Viscosity testing by first-order jets, and closure of the jet inequality

Test functions produce jets. Suppose $\varphi \in C^1(U)$ and $u-\varphi$ has a local maximum at $z_0$; then for $z$ near $z_0$ we have $u(z)-u(z_0)\le \varphi (z)-\varphi (z_0)=⟨D\varphi (z_0),z-z_0⟩+o(\mathrm{\mid }z-z_0\mathrm{\mid })$ by [F2], so $D\varphi (z_0)\in D^{+}u(z_0)$. If instead $v-\varphi$ has a local minimum at $z_0$, the same computation with the inequality reversed gives $D\varphi (z_0)\in D^{-}v(z_0)$. Hence the jet inequality for all jets implies the test-function inequality of [F1], in both the sub- and the supersolution case.

Jets produce test functions. Assume $u$ is an upper semicontinuous subsolution in the test-function sense and let $p\in D^{+}u(z_0)$. Choose $\rho >0$ with $\bar{B}(z_0,\rho )\subseteq U$ and put $R(h):=u(z_0+h)-u(z_0)-⟨p,h⟩$ for $\mathrm{\mid }h\mathrm{\mid }\le \rho$. By the definition of $D^{+}u(z_0)$, ${\mathrm{lim\hspace{0.17em}sup}}_{h\to 0}R(h)\mathrm{/}\mathrm{\mid }h\mathrm{\mid }\le 0$, so $r^{+}(h):=\max \{R(h),0\}=o(\mathrm{\mid }h\mathrm{\mid })$. Define $\mu (s):=\sup \{r^{+}(h)\mathrm{/}\mathrm{\mid }h\mathrm{\mid }:0<\mathrm{\mid }h\mathrm{\mid }\le s\}$ for $0<s\le \rho$. This supremum is finite: the jet condition bounds the quotient near $h=0$, and on every remaining closed annulus $u(z_0+h)$ is bounded above by upper semicontinuity and compactness [F4]. Thus $\mu$ is finite and nondecreasing, $\mu (s)\to 0$ as $s\downarrow 0$, and $r^{+}(h)\le \mu (\mathrm{\mid }h\mathrm{\mid })\mathrm{\mid }h\mathrm{\mid }$. The function $\mu$ need not be continuous, so we first construct a continuous majorant. Put $r_k:=2^{-k}\rho$ and $a_k:=\mu (r_k)$ for $k\ge 0$. Define $\nu (0):=0$, set $\nu (r_k):=a_{k-1}$ for $k\ge 1$, interpolate linearly on each $[r_{k+1},r_k]$, and set $\nu (s):=a_0$ on $[\rho \mathrm{/}2,\rho ]$. The definitions agree at $r_1=\rho \mathrm{/}2$, $\nu$ is continuous with $\nu (s)\to 0$ at $0$, and $\nu (s)\ge \mu (s)$: on $[r_{k+1},r_k]$, $\nu (s)\ge a_k=\mu (r_k)\ge \mu (s)$, while on $[\rho \mathrm{/}2,\rho ]$ it equals $\mu (\rho )$. Let $\sigma (s):=\max \{0,\min \{1,2(\rho -s)\mathrm{/}\rho \}\}$, continuous with $\sigma =1$ on $[0,\rho \mathrm{/}2]$ and $\sigma (\rho )=0$, and put $g(s):=2\nu (\min \{2s,\rho \})\sigma (s)$; this is continuous on $[0,\rho ]$, with $g(0)=g(\rho )=0$. Define $\lambda (r):={\int }_0^{r}g(s)ds$ for $0\le r\le \rho$ and $\lambda (r):=\lambda (\rho )$ for $r>\rho$. By [F3], ${\lambda }^{\mathrm{\prime }}(r)=g(r)$ on $[0,\rho ]$, and $g(\rho )=0$ makes the constant extension $C^1$. For $0<r\le \rho \mathrm{/}2$, $\lambda (r)\ge {\int }_{r\mathrm{/}2}^{r}2\nu (2s)ds\ge {\int }_{r\mathrm{/}2}^{r}2\mu (r)ds=r\mu (r),$ because $2s\ge r$, $\nu (2s)\ge \mu (2s)\ge \mu (r)$ and $\sigma =1$ there. Hence $\lambda (\mathrm{\mid }h\mathrm{\mid })\ge r^{+}(h)\ge R(h)$ for $0<\mathrm{\mid }h\mathrm{\mid }\le \rho \mathrm{/}2$. Define $\varphi (z):=u(z_0)+⟨p,z-z_0⟩+\lambda (\mathrm{\mid }z-z_0\mathrm{\mid })$ on $U$. Then $u-\varphi$ has a local maximum at $z_0$, and $D\varphi (z_0)=p$ because $\lambda (r)\mathrm{/}r\le {\max }_{0\le s\le r}g(s)\to 0$. The radial term has gradient $g(\mathrm{\mid }z-z_0\mathrm{\mid })(z-z_0)\mathrm{/}\mathrm{\mid }z-z_0\mathrm{\mid }$ off $z_0$, which tends to $0$ there; thus $\varphi \in C^1(U)$. The subsolution inequality [F1] gives $p_t+H(z_0,p_x)\le 0$. The subjet case applies this construction to $-v$ and $-p$, then negates the resulting test function, giving $p_t+H(z_0,p_x)\ge 0$ for every $p\in D^{-}v(z_0)$.

Parts (1) and (2). Step 1.1 shows that the jet inequalities imply the test-function inequalities. Step 1.2 proves the reverse implication by constructing a $C^1$ test for every prescribed jet, with signs reversed for subjets. Hence both equivalences (1) and (2) hold.

Closure. Let $u$ be a subsolution in the test-function sense, let $z_k\to z_0$ in $U$, and let $p_k\in D^{+}u(z_k)$ with $p_k\to p$. Fix $k$. By step 2.1, part (1), applied at $z_k$, we have $p_k^t+H(z_k,p_k^x)\le 0$. Since $(z_k,p_k)\to (z_0,p)$ and $H$ is continuous, passing to the limit in the inequality gives $p_t+H(z_0,p_x)\le 0$, which is the closure statement for subsolutions; the same argument with the inequalities reversed and subjets in place of superjets gives the supersolution statement. The limit $p$ need not lie in $D^{+}u(z_0)$, and this is not used.