3.4 Subharmonic functions0BBC054FD2E1

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 $K\subset \Omega '$. $\square$

Proof. By proposition 2.2.2, $\nu $ is a Radon measure; let $L$ be its compact support, so that $\nu (\mathbb R^ d\setminus L)=0$ (section 2.2). For each $x$, the function $E_ d(x-\cdot )$ is bounded above on $L$, so $p(x)$ is well defined in $[-\infty ,\infty )$. Next, $E_ d$ is subharmonic on $\mathbb R^ d$. For $d\geq 2$ it is harmonic on $\mathbb R^ d\setminus \{ 0\} $; it is upper semicontinuous at the origin because $E_ d(x)\to -\infty $ as $x\to 0$, and the submean inequality at the origin holds because $E_ d(0)=-\infty $. Since subharmonicity is local, $E_ d\in \mathrm{SH}(\mathbb R^ d)$. For $d=1$, the triangle inequality gives $|x|\leq \frac12(|x+r|+|x-r|)$, which is the submean inequality for $E_1=|x|/2$.

To show that $p$ is subharmonic, we truncate the kernel. The functions $E_ d^{(m)}=\max \{ E_ d,-m\} $, $m\in \mathbb N$, are continuous and subharmonic, and they decrease to $E_ d$. Put $p_ m=E_ d^{(m)}*\nu $. For $x$ in a compact set $K$, the differences $x-y$ with $y\in L$ lie in the compact set $K-L$, on which $E_ d^{(m)}$ is uniformly continuous; hence $p_ m$ is continuous. By theorem 3.4.5, each translate $E_ d^{(m)}(\cdot -y)$ satisfies the submean inequality on every sphere. Integrating in $y$ and using Fubini's theorem for the bounded continuous integrand gives $p_ m(x)\leq M_{p_ m}(x,r)$ for all $x$ and $r>0$, so $p_ m\in \mathrm{SH}(\mathbb R^ d)$. For fixed $x$, the functions $E_ d^{(m)}(x-y)$ decrease to $E_ d(x-y)$ and are bounded above, uniformly in $m$ and $y\in L$, by the finite constant $\max \{ 0,\sup _{y\in L}E_ d(x-y)\} $. Monotone convergence, applied to the differences from this constant, gives $p_ m(x)\downarrow p(x)$. Hence $p$ is subharmonic by theorem 3.4.9.

To prove local integrability, let $K$ be compact and choose $R>0$ with $K-L\subset B(0,R)$. Since $E_ d$ is locally integrable, Tonelli's theorem gives

Hence $p$ is locally integrable. Finally, let $\chi $ be a test function. The preceding estimate with $K=\operatorname{Supp}\chi $ justifies the use of Fubini's theorem, and the identity $\Delta E_ d=\delta _0$, translated to the point $y$, gives

This is the distributional equation $\Delta p=\nu $. $\square$

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

Proof. If $\eta $ vanishes outside the compact set $C\subset \Omega $, then $\eta \mu _ u$ is a positive Borel measure on $\mathbb R^ d$ with support in $C$ and total mass at most $\sup \eta \cdot \mu _ u(C)<\infty $. Let $p=E_ d*(\eta \mu _ u)$. By proposition 3.4.11, $p$ is subharmonic and locally integrable on $\mathbb R^ d$, and $\Delta p=\eta \mu _ u$. If $\chi $ is a test function with $\operatorname{Supp}\chi \subset V$, then $\int \chi \eta \, \mathrm d\mu _ u=\int \chi \, \mathrm d\mu _ u$ because $\eta =1$ on $V$; hence $\Delta p=\Delta u$ on $V$. By Weyl's lemma (theorem 3.1.11), the real distribution $u-p$ is represented on $V$ by a smooth harmonic function $h$; since $\operatorname{Re}h$ also represents $u-p$, the uniqueness in theorem 3.1.11 shows that $h=\operatorname{Re}h$ is real. Then $p+h$ is a locally integrable subharmonic function on $V$ that agrees with $u$ almost everywhere on $V$ (proposition 2.5.3), and theorem 3.4.12(2), applied on $V$, gives $u=p+h$ at every point of $V$. $\square$

