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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.

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    $u^+_n\rightharpoonup u^+$ 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.2012-11-15
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    Yes @Elias, it is.2012-11-15

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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 $n\ge 1$.

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    Can you give the proof that $u_n^+$ doesn't converge weakly to $0$?2012-11-15
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    @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?2012-11-15
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    @DavideGiraudo: I thought this is more or less obvious, so I omitted the proof. Is it necessary to fill up the details?2012-11-15
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    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.2012-11-15
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    If $1_{A_n}\rightharpoonup 1_A$ and $u_n\rightharpoonup u$ is not true that $u^+ \rightharpoonup u^+$? Here $A=\{x \in \Omega : u(x)\geq 0\}$ and $A_n=\{x \in \Omega : u_n(x)\geq 0\}$.2012-11-15
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    @DavideGiraudo: Please see my updated answer.2012-11-15
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    How did you compute the integral? (I got, writing $|\sin|\geq \sin^2$ that the integral is $\geq \pi$, which is enough to conclude)2012-11-15
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    @Elias: I am not sure what you mean. Are you asking that if both $u_n\rightharpoonup u$ and $\mathbf{1}_{A_n}\rightharpoonup \mathbf{1}_A$ hold, can we conclude $u_n^+\rightharpoonup u^+$?2012-11-15
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    @DavideGiraudo: Yes, you are right. However, since the integral is easy to be evaluated, I prefer to use equality. Note that $|\sin nx|$ is of period $\frac{1}{n}$, or let $y=nx$ be a new variable.2012-11-15
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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.

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    Your answer does not make sense Elias.2012-11-15
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    @Tomás I edit may answer.2012-11-15