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If $v_n \to 0$ in $L^2$, does $v_n \to 0$ in $L^8$? Suppose the domain is a compact surface in $\mathbb{R}^n$. For example it could be a sphere.

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    Are you talking about a bounded domain in $\mathbb R^n$ with Lebesgue measure?2012-12-09
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    @JonasMeyer Yeah a bounded compact domain in $\mathbb{R}^n$, it can be a surface for example. I assume the measure is the usual Lebesgue measure.2012-12-09
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    I presume you are speaking of sequences $(v_n)$?2012-12-09
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    @maximumtag That would be some good information to add to your original post...2012-12-09
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    $L^2$ is not a subset of $L^8$ in that case. Consider, on $[0,1]$, $v_n=n\chi_{[0,1/n^4]}$. The converse can be proven for bounded domains using Hölder's inequality.2012-12-09
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    @DavidMitra You're right, I was very careless. Sorry all. I edited the post.2012-12-09
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    @JonasMeyer Thanks. Is it ever true with Lebesgue measure?2012-12-09
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    @maximumtag: The only thing that matters about the interval $[0,1/n^4]$ in that example is its measure. You can generalize to other domains.2012-12-09

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