Proof. We first record a local estimate. Let $a\in \Omega $ and $r>0$ with $\overline{B(a,4r)}\subset \Omega $, and let $M\geq 0$ be a common upper bound of the functions $u_ j$ on $B(a,4r)$. Suppose that $u_ j(z)\geq -C$ for some index $j$, some point $z\in B(a,r)$ and some constant $C$. Then $B(a,r)\subset B(z,2r)$ and $\overline{B(z,2r)}\subset B(a,4r)$. Since $M-u_ j\geq 0$ on $B(z,2r)$, the ball submean inequality of theorem 3.4.5 gives

and since $|u_ j|\leq (M-u_ j)+M$, we obtain

Passing to a subsequence, we may assume that $x_ j\to x_0\in K$. Choose $r>0$ with $\overline{B(x_0,4r)}\subset \Omega $. Discarding finitely many terms, we may assume that $|x_ j-x_0|<r$ for all $j$. Then (3.4.18), with $a=x_0$, $z=x_ j$ and $C=C_0$, bounds $\int _{B(x_0,r)}|u_ j|\, \mathrm d x$ uniformly in $j$.

Let $Y$ be the set of points $y\in \Omega $ that have a neighborhood $U$ with $\sup _ j\int _ U|u_ j|\, \mathrm d x<\infty $. Then $Y$ is open and contains $x_0$. Let $y\in \Omega $ lie in the closure of $Y$. Choose $r>0$ with $\overline{B(y,4r)}\subset \Omega $ and an open ball $A$ with $\overline A\subset Y\cap B(y,r)$. Covering the compact set $\overline A$ by finitely many of the neighborhoods in the definition of $Y$, we find $C_ A=\sup _ j\int _ A|u_ j|\, \mathrm d x<\infty $. Put $C=C_ A/|A|$. For each $j$ there is a point $z_ j\in A$ with $u_ j(z_ j)\geq -C$, since otherwise $\int _ A|u_ j|\, \mathrm d x\geq \int _ A(-u_ j)\, \mathrm d x>C|A|=C_ A$. By (3.4.18), with $a=y$ and $z=z_ j$, the integrals $\int _{B(y,r)}|u_ j|\, \mathrm d x$ are bounded uniformly in $j$, so $y\in Y$. Thus $Y$ is closed in $\Omega $, and $Y=\Omega $ because $\Omega $ is connected. Finally, every compact subset of $\Omega $ is covered by finitely many of the neighborhoods in the definition of $Y$. $\square$

Proof. We prove (2) first, because (1) reduces to it. As observed after theorem 3.4.12, $T$ is real, $\Delta T\geq 0$, and $T$ has a canonical representative $u$. We begin by showing that $(u_ j)$ is bounded in $L^1(L)$ for every compact set $L\subset \Omega $. Fix such a set $L$, let $\chi $ be a nonnegative test function equal to one near $L$, and let $M\geq 0$ bound all $u_ j$ from above on $\operatorname{Supp}\chi $. Then $|u_ j|\leq 2M-u_ j$ on $\operatorname{Supp}\chi $, so

which is bounded because $\int u_ j\chi \, \mathrm d x$ converges.

