Weak Convergence of Positive Part

Suppose $\Omega\subset\mathbb{R}^n$ is a bounded domain and $p\in (1,\infty)$. Suppose $u_n\in L^p(\Omega)$ is such that $u_n\rightharpoonup u$ in $L^p(\Omega)$. Define the positive part of $u$ by $u^+=\max(u,0)$. Is it true that $$u^+_n\rightharpoonup u^+\,?$$

Thanks.

-
$u^+_n\rightharpoonup u^+$ if, only if, $F(u^+_n)\to F(u^+)$ for all linear functinal $F:L^p(\Omega)\to\mathbb{R}$ continuls. I supose that you call of weak topology. –  Math_overview Nov 15 '12 at 15:39
Yes @Elias, it is. –  Tomás Nov 15 '12 at 15:47

It is false. For example, $n=1$, $\Omega=[0,2\pi]$, $p=2$, $u_n(x)=\sin nx$ and $u=0$.

Remark: Suppose that $u_n^+\rightharpoonup u^+$, i.e $u_n^+\rightharpoonup 0$. Then $|u_n|=2u_n^+-u_n\rightharpoonup 0$, which is absurd, because $\int_0^{2\pi}|\sin nx|dx=4$ for every $n\ge 1$.

-
Can you give the proof that $u_n^+$ doesn't converge weakly to $0$? –  Davide Giraudo Nov 15 '12 at 16:03
@richard If $A_n=\{ x \in [0,2\pi] : u_n(x)=\sin(nx)\geq 0\}$ if $1_{A_n}$ converge weakly to $1_\{ x \in [0,2\pi] : u_(x)\geq 0\}$ this not true? –  Math_overview Nov 15 '12 at 16:12
@DavideGiraudo: I thought this is more or less obvious, so I omitted the proof. Is it necessary to fill up the details? –  23rd Nov 15 '12 at 16:13
I'm not sure it's obvious, except I'm missing something. So at least give an argument showing weak convergence of $\{u_n^+\}$ to $0$ doesn't hold. –  Davide Giraudo Nov 15 '12 at 16:15
@DavideGiraudo: Please see my updated answer. –  23rd Nov 15 '12 at 16:23

Let's $A_n=\{ x\in\Omega : u_n(x) \geq 0\}$ and $A=\{ x\in\Omega : u(x) \geq 0\}$. We have $u_n^+(x)=u_n(x)\cdot 1_{A_n}$ and $u^+(x)=u(x)\cdot 1_A$.

Then $u_n\rightharpoonup u$ and $1_{A_n}\rightharpoonup 1_A$ implies $u_n^+\rightharpoonup u^+$. By cause in all metric space if $g_n\to g$ and $h_n\to h$ we have $g_n\cdot h_n \to g\cdot h$.

Now if we not have $1_{A_n}\rightharpoonup 1_A$ then $weak\,lim 1_{A_n}\neq 1_A$ end $|weak\,lim 1_{A_n}-1_A|=1$. And for all linear fuctional $F:L^p\to \mathbb{R}$ we have $$|F(u_n^+)-F(u^+)|=|F(u_n^{+})-F(u_n\cdot 1_{A})+F(u_n\cdot 1_{A})-F(u^+)|$$ and implies for $N$ big $$|F(u_n^+)-F(u^+)|\geq |F(u_n\cdot 1_{A_n})-F(u_n\cdot 1_A)|= |F(u_n)|\cdot| 1_A-1_{A_n}|, \quad \forall n> N$$ If $weak\,lim \inf\{u_n(x): x\in\Omega\}\neq 0$ not have convergence.

-
Your answer does not make sense Elias. –  Tomás Nov 15 '12 at 16:49
@Tomás I edit may answer. –  Math_overview Nov 15 '12 at 17:06