Half-relaxed limits of sub- and supersolutions with vanishing perturbations

Case (a): the strict-contact case, choice-free. Let $\varphi \in C^1(Z)$ and let $\bar{u}-\varphi$ have a strict local maximum at $z_0=(x_0,t_0)\in Z$. Choose $\rho >0$ with $\bar{B}(z_0,\rho )\subseteq Z$ and strict inequality away from $z_0$. Assume for contradiction that $\theta :={\varphi }_t(z_0)+H(z_0,D\varphi (z_0))>0$. By continuity there is $r\in (0,\rho \mathrm{/}2)$ such that ${\varphi }_t(z)+2(t-t_0)+H(z,D\varphi (z)+2(x-x_0))>3\theta \mathrm{/}4$ for $\mathrm{\mid }z-z_0\mathrm{\mid }\le r$. The compact annulus $A_r:=\{z:r\mathrm{/}2\le \mathrm{\mid }z-z_0\mathrm{\mid }\le r\}$ has a strict gap $g:=\bar{u}(z_0)-\varphi (z_0)-{\max }_{A_r}(\bar{u}-\varphi )>0$ by [F1, F4]. Write $z=(x,t)$ and $z_0=(x_0,t_0)$. For each $z\in A_r$, the defining infimum for $\bar{u}(z)$ gives ${\delta }_z>0$ such that $u_{\epsilon }(y)<\bar{u}(z)+g\mathrm{/}8$ whenever $0<\epsilon <{\delta }_z$ and $\mathrm{\mid }y-z\mathrm{\mid }<{\delta }_z$; shrink ${\delta }_z$ so also $\mathrm{\mid }\varphi (y)+\mathrm{\mid }y-z_0{\mathrm{\mid }}^2-\varphi (z)-\mathrm{\mid }z-z_0{\mathrm{\mid }}^2\mathrm{\mid }<g\mathrm{/}8$ there. Use the collection of all pairs $(z,\delta )$ satisfying these bounds; their balls $B(z,\delta \mathrm{/}2)$ cover $A_r$ without choosing one radius at each point. By compactness and [F6], finitely many such balls $B(z_i,{\delta }_i\mathrm{/}2)$ cover $A_r$; let ${\epsilon }_A:={\min }_i{\delta }_i$. For every $0<\epsilon <{\epsilon }_A$ and $y\in A_r$, these bounds give $u_{\epsilon }(y)-\varphi (y)-\mathrm{\mid }y-z_0{\mathrm{\mid }}^2<\bar{u}(z_i)-\varphi (z_i)-\mathrm{\mid }{z}_i-z_0{\mathrm{\mid }}^2+g\mathrm{/}4\le \bar{u}(z_0)-\varphi (z_0)-3g\mathrm{/}4$ for some $i$. Uniform convergence $H_{\epsilon }\to H$ on compact subsets and local uniform convergence $c_{\epsilon }\to 0$ provide ${\epsilon }_H>0$ such that for $\epsilon <{\epsilon }_H$ and $\mathrm{\mid }z-z_0\mathrm{\mid }\le r$, the corresponding upper-test residual with gradient $D\varphi (z)+2(x-x_0)$ is $>\theta \mathrm{/}2$ and $c_{\epsilon }(z)<\theta \mathrm{/}2$. Choose $0<\eta <g\mathrm{/}12$ and then $\delta <\min \{r\mathrm{/}4,{\epsilon }_A,{\epsilon }_H\}$ so that $\mathrm{\mid }\varphi (y)-\varphi (z_0)\mathrm{\mid }<\eta$ and $\mathrm{\mid }y-z_0{\mathrm{\mid }}^2<\eta$ when $\mathrm{\mid }y-z_0\mathrm{\mid }<\delta$. By [F1] there is one pair $(\epsilon ,y)$ with $0<\epsilon <\delta$, $\mathrm{\mid }y-z_0\mathrm{\mid }<\delta$ and $u_{\epsilon }(y)>\bar{u}(z_0)-\eta$. Let $z_{\epsilon }$ maximise the upper semicontinuous function $u_{\epsilon }(z)-\varphi (z)-\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$ on the compact ball $\bar{B}(z_0,r)$, possible by [F4]. Its value is $>\bar{u}(z_0)-\varphi (z_0)-3\eta$, so the uniform annulus bound forces $\mathrm{\mid }{z}_{\epsilon }-z_0\mathrm{\mid }<r\mathrm{/}2$. Thus $u_{\epsilon }-(\varphi +\mathrm{\mid }z-z_0{\mathrm{\mid }}^2)$ has a local maximum at $z_{\epsilon }$. Writing $z_0=(x_0,t_0)$ and $z_{\epsilon }=(x_{\epsilon },t_{\epsilon })$, the test has time derivative ${\varphi }_t(z_{\epsilon })+2(t_{\epsilon }-t_0)$ and spatial gradient $D\varphi (z_{\epsilon })+2(x_{\epsilon }-x_0)$; its residual is $>\theta \mathrm{/}2$ while $c_{\epsilon }(z_{\epsilon })<\theta \mathrm{/}2$, contradicting the assumed subsolution inequality. Hence ${\varphi }_t(z_0)+H(z_0,D\varphi (z_0))\le 0$ at every strict local maximum.