Next we show that the mollifications $u_ j*\rho _\varepsilon $ converge locally uniformly. Fix $\varepsilon >0$. For $x\in \Omega _\varepsilon $, the function $\rho _\varepsilon (x-\cdot )$ is a test function, so $u_ j*\rho _\varepsilon (x)\to \langle T,\rho _\varepsilon (x-\cdot )\rangle =u*\rho _\varepsilon (x)$. Let $L\subset \Omega _\varepsilon $ be compact; then $L'=\{ y:\operatorname {dist}(y,L)\leq \varepsilon \} $ is a compact subset of $\Omega $, and the mean value theorem, applied to $\rho _\varepsilon $, gives

Thus the functions $u_ j*\rho _\varepsilon $ are $A$-Lipschitz on $L$, where $A=\sup |\nabla \rho _\varepsilon |\cdot \sup _ j\int _{L'}|u_ j|\, \mathrm dy$, and so is their pointwise limit $u*\rho _\varepsilon $. Given $\tau >0$, cover $L$ by finitely many balls $B(x_ i,\tau /(3A+1))$ with $x_ i\in L$, and choose $j_0$ such that $|u_ j*\rho _\varepsilon (x_ i)-u*\rho _\varepsilon (x_ i)|<\tau /3$ for all $i$ and all $j\geq j_0$. Then $|u_ j*\rho _\varepsilon -u*\rho _\varepsilon |<\tau $ on $L$ for $j\geq j_0$, so the convergence is uniform on $L$.

Now let $\chi $ be a nonnegative test function, choose $\varepsilon $ so small that $\operatorname{Supp}\chi \subset \Omega _\varepsilon $, let $\delta >0$, and put $g=u*\rho _\varepsilon +\delta $. By (3.4.13), $u_ j\leq u_ j*\rho _\varepsilon $ and $u\leq u*\rho _\varepsilon $ on $\Omega _\varepsilon $. By the uniform convergence just proved, $u_ j\leq g$ on $\operatorname{Supp}\chi $ for all large $j$; also $u\leq g$. Hence $|u_ j-u|\leq (g-u_ j)+(g-u)$ almost everywhere on $\operatorname{Supp}\chi $, and since $\int u_ j\chi \, \mathrm d x\to \int u\chi \, \mathrm d x$,

Let $\delta \downarrow 0$, and then $\varepsilon \downarrow 0$; since $u*\rho _\varepsilon \to u$ in $L^1$ on $\operatorname{Supp}\chi $ (lemma 2.5.1), the right-hand side tends to zero. Choosing $\chi $ equal to one on a prescribed compact set proves $L^1$ convergence on that set. This proves (2).

(1) Suppose that $u_ j$ does not tend to $-\infty $ uniformly on compact subsets. Then there are a compact set $K\subset \Omega $, a constant $C_0$ and infinitely many indices $j$ with $\sup _ Ku_ j>-C_0$, and for these $j$ we can choose $x_ j\in K$ with $u_ j(x_ j)>-C_0$. By lemma 3.4.17, a subsequence, which we again denote by $(u_ j)$, is bounded in $L^1(L)$ for every compact set $L\subset \Omega $; in particular, every $u_ j$ is locally integrable. By (2), it suffices to show that a further subsequence converges in the sense of distributions.

Choose open sets $U_ m\Subset \Omega $ increasing to $\Omega $, for instance the interiors of the sets of a compact exhaustion (section 2.2), and constants $M_ m\geq 0$ with $u_ j\leq M_ m$ on $\overline{U_ m}$ for all $j$. On $U_ m$, the Borel measures $(M_ m-u_ j)\, \mathrm d x$ are positive, and their masses on a compact set $L\subset U_ m$ are at most $M_ m|L|+\sup _ j\int _ L|u_ j|\, \mathrm dx$, uniformly in $j$; in particular, they are Radon measures by proposition 2.2.2. By theorem 2.6.5 and a diagonal extraction over $m$, we may assume that for each $m$ these measures converge vaguely on $U_ m$. Every $\chi \in C_ c(\Omega )$ has its support in some $U_ m$, so

converges. Let $T(\chi )$ denote the limit when $\chi $ is a test function. If $\operatorname{Supp}\chi \subset L$ with $L$ compact, then $|T(\chi )|\leq \sup _ j\int _ L|u_ j|\, \mathrm d x\cdot \sup |\chi |$. Hence $T$ is a distribution of order zero, and $u_ j\to T$ in the sense of distributions. $\square$

Proof. Convergence in $L^1_{\mathrm{loc}}$ implies convergence in the sense of distributions, because $|\int (u_ j-v)\chi \, \mathrm dx|\leq \sup |\chi |\int _{\operatorname{Supp}\chi }|u_ j-v|\, \mathrm dx$. Hence, as observed after theorem 3.4.12, the distribution defined by $v$ is real, has positive Laplacian and has a canonical representative $u$.

