## Convergence of Cauchy sequence of $L_1$-functions [closed]

Let $\Omega \subset \mathbb{R}^n$ be an open subset. Given a Cauchy sequence $u: \mathbb{N} \rightarrow C^2(\Omega)$ of harmonic functions on $\Omega$, one has $\forall x \in \Omega , B(x,r) \subset \Omega: |u_j - u_m|(x) \leq \frac{1}{\omega _n r^n}\|u_j - u_m\| _{L_1(\Omega)}$, so $u_j (x)$ is a Cauchy sequence for all $x \in \Omega$ and we can define $u = \lim u_j$ as a pointwise limit. Why is $u \in L_1$ and why does $u_j \rightarrow u$ in $L_1$? This is used in a proof in PDE and seems kind of obvious, but I couldn't find a simple argument.

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You probably forgot to assume that $\Omega$ is bounded or that $u_j \in C^2 \cap L^1$, otherwise you can take constant functions. I recommend you to ask this question on math.stackexchange.com where you'll certainly get a good answer. Here it is a bit off-topic, as this site is intended for "research level" questions. – Theo Buehler Jul 25 2011 at 7:52
Cauchy in which norm? – Spencer Jul 25 2011 at 12:05
Sorry, $u$ is a Cauchy sequence in the $L_1(\Omega)$ norm (so the $u_j$ are automatically $L_1$). – Karl Wedel Jul 25 2011 at 16:45
then where's the problem? You may see it like that: any Cauchy sequence in $L^1(\Omega)$ certainly converges in $L^1$ to a function $v\in L^1(\Omega)$ by completeness, and up to subsequences, also a.e. So the limit coincides a.e. with any other point-wise limit $u$. – Pietro Majer Jul 25 2011 at 19:32