convex in the finite-dimensional [inner product space](/page/Inner%20Product%20Space) $(E_{x_*},h(t_*))$, the [Existence and Uniqueness of the Projection Operator](/theorems/86) gives a point $v_*\in\mathcal K_{x_*}$ satisfying \begin{align*} |s(x_*,t_*)-v_*|_{h(t_*)}=D(t_*). \end{align*} Define the covector $\ell_*\in E_{x_*}^*$ by \begin{align*} \ell_*(w):=\frac{h(t_*)_{x_*}(s(x_*,t_*)-v_*,w)}{|s(x_*,t_*)-v_*|_{h(t_*)}}. \end{align*} Then $\|\ell_*\|_{h(t_*)^*}=1$, and the variational inequality for the metric projection onto a closed convex subset of a finite-dimensional [inner product](/page/Inner%20Product) space gives $\ell_*(z-v_*)\le 0$ for every $z\in\mathcal K_{x_*}$. Choose a coordinate ball $U\subset M$ around $x_*$ such that every point of $U$ is joined to $x_*$ by a unique radial curve contained in $U$. Define the local section $v:U\to E|_U$ by $\nabla^{t_*}$-parallel transport of $v_*$ along these radial curves. Define the local dual section $\ell:U\to E^*|_U$ by $\nabla^{t_*}$-parallel transport of $\ell_*$ using the dual connection. Compatibility of $\nabla^{t_*}$ with $h(t_*)$ gives $\|\ell(x)\|_{h(t_*)^*}=1$ for every $x\in U$. Parallel-transport invariance of $\mathcal K$ gives $v(x)\in\mathcal K_x$ for every $x\in U$ and transports the supporting inequality to $\ell(x)(z-v(x))\le 0$ for every $x\in U$ and every $z\in\mathcal K_x$. Define $f:U\times[0,T]\to\mathbb R$ by \begin{align*} f(x,t):=\ell(x)(s(x,t)-v(x)). \end{align*} For every $x\in U$, \begin{align*} f(x,t_*)\le d_x^{t_*}(s(x,t_*))\le D(t_*)=f(x_*,t_*). \end{align*} Thus $x_*$ is a spatial maximum of $f(\cdot,t_*)$ on $U$. The scalar second-derivative test in local coordinates gives $\Delta_{g(t_*)}f(x_*,t_*)\le 0$. We justify the Laplacian identity at the centre of the radial gauge. Choose $g(t_*)$-[normal coordinates](/theorems/2713) $(x_1,\dots,x_n)$ centred at $x_*$ and use the $\nabla^{t_*}$-radial parallel frame of $E$ determined by a basis of $E_{x_*}$. In this frame the connection one-form $A_i$ satisfies the radial gauge identity \begin{align*} \sum_i x_i A_i(x)=0. \end{align*} Differentiating this identity in the coordinate direction $x_j$ gives \begin{align*} A_j(x)+\sum_i x_i\partial_{x_j}A_i(x)=0. \end{align*} Evaluating at $x_*$ gives $A_j(x_*)=0$ for every $j$. Differentiating the same displayed identity once more in the direction $x_j$, then evaluating at $x_*$ and summing over $j$, gives \begin{align*} \sum_{j=1}^n\partial_{x_j}A_j(x_*)=0. \end{align*} Hence any radial-parallel constant section $q$ satisfies \begin{align*} \sum_{i=1}^n (\nabla^{t_*}_{\partial_{x_i}}\nabla^{t_*}_{\partial_{x_i}}q)(x_*)=0, \end{align*} and the same statement holds for the dual connection. Applying this to $q=v$ and $q=\ell$, and using that the Christoffel symbols of $g(t_*)$ vanish at $x_*$ in normal coordinates, the product rule for the connection Laplacian gives \begin{align*} \Delta_{g(t_*)}f(x_*,t_*)=\ell_*(\Delta_{t_*}s(x_*,t_*)). \end{align*} Therefore $\ell_*(\Delta_{t_*}s(x_*,t_*))\le 0$. [/step] [step:Convert ODE