Let $x_ j\to x$, and choose $\varepsilon >0$ with $\overline{B(x,2\varepsilon )}\subset \Omega $. For $y\in B(x,\varepsilon )$ we have

so $u_ j*\rho _\varepsilon \to u*\rho _\varepsilon $ uniformly on $B(x,\varepsilon )$. For large $j$ we have $x_ j\in B(x,\varepsilon )$. Since $u_ j\leq u_ j*\rho _\varepsilon $ by (3.4.13) and $u*\rho _\varepsilon $ is continuous, we obtain

Letting $\varepsilon \downarrow 0$ and using (3.4.13) again proves the first assertion.

For the second assertion we may assume $K\neq \varnothing $. Since $u$ and $-f$ are upper semicontinuous on the compact set $K$, they are bounded above there, so $s=\sup _ K(u-f)$ lies in $[-\infty ,\infty )$. Suppose that $\varlimsup _ j\sup _ K(u_ j-f)>c>s$ for some real $c$. Then $\sup _ K(u_ j-f)>c$ for infinitely many $j$, so along a subsequence there are points $x_ j\in K$ with $u_ j(x_ j)-f(x_ j)>c$, and after passing to a further subsequence we may assume $x_ j\to x\in K$. Since $\varliminf _ jf(x_ j)\geq f(x)$ by lower semicontinuity, the first assertion, applied to this subsequence, gives

a contradiction. Finally, if $u<f$ on $K$, then the upper semicontinuous function $u-f$ attains a negative maximum on $K$, so $\sup _ K(u-f)<0$, and the second assertion gives $\sup _ K(u_ j-f)<0$ for all large $j$. $\square$

Proof. Call $B(q,s)$ a rational ball if $q\in \mathbb Q^ d$ and $s$ is a positive rational number. Enumerate the rational balls $B$ with $\overline B\subset \Omega $ as $B_1,B_2,\ldots $, in such a way that each of them occurs infinitely often. Since $F$ is bounded above on the compact set $\overline{B_ j}$, we have $\sup _{B_ j}F<\infty $. If $\sup _{B_ j}F>-\infty $, choose $v_ j\in \mathcal F$ with $\sup _{B_ j}v_ j>\sup _{B_ j}F-1/j$; otherwise let $v_ j$ be any member of $\mathcal F$. Put $v=\sup _ jv_ j\leq F$. If $B$ is a rational ball with $\overline B\subset \Omega $, then $B=B_ j$ for infinitely many $j$, so $\sup _ Bv=\sup _ BF$. Every ball $B(x,r)$ contains a rational ball that contains $x$: choose $q\in \mathbb Q^ d$ with $|q-x|<r/3$ and a rational number $s\in (r/3,2r/3)$; then $x\in B(q,s)\subset B(x,r)$. Conversely, a rational ball $B(q,s)$ containing $x$ contains $B(x,s-|q-x|)$. Hence $F^*(x)$ is the infimum of $\sup _ BF$ over the rational balls $B\ni x$ with $\overline B\subset \Omega $, and similarly for $v^*(x)$. Therefore $v^*=F^*$. $\square$

Proof. If the family is empty, then $F=F^*=-\infty $. Otherwise, we replace the family by an increasing sequence, using Choquet's lemma, and identify the regularized limit of this sequence with a canonical representative. The connected components of $\Omega $ are open, because balls are connected; hence they are pairwise disjoint sets of positive Lebesgue measure, and there are at most countably many of them by lemma 2.2.5. As a countable union of null sets is null, it suffices to prove the assertions on each component $\Omega '$. On $\Omega '$ the regularization $F^*$ coincides with the regularization of $F|_{\Omega '}$ in $\Omega '$, because small balls about a point of $\Omega '$ lie in $\Omega '$. If every $u_ t$ is identically $-\infty $ on $\Omega '$, then $F=F^*=-\infty $ there.

