Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$u^+_n\rightharpoonup u^+\,?$$


share|cite|improve this question
$u^+_n\rightharpoonup u^+$ 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. – MathOverview Nov 15 '12 at 15:39
Yes @Elias, it is. – Tomás Nov 15 '12 at 15:47
up vote 1 down vote accepted

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 $n\ge 1$.

share|cite|improve this answer
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? – MathOverview 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.

share|cite|improve this answer
Your answer does not make sense Elias. – Tomás Nov 15 '12 at 16:49
@Tomás I edit may answer. – MathOverview Nov 15 '12 at 17:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.