Theorem 3.4.5384B48086228

Proof. We begin with (1). Fix $\varepsilon >0$ and consider the upper semicontinuous function $w(y)=u(y)-h(y)+\varepsilon |y|^2$ on $D$. For $x\in D$ and small $r>0$ we have $M_ h(x,r)=h(x)$ by proposition 3.1.5, and $M_{|\cdot |^2}(x,r)=|x|^2+r^2$ by the identity for spherical means of $|y|^2$ proved in subsection 3.1.23. Together with (3.4.3) this gives

for every $x\in D$ and every sufficiently small $r>0$. Consequently, $w$ cannot attain a finite maximum over $D$ at a point $x\in D$, since we would then have $M_ w(x,r)\leq w(x)$ for small $r$.

Now let $\delta >0$. Because $h$ is continuous on $\overline D$, the boundary hypothesis gives $\varlimsup _{D\ni x\to \zeta }(u-h)(x)\leq 0$, so every $\zeta \in \partial D$ has an open neighborhood $N_\zeta $ with $u-h<\delta $ on $N_\zeta \cap D$. Finitely many of these neighborhoods cover the compact set $\partial D$. If $N$ denotes their union, then $K=\overline D\setminus N$ is a compact subset of $D$. Put $c=\delta +\varepsilon \sup _{y\in \overline D}|y|^2$, so that $w<c$ on $D\setminus K$. If $w$ took a value larger than $c$, then $\sup _ Dw=\sup _ Kw$ would be a finite maximum attained at a point of $K$, which we have just excluded. Hence $u-h\leq w\leq c$ on $D$, and letting $\varepsilon \downarrow 0$ and $\delta \downarrow 0$ gives $u\leq h$.

For (2), let $B=B(x,R)$. By lemma 3.4.4 there are continuous functions $f_ n$ on $\partial B$ that decrease to $u|_{\partial B}$. Let $H_ n=H_ Bf_ n$ be their Poisson integrals; by theorem 3.1.18, $H_ n$ is harmonic on $B$ and continuous on $\overline B$ with boundary values $f_ n$. For $\zeta \in \partial B$, upper semicontinuity gives $\varlimsup _{B\ni y\to \zeta }u(y)\leq u(\zeta )\leq f_ n(\zeta )$, so (1) yields $u\leq H_ n$ on $B$. The value of a Poisson integral at the center of the ball is the mean of its boundary values, because $P_ B(x,\cdot )$ is the constant $(s_{d-1}R^{d-1})^{-1}$ by (3.1.17). Using also the mean-value property of $H_ n$, we obtain for $0<r<R$

By lemma 3.4.4, applied to surface measure on $\partial B$, $M_{f_ n}(x,R)$ decreases to $M_ u(x,R)$. Therefore $u(x)\leq M_ u(x,R)$, and $M_ u(x,r)\leq M_ u(x,R)$ for $0<r<R$. As $R$ may be replaced by any smaller radius, $M_ u(x,\cdot )$ is increasing on $(0,R]$. Moreover, $u(x)\leq M_ u(x,r)\leq \sup _{\partial B(x,r)}u$, and the upper limit of the right-hand side as $r\downarrow 0$ is at most $u(x)$ by upper semicontinuity; hence $M_ u(x,r)$ decreases to $u(x)$ as $r\downarrow 0$.

Finally, let $M$ be a real upper bound for $u$ on $\overline B$. Integration in polar coordinates (3.1.2), applied to the nonnegative function $M-u$, and the substitution $r=Rt$ give

where all integrals lie in $[-\infty ,\infty )$; here we used $|B|=s_{d-1}R^ d/d$. Since $M_ u(x,r)\geq u(x)$ for $0<r\leq R$ and $d\int _0^1t^{d-1}\, \mathrm dt=1$, the right-hand side is at least $u(x)$.

The proof of (2) used the subharmonicity of $u$ only through (1) for the ball $B$. Hence, if $u$ is upper semicontinuous and satisfies (1) for all balls $B$ with $\overline B\subset \Omega $, then $u(x)\leq M_ u(x,R)$ whenever $\overline{B(x,R)}\subset \Omega $, so $u$ is subharmonic. $\square$