Otherwise fix $t_0$ such that $u_{t_0}$ is nontrivial, hence locally integrable, on $\Omega '$. By lemma 3.4.22, applied on $\Omega '$, there are indices $t_1,t_2,\ldots $ with $(\sup _{k\geq 1}u_{t_ k})^*=F^*$ on $\Omega '$. Put $v_ j=\max \{ u_{t_0},u_{t_1},\ldots ,u_{t_ j}\} $ and $v=\sup _ jv_ j$. Since $\sup _{k\geq 1}u_{t_ k}\leq v\leq F$, we still have $v^*=F^*$ on $\Omega '$. The functions $v_ j$ are subharmonic, increase with $j$, are uniformly bounded above on compact sets and satisfy $v_ j\geq u_{t_0}$; in particular, they are locally integrable on $\Omega '$. If $K\subset \Omega '$ is compact and $M$ bounds all $v_ j$ from above on $K$, then $0\leq M-v_ j\leq M-u_{t_0}$ on $K$ with $M-u_{t_0}$ integrable, and $M-v_ j$ decreases to $M-v$. Dominated convergence shows that $v$ is locally integrable and $v_ j\to v$ in $L^1_{\mathrm{loc}}(\Omega ')$, hence in the sense of distributions. As observed after theorem 3.4.12, $v$ therefore has a canonical representative $w$: a subharmonic function on $\Omega '$ with $w=v$ almost everywhere.

We claim that $w=v^*$. For each $j$ we have $v_ j\leq v$, so $v_ j*\rho _\varepsilon \leq v*\rho _\varepsilon =w*\rho _\varepsilon $, and letting $\varepsilon \downarrow 0$ in (3.4.13) gives $v_ j\leq w$ everywhere. Hence $v\leq w$, and $v^*\leq w$ because $w$ is upper semicontinuous. Conversely, by (3.4.13), $w(x)$ is the limit of the averages of $w$ over $B(x,r)$; these coincide with the averages of $v$, which are at most $\sup _{B(x,r)}v$. Letting $r\downarrow 0$ gives $w(x)\leq v^*(x)$. Thus $F^*=v^*=w$ is subharmonic on $\Omega '$.

Finally, $v\leq F\leq F^*=w$ on $\Omega '$, and $v=w$ almost everywhere. The functions $v$ and $w$ are Borel measurable, so the set where $F\neq F^*$ is contained in the Borel null set $\{ v\neq w\} $. Since Lebesgue measure is complete, $F$ is Lebesgue measurable and $F=F^*$ almost everywhere. $\square$

Proof. Put $\widetilde u=u$ on $\Omega \setminus A$, and define $\widetilde u$ on $A$ by the displayed formula. Since $A$ has Lebesgue measure zero, it has no interior points, so every point of $A$ is a limit of points of $\Omega \setminus A$; together with the local upper bound for $u$, this shows that $\widetilde u(x)\in [-\infty ,\infty )$ is well defined for $x\in A$. We first show that $\widetilde u$ is upper semicontinuous and locally bounded above. On the open set $\Omega \setminus A$ both properties hold because $\widetilde u=u$ there. Let $x\in A$ and $\overline{B(x,r)}\subset \Omega $. For $y\in A\cap B(x,r)$, the definition gives $\widetilde u(y)\leq \sup _{B(x,r)\setminus A}u$, hence

The right-hand side is bounded above by hypothesis and tends to $\widetilde u(x)$ as $r\downarrow 0$, which proves both claims at $x$.

