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Suppose that $f,u,u_n\in L^p(\Omega)$ where $\Omega\subset\mathbb{R}^N$ is a bounded domain and $f,u,u_n\geq 0$. Suppose $$\|u+u_n\|_p\rightarrow \|u+f\|_p,$$ $$u_n\rightarrow f\text{ a.e. in } \Omega,$$ and $$u_n>f.$$

Does this implies that $$\|u_n-f\|_p\rightarrow0\,?$$

Thanks.

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Do you mean $\|u_n\|_p\to\|f\|_p$? –  23rd Nov 12 '12 at 17:41
    
yes its @richard –  Tomás Nov 12 '12 at 17:41

1 Answer 1

up vote 2 down vote accepted

If $\|u\|_p=\infty$, it is not necessarily true. For example, $N=1$, $\Omega=[0,1]$, $u_n=n\cdot\mathbf{1}_{[0,\frac{1}{n}]}$ and $f=0$.

If $\|u\|_p<\infty$, it it true without assuming that $u_n>f$. The proof is as follows. Let $$g_n=2^{p-1}(|u+u_n|^p+|u+f|^p)-|u_n-f|^p\ge 0,$$ by Fatou's lemma and your assumptions,

$$2^p\|u+f\|_p^p=\int_\Omega \liminf_{n\to\infty} ~g_n\le \liminf_{n\to\infty}\int_\Omega g_n=2^p\|u+f\|_p^p-\limsup_{n\to\infty}\int_\Omega |u_n-f|^p.$$

The conclusion follows.

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Interesting. Now correct if im worng. With the assumption $\|u+u_n\|\rightarrow\|u+f\|$ and $u_n>f$, can i conclude that $u_n\rightarrow u$ almost everywhere? –  Tomás Nov 12 '12 at 18:26
    
@Tomás: I think you actually mean $u_n\to f$ a.e., but you cannot conclude that, because even $u_n>f$ and $\|u_n-f\|_p\to 0$ cannot imply a.e. convergence. Instead, you can conclude that $u_n\to f$ in measure. –  23rd Nov 12 '12 at 18:42
    
What is worng with this demonstration: Suppose $u_n$ does not converge to $f$ a.e.. Then there is a set $U$ of positive measure and a subsequence of $u_n$ (not relabeled) such that for some $\delta>0$, $u_n>\delta+f$. Hence, there exist some Constant $C_\delta>0$ depending on $\delta$, such that $(u+u_n)^p>|u+\delta+f|^p>C_\delta+(u+f)^p$ in $U$, therefore $\int_{\Omega}(u+u_n)^p=\int_{U}(u+u_n)^p+\int_{U^c}(u+u_n)^p>C+\int_{\Omega}(u+‌​f)^p$ whre $C>0$. Now using the hypothesis we get $\int_{\Omega}(u+f)^p\geq C+\int_{\Omega}(u+f)^p$, absurd. –  Tomás Nov 12 '12 at 18:56
    
@Tomás: $u_n>\delta+f$ does not hold. –  23rd Nov 13 '12 at 3:07
    
Yes you are right, thank you @richard –  Tomás Nov 13 '12 at 10:51

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