Is it true that if $\Omega$ is an open, bounded subset of $R^{N}$, $u_{n} \to u$ almost everywhere, $1 , 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
Does almost everywhere convergence imply convergence in $L^p$?
-
0If we add the hypothesis that $\lim_{n \rightarrow \infty} \Vert u_n \Vert_{p} = \Vert u \Vert_{p}$, can we build a dominating function? ā 2014-04-14
2 Answers
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 $\Omega$.
-
0@Matt thanks much ā 2011-12-11
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 $x$.
-
0Let $g$ 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. ā 2011-12-11