3
$\begingroup$

If $\{u_n:\Omega\to \mathbb{R}\}$ is a continuous function sequence such that $u_n>0$ for all $n\in \mathbb{N}$ and suppose that $u_n$ converges weak to $u \in L^2(\Omega)$. Can I obtain $u>0$?

$\endgroup$

2 Answers 2

8
$\begingroup$

No, just take $\Omega=(0,1)$ and $u_n=\frac{1}{n}$. Then $$ ||u_n-0||_{L^2}^2=\int_0^1 \frac{1}{n^2}dx=\frac{1}{n^2} \to 0 $$ and hence $u_n \to 0$ strongly, and hence weakly, in $L^2$.

$\endgroup$
6
$\begingroup$

Not really. For example, let $$ u_n=x^n, x\in(0,1) $$ and hence $$ \|u_n-0\|_2^2=\int_0^1|x^n-0|^2=\frac1{2n+1}\to 0. $$ Namely $u_n$ converges to $0$ strongly in $L^2(0,1)$ and hence weakly to $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .