Theorem 3.1.25537ACE84788C

Proof. The bounds for $U_ f$ were proved above. To prove harmonicity, fix a ball $B$ with $\overline B\subset \Omega $ and a point $x_0\in B$; we show that $U_ f$ agrees on $B$ with a harmonic function. If $v\in \mathcal S_ f$, its harmonic lifting $v^ B$ on $B$ is a continuous subsolution by lemma 3.1.24, and it coincides with $v$ near $\partial \Omega $ because $\overline B$ is a compact subset of $\Omega $; hence $v^ B\in \mathcal S_ f$.

Choose $v_ j\in \mathcal S_ f$ with $v_ j(x_0)\to U_ f(x_0)$, and put $w_ j=\max \{ v_1,\ldots ,v_ j,\min _{\partial \Omega }f\} $. Then $(w_ j)$ is an increasing sequence in $\mathcal S_ f$ with values in $[\min _{\partial \Omega }f,\max _{\partial \Omega }f]$, and $w_ j(x_0)\to U_ f(x_0)$ because $v_ j\leq w_ j\leq U_ f$. The liftings $w_ j^ B$ belong to $\mathcal S_ f$ and satisfy $w_ j\leq w_ j^ B\leq U_ f$. On $B$ they are harmonic, increase with $j$ because $H_ B$ preserves order, and are bounded by $\max _{\partial \Omega }f$. By part (2) of corollary 3.1.21, applied on the domain $B$ to this sequence, which is bounded above, they converge on $B$ to a harmonic function $h$, and $h\leq U_ f$ on $B$, while $h(x_0)=U_ f(x_0)$.

Now let $v\in \mathcal S_ f$ be arbitrary. The same argument applied to the increasing sequence $\max \{ w_ j,v\} \in \mathcal S_ f$ shows that the liftings $(\max \{ w_ j,v\} )^ B$ increase on $B$ to a harmonic function $k$. Since these liftings dominate $w_ j^ B$ and $v$ and lie in $\mathcal S_ f$, we have $h\leq k\leq U_ f$ and $k\geq v$ on $B$. In particular $k(x_0)\leq U_ f(x_0)=h(x_0)$, so the nonnegative harmonic function $k-h$ on $B$ vanishes at $x_0$ and therefore vanishes identically, by proposition 3.1.19(2). Hence $v\leq h$ on $B$. Taking the supremum over $v\in \mathcal S_ f$ gives $U_ f\leq h$, so $U_ f=h$ on $B$. Since $B$ was arbitrary, $U_ f$ is harmonic on $\Omega $.

Finally, let $H$ be harmonic on $\Omega $ and continuous on $\overline\Omega $ with $H=f$ on $\partial \Omega $. Then $H\in \mathcal S_ f$, so $H\leq U_ f$. Conversely, for $v\in \mathcal S_ f$ the function $v-H$ is a continuous subsolution with $\varlimsup _{x\to \zeta }(v-H)(x)\leq 0$ at every $\zeta \in \partial \Omega $, so $v\leq H$ by lemma 3.1.24. Hence $U_ f=H$. $\square$