# Convergence of a sequence of solutions

Suppose that $(v_n)$ is a sequence of solutions of $$\begin{cases}-\Delta v_n = f_n&\text{in }\Omega\\ v_n = 0&\text{on }\partial \Omega,\end{cases}$$ where $\Omega \subset \mathbb{R}^2$ is a bounded domain and $f_n$ is a sequence of measurable functions converging to $0$ in measure. Can one infer that $v_n$ converges uniformly to $0$?

• Why 0? Wouldn't you except rather that they converge to the solution of $-\Delta v= f$ with $v=0$ on $\partial\Omega$? – Fabian Sep 5 '12 at 16:54
• The question is rather vague. How do you define $v_n$ to be a solution? I guess you need some elliptic regularity, but you give no starting point. Something more should be known about $f_n$. – Siminore Sep 5 '12 at 17:08
• For a sequence of solutions, I mean a sequence $(v_n)\subset L^1_{loc}(\Omega)$ such that $-\int_{\Omega}{v_n\Delta \varphi} = \int{f_n \varphi}$ for all $\varphi \in C^{\infty}_0(\Omega)$ – Jaques Sep 5 '12 at 17:22
• And I wrote $f$ instead of $0$... Now I've corrected it – Jaques Sep 5 '12 at 17:25

I am guessing the answer is No. Consider the Dirac delta as $f_n$'s limit.
If $v_n$ converges uniformly to $0$, then
$$\left|\int_{\Omega} v_n \Delta \varphi\right| \leq \sup_{x\in \Omega} |v_n(x)| \int_{\Omega} |\Delta \varphi|\rightarrow 0$$
But the right side, for $y\in \Omega$ let $f_n(x) = n^2\max\{1-n|x-y|,0\}$, this converges to $0$ in measure, also converges to Dirac delta in measure, and the approximating to identity construction would gave us:
$$\lim_{n\to \infty} \int_{\Omega} f_n\varphi = \varphi(y)$$ which implies the right side limit may not be zero, contradiction.