3.4.16 Compactness and Hartogs' lemmaAA374DAE79FD

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 $j$. $\square$