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 $0$?
Convergence of a sequence of solutions
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real-analysis
pde
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0And I wrote $f$ instead of $0$... Now I've corrected it – 2012-09-05
1 Answers
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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.