Next we show that $\widetilde u$ is subharmonic near each point $x\in A$; near points of $\Omega \setminus A$ this holds because $\widetilde u=u$. The idea is to approximate $u$ from below by functions $u+\varepsilon \psi $ that tend to $-\infty $ at $A$. Let $V$ and $\psi $ be as in definition 3.4.26. By proposition 3.4.8, $\psi $ is locally integrable on the connected set $V$, so it is not identically $-\infty $ on any ball in $V$, and its pole set $\{ \psi =-\infty \} $ is a null set. Replacing $V$ by a ball centered at $x$ whose closure lies in $V$, and $\psi $ by $\psi $ minus an upper bound on that ball, we may therefore assume that $V$ is a ball, that $\psi \leq 0$ on $V$, and that $\psi $ is nontrivial on $V$.

For $\varepsilon >0$ define $v_\varepsilon =u+\varepsilon \psi $ on $V\setminus A$ and $v_\varepsilon =-\infty $ on $A\cap V$. On the open set $V\setminus A$, $v_\varepsilon $ is subharmonic. At a point $y\in A\cap V$, $u$ is bounded above near $y$, and $\psi (z)\to -\infty $ as $z\to y$ because $\psi $ is upper semicontinuous with $\psi (y)=-\infty $; hence $v_\varepsilon (z)\to -\infty $ as $z\to y$. So $v_\varepsilon $ is upper semicontinuous at $y$, and the submean inequality at $y$ holds because $v_\varepsilon (y)=-\infty $. Thus $v_\varepsilon \in \mathrm{SH}(V)$.

As $\varepsilon \downarrow 0$, the functions $v_\varepsilon $ increase because $\psi \leq 0$, and they are bounded above by $u$ on $V\setminus A$, hence uniformly bounded above on compact subsets of $V$. By theorem 3.4.23, the regularized supremum $w=(\sup _{\varepsilon >0}v_\varepsilon )^*$ is subharmonic on $V$ and equals $\sup _{\varepsilon >0}v_\varepsilon $ almost everywhere. This supremum equals $u$ on $V\setminus (A\cup \{ \psi =-\infty \} )$, so $w=u$ almost everywhere on $V\setminus A$. Since $w$ and $u$ are subharmonic on the open set $V\setminus A$, they agree everywhere there by theorem 3.4.12(2).

It remains to show that $w=\widetilde u$ on $A\cap V$. Let $y\in A\cap V$. Since $w$ is upper semicontinuous and $w=u$ on $V\setminus A$, we have $w(y)\geq \varlimsup _{V\setminus A\ni z\to y}u(z)=\widetilde u(y)$. Conversely, by (3.4.13), $w(y)$ is the limit of the averages of $w$ over $B(y,r)$. Since $A$ is null and $w=u$ off $A$, these averages are at most $\sup _{B(y,r)\setminus A}u$, which tends to $\widetilde u(y)$ as $r\downarrow 0$. Hence $\widetilde u=w$ on $V$, and $\widetilde u$ is subharmonic on $V$. This proves that $\widetilde u\in \mathrm{SH}(\Omega )$.

For uniqueness, let $u'\in \mathrm{SH}(\Omega )$ agree with $u$ on $\Omega \setminus A$. Since $A$ is null, $u'=\widetilde u$ almost everywhere on $\Omega $, and hence everywhere by theorem 3.4.12(2).

Finally, let $\Omega $ be a domain, and suppose that $\Omega \setminus A$ is the union of two disjoint nonempty open sets $W_0$ and $W_1$. The function $g$ equal to $0$ on $W_0$ and to $1$ on $W_1$ is harmonic, hence subharmonic, and bounded on $\Omega \setminus A$. Its subharmonic extension $\widetilde g$ to $\Omega $ takes values in $[0,1]$ and equals $1$ on $W_1$, so it attains the finite maximum $1$. By proposition 3.4.7, $\widetilde g$ is identically $1$ on $\Omega $, contradicting $\widetilde g=0$ on $W_0$. $\square$