Theorem 3.4.12E76E281548A5

Proof. (1) By proposition 3.4.8, $u$ is locally integrable. To show that $\Delta u\geq 0$, fix $\varepsilon >0$ and $x\in \Omega _\varepsilon $, with $\Omega _\varepsilon $ as in section 2.5. For sufficiently small $r>0$, all the closed balls $\overline{B(x-z,r)}$ with $|z|\leq \varepsilon $ lie in $\Omega $, so $u(x-z)\leq M_ u(x-z,r)$ for these $z$ by theorem 3.4.5. We multiply by $\rho _\varepsilon (z)$ and integrate over $z$. Since $u$ is bounded above on the compact set $\overline{B(x,\varepsilon +r)}\subset \Omega $, Fubini's theorem applies to the difference between an upper bound and $u$; as $\overline{B(x,r)}\subset \Omega _\varepsilon $, we obtain

The function $v=u*\rho _\varepsilon $ is smooth and real on $\Omega _\varepsilon $ by proposition 2.5.3, and we have just shown that $v(x)\leq M_ v(x,r)$ for every $x\in \Omega _\varepsilon $ and every sufficiently small $r>0$. This forces $\Delta v\geq 0$ on $\Omega _\varepsilon $. Indeed, suppose that $\Delta v(x)<0$. Since $\Delta v$ is continuous, there is $R>0$ such that $\overline{B(x,R)}\subset \Omega _\varepsilon $, $\Delta v<0$ on $B(x,R)$ and $v(x)\leq M_ v(x,r)$ for $0<r<R$. By (3.1.6), the function $r\mapsto M_ v(x,r)$ has negative derivative on $(0,R)$, so it is strictly decreasing there. Moreover, $M_ v(x,s)\to v(x)$ as $s\downarrow 0$, because $|M_ v(x,s)-v(x)|\leq \max _{|y-x|=s}|v(y)-v(x)|$ and $v$ is continuous at $x$. Hence $M_ v(x,r)<M_ v(x,r/2)\leq \lim _{s\downarrow 0}M_ v(x,s)=v(x)$ for $0<r<R$, a contradiction.

Now let $\chi $ be a nonnegative test function. Then $\operatorname{Supp}\chi \subset \Omega _\varepsilon $ for small $\varepsilon $, and $u*\rho _\varepsilon \to u$ in $L^1$ on $\operatorname{Supp}\chi $ by lemma 2.5.1. Hence

By theorem 2.5.6, the positive distribution $\Delta u$ is a positive Radon measure.

(2) Fix $x\in \Omega $ and $0<\varepsilon <\operatorname {dist}(x,\mathbb R^ d\setminus \Omega )$, let $C$ be a real upper bound for $u$ on $\overline{B(x,\varepsilon )}$, and let $e_1$ be the first coordinate unit vector. Since $\rho $ is radial, $\rho _\varepsilon (r\omega )=\varepsilon ^{-d}\rho ((r/\varepsilon )e_1)$ for $\omega \in S^{d-1}$, and the spherical mean of $\omega \mapsto u(x-r\omega )$ equals $M_ u(x,r)$ by the symmetry $\omega \mapsto -\omega $. Integration in polar coordinates (3.1.2), applied to the nonnegative function $(C-u(x-y))\rho _\varepsilon (y)$ on $B(0,\varepsilon )$, and the substitution $r=\varepsilon t$ therefore give, with $w(t)=s_{d-1}\rho (te_1)t^{d-1}$,

The same computation, applied to $\rho _\varepsilon $ alone, shows that $\int _0^1w\, \mathrm dt=\int \rho _\varepsilon \, \mathrm dy=1$, so $w(t)\, \mathrm dt$ is a probability measure on $(0,1)$, and we obtain

By theorem 3.4.5, $M_ u(x,\varepsilon t)$ increases with $\varepsilon $ and is at least $u(x)$. Hence $u*\rho _\varepsilon (x)$ decreases as $\varepsilon \downarrow 0$ and is at least $u(x)$. On the other hand, $u*\rho _\varepsilon (x)\leq \sup _{B(x,\varepsilon )}u$, and the upper limit of the right-hand side as $\varepsilon \downarrow 0$ is at most $u(x)$ by upper semicontinuity, also when $u(x)=-\infty $. Thus $u*\rho _\varepsilon (x)\downarrow u(x)$. The ball averages are treated in the same way: by (3.4.6), the average of $u$ over $B(x,r)$ is the average of $M_ u(x,rt)$, $0<t<1$, with respect to the probability measure $d\, t^{d-1}\, \mathrm d t$, so it lies between $u(x)$ and $\sup _{B(x,r)}u$ and tends to $u(x)$. This proves (3.4.13). The ball averages in (3.4.13) depend only on the almost-everywhere class of $u$; hence two subharmonic functions that agree almost everywhere agree everywhere.

(3) Since a distribution determines a locally integrable function representing it up to null sets (proposition 2.5.3), (2) shows that $T$ has at most one locally integrable subharmonic representative. It remains to construct one; we do this on balls and then glue.

By theorem 2.5.6, $\mu =\Delta T$ is a positive Radon measure. Let $B$ be a ball with $\overline B\subset \Omega $, and choose a nonnegative cutoff $\eta \in C_ c^\infty (\Omega )$ equal to one on an open set $U\supset \overline B$ (lemma 2.5.1). The measure $\eta \mu $, extended by zero to $\mathbb R^ d$, is finite and has compact support, so by proposition 3.4.11 its fundamental potential $p=E_ d*(\eta \mu )$ is subharmonic and locally integrable, and $\Delta p=\eta \mu =\mu $ on $U$. Thus $\Delta (T-p)=0$ on $U$, and by Weyl's lemma (theorem 3.1.11) the real distribution $T-p$ is represented on $U$ by a smooth harmonic function $h$; since $\operatorname{Re}h$ also represents $T-p$, the uniqueness in theorem 3.1.11 shows that $h=\operatorname{Re}h$ is real. Then $u_ B=p+h$ is a locally integrable subharmonic function on $B$ representing $T$ there.

If $B'$ is another such ball, $u_ B$ and $u_{B'}$ both represent $T$ on $B\cap B'$, so they agree almost everywhere there (proposition 2.5.3), and hence everywhere on $B\cap B'$ by (2). Hence there is a function $u$ on $\Omega $ with $u=u_ B$ on every ball $B$ with $\overline B\subset \Omega $. It is subharmonic, because subharmonicity is local, and it is locally integrable.

To see that $u$ represents $T$ on all of $\Omega $, let $T_\varepsilon (x)=\langle T,\rho _\varepsilon (x-\cdot )\rangle $ be the mollification of proposition 2.5.3. If $x\in \Omega _\varepsilon $, then $\overline{B(x,\varepsilon ')}\subset \Omega $ for some $\varepsilon '>\varepsilon $, and the test function $\rho _\varepsilon (x-\cdot )$ is supported in the ball $B=B(x,\varepsilon ')$, on which $u=u_ B$ represents $T$; hence $T_\varepsilon (x)=u*\rho _\varepsilon (x)$. For every test function $\chi $, proposition 2.5.3, lemma 2.5.1 now give