invariance into the inward pointing inequality] We claim that \begin{align*} \ell_*(\Phi(v_*,t_*))\le 0. \end{align*} Indeed, let \begin{align*} \sigma:[t_*,t_*+\varepsilon)\to E_{x_*} \end{align*} be the solution of \begin{align*} \frac{d\sigma}{dt}(t)=\Phi(\sigma(t),t),\qquad \sigma(t_*)=v_*. \end{align*} For sufficiently small $\varepsilon>0$, the ODE exists by [Picard Lindelof](/theorems/69) applied in the finite-dimensional vector space $E_{x_*}$, using local Lipschitz continuity of $\Phi$ in the fiber variable and continuity in time. By the ODE invariance hypothesis, $\sigma(t)\in\mathcal K_{x_*}$ for $t\in[t_*,t_*+\varepsilon)$. The supporting inequality gives \begin{align*} \ell_*(\sigma(t)-v_*)\le 0. \end{align*} Dividing by $t-t_*>0$ and taking $t\downarrow t_*$ yields \begin{align*} \ell_*(\Phi(v_*,t_*))=\ell_*\left(\frac{d\sigma}{dt}(t_*)\right)\le 0. \end{align*} [/step] [step:Apply a global exponential barrier to rule out positive distance] Fix a constant $\mu\ge0$ controlling the time variation of the dual metrics on the compact set under consideration: for every fixed covector $\lambda\in E_x^*$, the function $a_\lambda:[0,T]\to[0,\infty)$ defined by $a_\lambda(t)=\|\lambda\|_{h(t)^*}$ satisfies $|a_\lambda'(t)|\le\mu a_\lambda(t)$ whenever $a_\lambda(t)>0$. Such a $\mu$ exists because $h(t)$ is smooth on the compact manifold $M$ and the interval $[0,T]$ is compact. Define $C:=L+\mu+1$. We first prove the viscosity differential inequality $D'(t)\le CD(t)$ in the upper-test sense at interior times. Let $\tau\in(0,T)$ with $D(\tau)>0$. Choose $\eta>0$ small enough that $I_\eta:=(\tau-\eta,\tau+\eta)\cap[0,T]$ is a neighbourhood of $\tau$ in the time interval. Let $\varphi:I_\eta\to\mathbb R$ be a smooth function such that $D(t)-\varphi(t)$ has a local maximum at $\tau$ with equality at $\tau$. Choose $x_0\in M$ with $d_{x_0}^{\tau}(s(x_0,\tau))=D(\tau)$, choose a nearest point $v_0\in\mathcal K_{x_0}$, and choose the associated unit supporting covector $\ell_0\in E_{x_0}^*$ as in the preceding step. Parallel-transport $v_0$ and $\ell_0$ at the fixed time $\tau$ to obtain local sections $v$ and $\ell$, and define $f:U\times I_\eta\to\mathbb R$ by $f(x,t)=\ell(x)(s(x,t)-v(x))$. Define the scalar functions $a:I_\eta\to(0,\infty)$ and $S:I_\eta\to\mathbb R$ by \begin{align*} a(t):=\|\ell(x_0)\|_{h(t)^*},\qquad S(t):=a(t)^{-1}f(x_0,t). \end{align*} Since $v(x_0)\in\mathcal K_{x_0}$ and $\ell(x_0)$ supports $\mathcal K_{x_0}$ at $v(x_0)$, the finite-dimensional dual [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) gives $S(t)\le d_{x_0}^t(s(x_0,t))\le D(t)$. At $t=\tau$, $a(\tau)=1$ and $S(\tau)=D(\tau)$. Hence $S(t)-\varphi(t)$ has a local maximum at $\tau$, so $S'(\tau)=\varphi'(\tau)$. At the fixed time $\tau$, the support-function argument gives $\Delta_{g(\tau)}f(x_0,\tau)=\ell_0(\Delta_\tau s(x_0,\tau))\le0$. Using the equation, the inward-pointing inequality $\ell_0(\Phi(v_0,\tau))\le0$, and the Lipschitz bound for $\Phi$, we obtain $\partial_t f(x_0,\tau)\le Lf(x_0,\tau)$. Since $|a'(\tau)|\le\mu a(\tau)$ and $a(\tau)=1$, differentiating $S=a^{-1}f$ gives $\varphi'(\tau)=S'(\tau)\le(L+\mu)D(\tau)\le CD(\tau)$. This proves the claimed viscosity inequality. Fix $T'\in(0,T)$. For $\varepsilon>0$, define $Q_{\varepsilon,T'}:[0,T']\to\mathbb R$ by \begin{align*} Q_{\varepsilon,T'}(t):=e^{-Ct}D(t)-\varepsilon t. \end{align*} The initial hypothesis $s(x,0)\in\mathcal K_x$ for every $x\in M$ gives $D(0)=0$, hence $Q_{\varepsilon,T'}(0)=0$. If $Q_{\varepsilon,T'}$ were positive somewhere, continuity of $D$ and compactness of $[0,T']$ would give a positive global maximum at some $\bar t\in(0,T']$, and the positivity excludes $\bar t=0$. Since $T'<T$, this maximum time is an interior time for the original interval. Maximality gives, for $t$ near $\bar t$, \begin{align*} e^{-Ct}D(t)-\varepsilon t\le e^{-C\bar t}D(\bar t)-\varepsilon\bar t. \end{align*} Equivalently, \begin{align*} D(t)\le e^{C(t-\bar t)}D(\bar t)+\varepsilon e^{Ct}(t-\bar t). \end{align*} Thus the smooth function $\varphi(t):=e^{C(t-\bar t)}D(\bar t)+\varepsilon e^{Ct}(t-\bar t)$ touches $D$ from above at $\bar t$. The viscosity inequality gives $\varphi'(\bar t)\le CD(\bar t)$, but direct differentiation gives $\varphi'(\bar t)=CD(\bar t)+\varepsilon e^{C\bar t}$, a contradiction. Therefore $Q_{\varepsilon,T'}(t)\le0$ for every $t\in[0,T']$. Letting $\varepsilon\downarrow0$ gives $D(t)=0$ for every $t\in[0,T']$. Since $T'<T$ was arbitrary and $D$ is continuous on $[0,T]$, it follows that $D(t)=0$ for every $t\in[0,T]$. [guided] The delicate point is that the metric defining the fiber distance changes with time. A covector normalized at one time need not remain normalized at nearby times, so the support function must be corrected by the changing dual norm. Because the bundle metric $h(t)$ is smooth on compact $M\times[0,T]$, there is a constant $\mu\ge0$ such that every fixed covector $\lambda\in E_x^*$ has dual norm $a_\lambda(t)=\|\lambda\|_{h(t)^*}$ satisfying $|a_\lambda'(t)|\le\mu a_\lambda(t)$ whenever $a_\lambda(t)>0$. We set $C:=L+\mu+1$; the extra $1$ leaves a strict margin for the final barrier contradiction. We prove first that $D$ satisfies $D'(t)\le CD(t)$ in the viscosity sense at interior times. Let $\tau\in(0,T)$ with $D(\tau)>0$, and suppose a smooth function $\varphi$ touches $D$ from above at $\tau$, meaning $D(t)\le\varphi(t)$ near $\tau$ and $D(\tau)=\varphi(\tau)$. Choose $x_0\in M$ with $d_{x_0}^{\tau}(s(x_0,\tau))=D(\tau)$. By the finite-dimensional [projection theorem](/theorems/1985) for the closed convex set $\mathcal K_{x_0}$, there is a nearest point $v_0\in\mathcal K_{x_0}$. The corresponding unit covector $\ell_0\in E_{x_0}^*$ supports $\mathcal K_{x_0}$ at $v_0$, so $\ell_0(z-v_0)\le0$ for every $z\in\mathcal K_{x_0}$. The preceding ODE argument also gives the inward-pointing inequality $\ell_0(\Phi(v_0,\tau))\le0$. Choose $\eta>0$ small enough that $I_\eta:=(\tau-\eta,\tau+\eta)\cap[0,T]$ is the time interval on which the upper [test function](/page/Test%20Function) is defined. Parallel-transport $v_0$ and $\ell_0$ at the fixed time $\tau$ to local sections $v$ and $\ell$, and define $f:U\times I_\eta\to\mathbb R$ by $f(x,t)=\ell(x)(s(x,t)-v(x))$. At the point $x_0$, define $a:I_\eta\to(0,\infty)$ and $S:I_\eta\to\mathbb R$ by \begin{align*} a(t):=\|\ell(x_0)\|_{h(t)^*},\qquad S(t):=a(t)^{-1}f(x_0,t). \end{align*} Why divide by $a(t)$? Because this restores the correct comparison with the distance computed using $h(t)$. Since $v(x_0)\in\mathcal K_{x_0}$ and $\ell(x_0)$ supports $\mathcal K_{x_0}$ at $v(x_0)$, the finite-dimensional dual [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) gives $S(t)\le d_{x_0}^t(s(x_0,t))\le D(t)$ for nearby $t$. At $t=\tau$, $a(\tau)=1$ and $S(\tau)=D(\tau)$. Therefore $S-\varphi$ has a local maximum at $\tau$, and hence $S'(\tau)=\varphi'(\tau)$. Now the PDE bounds the derivative of this supporting function. At the fixed time $\tau$, the spatial support argument gives $\Delta_{g(\tau)}f(x_0,\tau)=\ell_0(\Delta_\tau s(x_0,\tau))\le0$. The equation $\partial_t s=\Delta_\tau s+\Phi(s,\tau)$, the inward-pointing inequality $\ell_0(\Phi(v_0,\tau))\le0$, and the Lipschitz estimate for $\Phi$ give $\partial_t f(x_0,\tau)\le Lf(x_0,\tau)$. Since $S=a^{-1}f$, $a(\tau)=1$, and $|a'(\tau)|\le\mu$, we get $S'(\tau)\le(L+\mu)D(\tau)\le CD(\tau)$. Thus every smooth upper test function satisfies $\varphi'(\tau)\le CD(\tau)$, which is the required viscosity inequality. Fix $T'\in(0,T)$ and define $Q_{\varepsilon,T'}:[0,T']\to\mathbb R$ by $Q_{\varepsilon,T'}(t)=e^{-Ct}D(t)-\varepsilon t$. The initial condition gives $D(0)=0$, so $Q_{\varepsilon,T'}(0)=0$. If $Q_{\varepsilon,T'}$ had a positive global maximum, compactness of $[0,T']$ and continuity of $D$ would give such a maximum at some $\bar t\in(0,T']$, and positivity rules out $\bar t=0$. Since $T'<T$, this is an interior time for the original interval. Maximality gives \begin{align*} e^{-Ct}D(t)-\varepsilon t\le e^{-C\bar t}D(\bar t)-\varepsilon\bar t. \end{align*} Multiplying by $e^{Ct}$ gives the upper test inequality \begin{align*} D(t)\le e^{C(t-\bar t)}D(\bar t)+\varepsilon e^{Ct}(t-\bar t). \end{align*} Thus $\varphi(t)=e^{C(t-\bar t)}D(\bar t)+\varepsilon e^{Ct}(t-\bar t)$ touches $D$ from above at $\bar t$. The viscosity inequality would imply $\varphi'(\bar t)\le CD(\bar t)$, while direct differentiation gives $\varphi'(\bar t)=CD(\bar t)+\varepsilon e^{C\bar t}$. This contradiction shows $Q_{\varepsilon,T'}\le0$ on $[0,T']$. Letting $\varepsilon\downarrow0$ gives $D(t)=0$ for every $t\in[0,T']$. Finally let $T'\uparrow T$ and use continuity of $D$ to obtain $D(t)=0$ on all of $[0,T]$. [/guided] [/step] [step:Conclude that the section remains in the convex set] The barrier argument gives $D(t)=0$ for every $t\in[0,T]$. Hence for every $x\in M$ and every $t\in[0,T]$, \begin{align*} d_x^t(s(x,t))=0. \end{align*} Since each fiber $\mathcal K_x$ is closed in $E_x$, vanishing distance implies $s(x,t)\in\mathcal K_x$. Thus $s(x,t)\in\mathcal K_x$ for every $x\in M$ and every $t\in[0,T]$, as required. [/step]