Forward orientation and the Lipschitz boundary graph

To apply [F4] to a sector at a boundary point $z$, choose $r>0$ with $\bar{B}(z,r)\subset {\mathrm{\Gamma }}_i$ and put $K=A_x\cap \bar{B}(z,r)$. This is globally closed and convex, and agrees with $A_x$ near $z$. It has nonempty interior: join $z$ to an interior point of $A_x$ and take points sufficiently close to $z$. Supporting planes at $z$ coincide for $K$ and $A_x$: a local supporting inequality extends along every segment from $z$ to a point of $A_x$. Any such plane also supports $A_z\subseteq A_x$ at $z$, so S2 makes it unique. Thus [F4] applies to $K$ and its interior conclusions hold for $A_x$ locally. For fixed $x=(u_x,v_x)$, the vertical section $I_x:=\{u:(u,v_x)\in A_x\}$ is a relatively closed convex interval by [F1]. The support plane at $x$ confines it to one side of $u_x$. Its inward normal has nonzero energy component, so [F4] gives a nontrivial interior vertical segment on that side.

Let $u,w$ be two $C^1$ graph solutions of $Dq=-P_i(q,v)$ on a common open domain, and suppose $u(v_0)=w(v_0)$. Since $P_i$ is locally Lipschitz, choose a convex neighborhood of $(u(v_0),v_0)$ on which one Lipschitz constant $L$ works. Continuity of $u,w$ lets us choose a convex ball about $v_0$ whose closure lies in the common domain and whose two graph images stay in that neighborhood; take any $v$ in a smaller concentric ball. Put $d=v-v_0$ and $\gamma (t)=v_0+td$ for $0\le t\le 1$. By [F8], $r(t)=u(\gamma (t))-w(\gamma (t))$ is $C^1$ and $r^{\mathrm{\prime }}(t)=[-P_i(u(\gamma (t)),\gamma (t))+P_i(w(\gamma (t)),\gamma (t))]\cdot d\mathrm{.}$ The local Lipschitz bound for $P_i$ gives $\mathrm{\mid }{r}^{\mathrm{\prime }}(t)\mathrm{\mid }\le L\mathrm{\parallel }d\mathrm{\parallel }\mathrm{\mid }r(t)\mathrm{\mid }$. The derivative is continuous, hence integrable by [F10]; [F9] gives $r(t)={\int }_0^t{r}^{\mathrm{\prime }}(s)ds$; hence $h(t):=\mathrm{\mid }r(t)\mathrm{\mid }\le L\mathrm{\parallel }d\mathrm{\parallel }{\int }_0^{t}h(s)ds$. Apply [F6] on $[0,1]$ to obtain $h=0$. Thus $u=w$ near $v_0$. Along any polygonal path in the common open domain, the set of path parameters at which equality holds is closed by continuity and open by this local result; since the path interval is connected and the set is nonempty, equality holds along the whole path. Every open connected subset of Euclidean space is polygonally connected: points reachable from a fixed point by finitely many segments in small balls form a nonempty relatively open and relatively closed set. Therefore equality extends throughout each common connected domain.

The convex, relatively closed interval $I_x$ contains a nontrivial segment on the inward side by step 1.1 and [F4]. Suppose it does not contain the whole open ${\mathrm{\Gamma }}_i$-fiber on that side. Its endpoint $b$ toward that side then lies inside ${\mathrm{\Gamma }}_i$; relative closedness puts $b$ in $I_x$, so the interval from $x$ to $b$ is a compact nondegenerate interval by [F7]. Let $x_1$ be its midpoint. Since $x_1\in A_x$, transitivity gives $A_{x_1}\subseteq A_x$. The supporting plane of $A_{x_1}$ has nonzero energy coefficient, so its vertical section lies wholly on one side of $x_1$; being contained in $I_x$, it lies in one half of $[x,b]$ and has at most half its length. It is nonempty and closed relative to $[x,b]$, hence compact. Repeat: take the midpoint $x_{k+1}$ of the current compact interval $I_{x_k}$; then $x_k\prec x_{k+1}$, so the next section is nested inside the current one and again has at most half its length. These intervals have nonempty total intersection: otherwise their open complements would cover the first compact interval, a finite subcover would do so by [F7], and nesting would make one interval empty. Their lengths tend to zero, so the intersection has a unique point $x_{\mathrm{\infty }}$, still in the first interval contained in ${\mathrm{\Gamma }}_i$. The midpoint recursion and this unique intersection require no selection from an arbitrary countable family. Relative closedness puts $x_{\mathrm{\infty }}$ in every $A_{x_k}$, hence $A_{x_{\mathrm{\infty }}}\subseteq A_{x_k}$ by transitivity. Its vertical section is therefore only $\{x_{\mathrm{\infty }}\}$, contradicting the nontrivial interior vertical segment at $x_{\mathrm{\infty }}$ supplied by [F4]. Thus $I_x$ contains the whole open fiber on its inward side.

