Theorem 3.4.20 (Hartogs' lemma).5784B80133BF Let $(u_ j)$ be a sequence of locally integrable subharmonic functions on $\Omega $ converging in $L^1_{\mathrm{loc}}(\Omega )$ to a function $v\in L^1_{\mathrm{loc}}(\Omega )$. Then $\Delta v\geq 0$ in the sense of distributions. Let $u$ be the canonical representative of $v$, that is, the locally integrable subharmonic function equal to $v$ almost everywhere. Then, for every $x\in \Omega $ and every sequence $x_ j\to x$ in $\Omega $,
\[ \varlimsup _ j u_ j(x_ j)\leq u(x). \]
For every compact set $K\subset \Omega $ and every lower semicontinuous function $f\colon K\to \mathbb R$, for instance every real continuous function on $K$,
\[ \varlimsup _ j\sup _ K(u_ j-f)\leq \sup _ K(u-f). \]
In particular, if $f\colon K\to \mathbb R$ is lower semicontinuous and $u<f$ on $K$, then $u_ j<f$ on $K$ for all sufficiently large $j$.
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
\[ |u_ j*\rho _\varepsilon (y)-u*\rho _\varepsilon (y)|\leq \sup \rho _\varepsilon \int _{B(x,2\varepsilon )}|u_ j-v|\, \mathrm d z, \]
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
\[ \varlimsup _ j u_ j(x_ j)\leq \varlimsup _ ju_ j*\rho _\varepsilon (x_ j)=u*\rho _\varepsilon (x). \]
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
\[ c\leq \varlimsup _ j\bigl (u_ j(x_ j)-f(x_ j)\bigr )\leq \varlimsup _ ju_ j(x_ j)-\varliminf _ jf(x_ j)\leq u(x)-f(x)\leq s<c, \]
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 $j$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.