3.4.1 Mean inequalities and comparisonABE9A23B5282

Proof. Since $f$ is bounded above, we can choose a positive integer $n_0>\sup _ Xf$. For $n\geq n_0$ and $x\in X$ put

Taking $y=x$ gives $f_ n(x)\geq \max \{ f(x),-n\} \geq f(x)$, and $f_ n\leq \max \{ \sup _ Xf,-n\} <n_0$. Thus $-n\leq f_ n<n_0$, so $f_ n$ is real-valued and bounded, and as a supremum of $n$-Lipschitz functions of $x$ it is $n$-Lipschitz. Both $\max \{ f(y),-n\} $ and $-n\operatorname {dist}(x,y)$ decrease as $n$ increases, so the sequence $(f_ n)$ is decreasing.

We show that $f_ n(x)\to f(x)$ for every $x\in X$. Otherwise there is a real number $c$ with $\lim _ nf_ n(x)>c>f(x)$. For each $n$ choose $y_ n\in X$ with $\max \{ f(y_ n),-n\} -n\operatorname {dist}(x,y_ n)>c$. Then $n\operatorname {dist}(x,y_ n)<n_0-c$, so $y_ n\to x$. Moreover $\max \{ f(y_ n),-n\} >c$, which forces $f(y_ n)>c$ as soon as $n>-c$. On the other hand, $\{ f<c\} $ is an open neighborhood of $x$ by upper semicontinuity, so it contains $y_ n$ for all large $n$. This contradiction proves that $f_ n\downarrow f$.

Finally, the nonnegative functions $n_0-f_ n$ increase to $n_0-f$, so the monotone convergence theorem gives $\int (n_0-f_ n)\, \mathrm d\mu \uparrow \int (n_0-f)\, \mathrm d\mu $. Subtracting these integrals from the finite number $n_0\mu (X)$ proves the assertion about the integrals of $f_ n$. $\square$

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. $\square$

Proof. Let $m=\max _\Omega u$. If $m=-\infty $, then $u$ is identically $-\infty $, so assume that $m$ is finite, and let $E=\{ u=m\} =\{ u\geq m\} $. This set is nonempty, and it is closed in $\Omega $ because $u$ is upper semicontinuous. Let $x\in E$. For every sufficiently small $r>0$, (3.4.3) gives $m=u(x)\leq M_ u(x,r)\leq m$, so

Suppose that $u(x+r\omega _0)<a<m$ for some $\omega _0\in S^{d-1}$ and some real number $a$. By upper semicontinuity, the set $A=\{ \omega \in S^{d-1}:u(x+r\omega )<a\} $ is relatively open in $S^{d-1}$, and it contains $\omega _0$. The cone $\{ t\omega :0<t<1,\ \omega \in A\} $ is the inverse image of $A$ under the continuous map $y\mapsto y/|y|$ on $B(0,1)\setminus \{ 0\} $, so it is a nonempty open set, and therefore $S(A)>0$ by the definition of surface measure in subsection 3.1.1. Since $m-u(x+r\omega )>m-a>0$ for $\omega \in A$, the integral above would be at least $(m-a)S(A)>0$, a contradiction. Hence $u=m$ on every sufficiently small sphere centered at $x$, and $E$ contains a ball around $x$. Thus $E$ is open and closed in $\Omega $, and $E=\Omega $ because $\Omega $ is connected.

The second assertion is the comparison principle on the bounded open set $D=\Omega $ with $h=0$, in the form noted after theorem 3.4.5. $\square$

Proof. Let $x\in \Omega $ with $u(x)>-\infty $, and let $\overline{B(x,r)}\subset \Omega $. If $M$ is a real upper bound for $u$ on $B(x,r)$, the ball submean inequality of theorem 3.4.5 gives

Since $M-u\geq 0$, it follows that $u$ is integrable on $B(x,r)$.

Now let $\Omega '$ be a connected component of $\Omega $ on which $u$ is not identically $-\infty $, and let $Y$ be the set of points of $\Omega '$ that have a neighborhood on which $u$ is integrable. Then $Y$ is open, and it is nonempty by the first paragraph. Let $y\in \Omega '$ lie in the closure of $Y$, and choose $r>0$ with $\overline{B(y,3r)}\subset \Omega $. The set $Y\cap B(y,r)$ contains a point $z$, and $u$ is integrable on an open neighborhood $U$ of $z$; hence $u$ is finite almost everywhere on the nonempty open set $U\cap B(y,r)$, which has positive measure, and there is a point $x\in B(y,r)$ with $u(x)>-\infty $. Since $|x-y|<r$, the ball $B(x,2r)$ contains $y$ and its closure lies in $\overline{B(y,3r)}\subset \Omega $. By the first paragraph, $u$ is integrable on $B(x,2r)$, hence $y\in Y$. Thus $Y$ is also closed in $\Omega '$, and $Y=\Omega '$ because $\Omega '$ is connected. Finally, a locally integrable function is finite almost everywhere. $\square$

Proof. The limit $u$ is upper semicontinuous, since $\{ u<a\} =\bigcup _ j\{ u_ j<a\} $ is open for every real $a$. Let $x\in \Omega $ and $\overline{B(x,r)}\subset \Omega $. By theorem 3.4.5, $u_ j(x)\leq M_{u_ j}(x,r)$ for every $j$. If $M$ is a real upper bound for $u_1$ on $\partial B(x,r)$, the nonnegative functions $M-u_ j$ increase to $M-u$ on this sphere, so the monotone convergence theorem gives $M_{u_ j}(x,r)\downarrow M_ u(x,r)$. Hence $u(x)\leq u_ j(x)\leq M_{u_ j}(x,r)$ for all $j$, and in the limit $u(x)\leq M_ u(x,r)$. Thus $u$ is subharmonic.

Let $\Omega '$ be a connected component of $\Omega $. By proposition 3.4.8, $u$ is either identically $-\infty $ or locally integrable on $\Omega '$. In the first case, fix a compact set $K\subset \Omega '$ and a real number $A$. The open sets $\{ u_ j<-A\} $ increase with $j$ and cover $\Omega '$, because $u_ j(x)\downarrow -\infty $ for every $x\in \Omega '$. Finitely many of them, and hence one of them, cover $K$; thus $u_ j<-A$ on $K$ for all large $j$. In the second case, $u_1\geq u$ is not identically $-\infty $ on $\Omega '$ and is therefore locally integrable there as well. Then $0\leq u_ j-u\leq u_1-u$ almost everywhere on $\Omega '$, and $u_ j-u\to 0$ wherever $u$ is finite, so dominated convergence gives $\int _ K(u_ j-u)\, \mathrm d x\to 0$ for every compact $K\subset \Omega '$. $\square$