All states on a fixed work-coordinate fiber have the same orientation. If $x$ is oriented upward and a state $z$ above it were oriented downward, the full upward section from step 2.1 gives $x\prec z$, and the full downward section from $z$ gives $z\prec w$ for a state $w$ below $x$, contradicting the support at $x$. If a state $y$ below an upward $x$ were oriented downward, choose $z$ above $x$ with $x\prec z$; the first collinear implication in [F1], applied to $x$ between $y$ and $z$, gives $y\prec x$, contradicting the downward support at $y$. The reversed argument handles a downward $x$.

By step 3.1, orientation is constant on each fixed-work fiber. The fiber orientation is locally constant in $v$. At any $x$, choose one vertical point just inside the supported half-space and one just outside; by [F4] the inside point can be chosen in $\mathrm{int}{A}_x$. Openness of ${\mathrm{\Gamma }}_i$ and of the interior and exterior regions preserves these memberships on nearby fibers. Each such fiber therefore meets $\mathrm{\partial }{A}_x$ between these two energy values. A support of $A_x$ at that boundary point also supports the point's own sector and so has its S2 orientation; the inside/outside placement fixes that orientation. Thus nearby fibers have the same sign as the fiber through $x$. Since ${\mathrm{\Gamma }}_i$ is convex, it is connected; a locally constant two-valued sign is constant.

Adopt the coordinate convention ${\epsilon }_i=1$. Step 2.1 and the common orientation give $x\prec y$ exactly for $u_y\ge u_x$ at fixed $v$, with strict comparison for $u_y>u_x$.

Put ${\rho }_x={\pi }_v(\mathrm{\partial }{A}_x)$. At a boundary point $z\in \mathrm{\partial }{A}_x$, [F3] supplies a support to $A_x$. Because $A_z\subseteq A_x$, that plane also supports $A_z$ and equals its unique S2 plane. Its finite slope gives a nonvertical inward direction; [F4] supplies nearby vertical points inside $A_x$, while the opposite side is outside by the support. Neighborhoods of one inside and one outside point show every nearby work fiber meets $\mathrm{\partial }{A}_x$. Hence ${\rho }_x$ is open. It is connected because S3 makes $\mathrm{\partial }{A}_x$ arcwise connected and projection is continuous. By step 5.1, every vertical fiber over ${\rho }_x$ has exactly one boundary energy; call it $u_x(v)$.

For $v,v^{\mathrm{\prime }}\in {\rho }_x$, convexity of $A_x$ puts every convex combination of their boundary states in $A_x$. On any sufficiently small convex ball in ${\rho }_x$, the unique-boundary property and upward orientation make $u_x$ the least energy in each fiber. Therefore $u_x(tv+(1-t)v^{\mathrm{\prime }})\le t{u}_x(v)+(1-t)u_x(v^{\mathrm{\prime }})$, so $u_x$ is finite and convex there. It is locally Lipschitz by [F5]. The inward support at $(u_x(v),v)$ gives $u_x(v^{\mathrm{\prime }})-u_x(v)\ge -P_i(u_x(v),v)\cdot (v^{\mathrm{\prime }}-v)\mathrm{.}$ The support at $(u_x(v^{\mathrm{\prime }}),v^{\mathrm{\prime }})$ gives $u_x(v^{\mathrm{\prime }})-u_x(v)\le -P_i(u_x(v^{\mathrm{\prime }}),v^{\mathrm{\prime }})\cdot (v^{\mathrm{\prime }}-v)\mathrm{.}$ Local Lipschitz continuity of $u_x$ and $P_i$ makes the difference between these linear bounds $O(\mathrm{\parallel }{v}^{\mathrm{\prime }}-v{\mathrm{\parallel }}^2)$. Thus $u_x$ is differentiable with derivative $-P_i(u_x(v),v)$, and that derivative is locally Lipschitz. Hence $u_x\in C^1$ with locally Lipschitz derivative.

If $y\in \mathrm{\partial }{A}_x$, then $A_y\subseteq A_x$. The boundary graphs agree at $v_y$ and solve the same differential equation. The set of $v\in {\rho }_y$ where $v\in {\rho }_x$ and $u_y(v)=u_x(v)$ is nonempty, open by step 1.2, and closed by continuity and relative closedness of $\mathrm{\partial }{A}_x$. Connectedness of ${\rho }_y$ makes it all of ${\rho }_y$. This shows ${\rho }_y\subseteq {\rho }_x$. The same argument, using relative closedness of $\mathrm{\partial }{A}_y$, shows ${\rho }_y$ is relatively closed in connected ${\rho }_x$; it is relatively open because both projections are open. Hence ${\rho }_x={\rho }_y$ and the boundary graphs agree. In particular $x\in \mathrm{\partial }{A}_y$, so $y\prec x$; together with $y\in A_x$ and transitivity this gives $A_x=A_y$.