Lemma 4.2.7A1CB0D81584C

Proof. The function $G$ is upper semicontinuous, because $\{ G<a\} =\{ g<a\} \times \mathbb R^{d-k}$ for every real $a$. Since subharmonicity is a local property, it suffices to show that $G$ is subharmonic on $D\times \mathbb R^{d-k}$ for every open ball $D=B(z_0,s)\subset \mathbb R^ k$ with $\overline{B(z_0,2s)}\subset V$. If $g$ is identically $-\infty $ on $B(z_0,2s)$, then so is $G$ on $B(z_0,2s)\times \mathbb R^{d-k}$, and there is nothing to prove. Otherwise $g$ is locally integrable on the ball $B(z_0,2s)$ by proposition 3.4.8.

For $0<\varepsilon <s$, the $k$-dimensional mollification $g_\varepsilon =g*\rho _\varepsilon $ is smooth on $D$ and satisfies $\Delta _ zg_\varepsilon =(\Delta g)*\rho _\varepsilon \geq 0$ there, by part (1) of theorem 3.4.12 and parts (1) and (3) of proposition 2.5.3; moreover, $g_\varepsilon $ decreases to $g$ on $D$ as $\varepsilon \downarrow 0$, by (3.4.13). The function $G_\varepsilon (z,y)=g_\varepsilon (z)$ is smooth on $D\times \mathbb R^{d-k}$, with $\Delta _ xG_\varepsilon (z,y)=\Delta _ zg_\varepsilon (z)\geq 0$. By (3.1.6), its sphere means in $\mathbb R^ d$ increase with the radius, and they tend to its value at the center as the radius tends to zero, because $G_\varepsilon $ is continuous. Hence $G_\varepsilon $ satisfies the submean inequality and is subharmonic on $D\times \mathbb R^{d-k}$. Along the sequence $\varepsilon _ j=s/(j+1)$, the functions $G_{\varepsilon _ j}$ decrease to $G$, which is therefore subharmonic on $D\times \mathbb R^{d-k}$ by theorem 3.4.9. $\square$