The initial trace. Assume the relaxed initial inequality and data convergence of the statement, and use the traces of [F7]. Fix $x\in O$ and $\eta >0$. By local equicontinuity and continuity of $u_0^{(\epsilon )}\to u_0$ there is a neighbourhood $V$ of $(x,0)$ such that, for all sufficiently small $\epsilon$ and $(y,s)\in V\cap Z$, $u_{\epsilon }(y,s)\le u_{\epsilon }^0(x)+\eta \le u_0^{(\epsilon )}(x)+\eta \le u_0(x)+2\eta \mathrm{.}$ Shrink to a neighbourhood $V^{\mathrm{\prime }}$ whose closure lies in $V$. For each $z^{\mathrm{\prime }}\in V^{\mathrm{\prime }}\cap Z$ sufficiently close to $(x,0)$, the neighborhoods in the definition of $\bar{u}(z^{\mathrm{\prime }})$ can be taken inside $V$, so the same bound gives $\bar{u}(z^{\mathrm{\prime }})\le u_0(x)+2\eta$. Therefore ${\mathrm{lim\hspace{0.17em}sup}}_{(y,s)\to (x,0),s>0}\bar{u}(y,s)\le u_0(x)+2\eta$; letting $\eta \downarrow 0$ proves the relaxed subsolution initial condition. The lower-limit argument is the dual one.

Case (b): the strict-contact case with near-maximiser extraction. Assume the exact limit-inferior condition and let $\varphi$ and $z_0$ be as in step 1.1. For each $j\ge 1$, consider triples $(\epsilon ,y,z)$ with $0<\epsilon <\min \{1,1\mathrm{/}j\}$, $y\in Z$, $\mathrm{\mid }y-z_0\mathrm{\mid }<\min \{\rho \mathrm{/}2,1\mathrm{/}j\}$, $u_{\epsilon }(y)>\bar{u}(z_0)-1\mathrm{/}j$, and $z$ a maximiser of $u_{\epsilon }(\cdot )-\varphi (\cdot )-\mathrm{\mid }\cdot -z_0{\mathrm{\mid }}^2$ on $\bar{B}(z_0,\rho )$. This set is nonempty by [F1] and [F4]; Countable Choice [F5] selects triples $({\epsilon }_j,y_j,z_j)$. Then ${\epsilon }_j\to 0$, $y_j\to z_0$ and the maximal values satisfy $u_{{\epsilon }_j}(z_j)-\varphi (z_j)-\mathrm{\mid }{z}_j-z_0{\mathrm{\mid }}^2\ge \bar{u}(z_0)-\varphi (z_0)-o(1)$. Fix any $r\in (0,\rho )$. The strict maximum of $\bar{u}-\varphi$ gives a positive gap on the compact annulus $r\mathrm{/}2\le \mathrm{\mid }z-z_0\mathrm{\mid }\le \rho$; the finite-cover argument of step 1.1 then bounds $u_{\epsilon }(z)-\varphi (z)-\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$ strictly below $\bar{u}(z_0)-\varphi (z_0)$ on this annulus for all sufficiently small $\epsilon$. Since ${\epsilon }_j\to 0$ and the maximizing values are at least that limit minus $o(1)$, eventually $\mathrm{\mid }{z}_j-z_0\mathrm{\mid }<r\mathrm{/}2$. As $r>0$ was arbitrary, $z_j\to z_0$. By taking the canonical strictly decreasing subsequence of $({\epsilon }_j)$ (at each stage use the least later index with smaller $\epsilon$, which exists because ${\epsilon }_j\to 0$) and relabelling, we may assume ${\epsilon }_j\downarrow 0$; then $z_j\to z_0$ still. Eventually $z_j$ is interior to $\bar{B}(z_0,\rho )$, so ${\psi }_j:=\varphi +\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$ is a local upper test for $u_{{\epsilon }_j}$ there. Thus ${\varphi }_t(z_j)+2(t_j-t_0)+H_{{\epsilon }_j}(z_j,D\varphi (z_j)+2(x_j-x_0))\le c_{{\epsilon }_j}(z_j)\mathrm{.}$ Writing $z_j=(x_j,t_j)$, the extra time derivative $2(t_j-t_0)$ tends to zero. Since the gradients converge and $c_{{\epsilon }_j}(z_j)\to 0$, taking the limit inferior of the displayed inequality and using (b) gives ${\varphi }_t(z_0)+H(z_0,D\varphi (z_0))\le 0$. Condition (a) implies (b) by uniform convergence on compact sets, so this proves the strict-contact case.

General contacts, the dual statement and conclusion. If $\bar{u}-\varphi$ merely has a local maximum at $z_0$, strictify with [F3] and apply the strict-contact conclusion of steps 1.1 or 2.1 to the strictified test; the perturbed test has the same value and first jet at $z_0$, so the resulting inequality is exactly ${\varphi }_t(z_0)+H(z_0,D\varphi (z_0))\le 0$. Hence $\bar{u}$ is a viscosity subsolution of the limit equation, and by step 1.2 it carries the initial datum when the additional hypotheses hold. The dual argument, replacing $u_{\epsilon }$ by $-u_{\epsilon }$ and local maxima by local minima, shows that $\underset{‾}{u}$ is a viscosity supersolution with the dual initial condition. The half-relaxed limits themselves are computed as infima and suprema over sets, and only step 2.1 involves a countable selection, so under hypothesis (a) no choice principle is used and under hypothesis (b) Countable Choice is used exactly as declared.