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Is it true that if $\Omega$ is an open, bounded subset of $R^{N}$, $u_{n} \to u$ almost everywhere, $1<p<\infty$, then $\|u_{n} - u\|_{L^{p}} \to 0$? Here sequence and the function are in $L^{p}$

My proof uses dominated convergence theorem to $v_{n}(x) = |u_{n}(x) - u(x)|^{p}.$ and the fact that $\Omega$ is bounded.

thanks

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How do you get a function which dominate your sequence? The the result you're trying to show is not true: put $\Omega:=(0,1)$ and $f_n(x)=n\mathbf 1_{(0,1/n)}$. We have the almost everywhere convergence to $0$, but $\lVert f_n\rVert_p=n^{2-\frac 1p}$. The converse is not true, but we can show that the $L^p$ convergence imply the existence of an almost everywhere converging subsequence. –  Davide Giraudo Dec 11 '11 at 19:48
    
If we add the hypothesis that $\lim_{n \rightarrow \infty} \Vert u_n \Vert_{p} = \Vert u \Vert_{p}$, can we build a dominating function? –  user142978 Apr 14 at 4:14

2 Answers 2

For $\Omega=(0,2)$ and $N=1$, take $u_n=n\chi_{(0,1/n]}$. Then $u_n$ converges to $0$ almost everywhere, but $ ||u_n-0||_p=n \cdot (1/n)^{1/p}=n^{1-{1\over p}}$ for all $n$.

Similar examples can be constructed for other $\Omega$.

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What is wrong with my proof: Obviously, $v_{n} ->0$ a.e., further, $v_{n} \leq 2^{p}(|u|^{p} + (\varepsilon + |u|)^{p})$ for n large enough. –  Terr Dec 11 '11 at 19:51
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See Davide's comment. You don't necessarily have a dominating function just knowing pointwise convergence. –  David Mitra Dec 11 '11 at 19:54
    
still unclear to me –  Terr Dec 11 '11 at 19:56
    
@DavidMitra I think the first $n$ in your first sentence should be $N$. –  Rudy the Reindeer Dec 11 '11 at 20:00
    
@Matt thanks much –  David Mitra Dec 11 '11 at 20:03

I will give a counter-example for an arbitrary open $\Omega$. Take $x_0\in\Omega$; for $n$ large enough $B(x_0,n^{-1})\subset \Omega$. Put $f_n(x):=n^{\frac Np+1}\mathbf 1_{B(x_0,n^{-1})}(x)$. $f_n(x)$ converges to $0$ except at $x_0$ which has $0$ measure, but $$\lVert f_n\rVert_p=n^{\frac Np+1}m(B(x_0,n^{-1}))^{\frac 1p}=n^{\frac Np+1}\left(\frac 1{n^N}\right)^{\frac 1p}v_N=nv_N,$$ where $v_n$ is the measure of the unit ball of $\mathbb R^N$, so we can 't have the $L^p$ convergence.

But there is a link between $L^p$ and almost everywhere convergences: if $\{f_n\}\subset L^p$ is a sequence which converges in $L^p$, we can extract an almost everywhere converging subsequence.

Your proof fails because in general we can't find an integrable function $g$ such that $|f_n|\leq g$ for all $n$. In my example, the we can try $g(x)=\sum_{n\geq 1}n^{\frac Np+1}\mathbf 1_{B(x_0,n^{-1})\setminus B(x_0,(n+1)^{-1})}$, but $$\int_{\Omega}|g¬^pd\lambda =\sum_{n\geq 1}n^{N+p}v_N\left(\frac 1{n^N}-\frac 1{(n+1)^N}\right)=v_N\sum_{n\geq 1}n^p\left(1-\frac 1n\right),$$ which is not convergent.

When you try to bound $|u_n-u|$, there is a problem, since the "$n$ large enough" depends on the point $x$. It's possible we can't take the same for almost every $x$.

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examples make sense, but i don't see why my proof fails –  Terr Dec 11 '11 at 20:12
    
Let $g$ a function such that $|f_n|\leq g$ almost everywhere and for all $n$. Then on the ball of radius $1$, we should have $g\geq 1$, on the ball of radius $2^{-1}$, $g\geq 2^{-1-\frac Np}$, ... so the smallest $g$ we can take is $\sum_{n=1}^{\infty}n^{\frac Np+1}\mathbf 1_{B(x_0,n^{-1})\setminus B(x_0,(n+1)^{-1})}$. Check that such a function is not integrable. So in general you cannot apply the dominated convergence theorem. –  Davide Giraudo Dec 11 '11 at 20:18

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