# 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.

#### Solutions Collecting From Web of "Weak Convergence of Positive Part"

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 $n\ge 1$.

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.