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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, 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

  • 3
    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.2011-12-11
  • 0
    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?2014-04-14

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