If $u_n$ (sequence of harmonic functions) converges weakly to $u$ in $L^2(\Omega)$, then $\Delta u = 0$ in $\Omega$.

Question

If a sequence of harmonic functions $u_n \rightharpoonup u$ (converges weakly) in $L^2(\Omega)$, then $\Delta u = 0$ in $\Omega$.

Recall a sequence of functions $f_n$ defined on an open set $\Omega$ is said to converge weakly in $L^2(\Omega)$ to a function $f$ if: $$\int f_n(x)\,g(x)\,dx \to \int f(x)\,g(x)\,dx \hspace{1cm} \forall g \in L^2 (\Omega).$$

My first thought is just to pass the limit using the Mean Value Theorem since if MVT holds that implies $u$ is harmonic. However, I don't think that works with 'weak convergence' with my definition above.

Answer 1

$$u_n \rightharpoonup u\implies \int_\Omega u_n(x)\phi(x)dx \to \int_\Omega u(x)\phi(x)dx \hspace{1cm} \forall \phi \in C^\infty_0 (\Omega).$$

in particular $$ \phi \in C^\infty_0 \implies \Delta \phi \in C^\infty_0 $$

Hence $$\int_\Omega u_n(x)\Delta\phi(x)dx \to \int_\Omega u(x)\Delta\phi(x)dx\Longleftrightarrow \int_\Omega (u_n(x)-u(x))\Delta\phi(x)dx \to 0\hspace{1cm} \forall \phi \in C^\infty_0 (\Omega).$$ But using integration by part or By derivative in the sense of distributions we know that $$\int_\Omega (u_n(x)-u(x))\Delta\phi(x)dx =\int_\Omega \Delta(u_n(x)-u(x))\phi(x)dx~~\\ =\int_\Omega -\Delta u(x)\phi(x)dx ~~\to 0~~~\forall~~ \phi \in C^\infty_0 (\Omega).$$ Since $$\Delta u_n =0$$

Hence, $$\int_\Omega -\Delta u(x)\phi(x)dx =0~~\forall~~ \phi \in C^\infty_0 (\Omega)\Longleftrightarrow \Delta u = 0~~\text{a.e on }~~\Omega$$

Answer 2

It uses the Wely's Lemma: https://en.wikipedia.org/wiki/Weyl%27s_lemma_(Laplace_equation) as follows:

If $u\in L^1_{loc}(\Omega)$ satisfies $\int_\Omega u \Delta \varphi dx=0$ for all $\varphi\in C_c^\infty(\Omega)$, then $u$ is smooth (possibly after redefining on a set of zero measure) and harmonic.

It suffices to show that $\int_\Omega u \Delta \varphi =0$. For any $\varphi$, we have $$ \int_\Omega u_n\Delta \varphi dx=0, $$ by integration and the property that $u_n$ is harmonic. Then by limiting above and using the weak convergence, we have $$ \int_\Omega u \Delta \varphi dx= \lim_{n\to\infty}\int_\Omega u_n\Delta \varphi dx=0. $$ For the proof of Wely's Lemma, please see: