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.
If $v_n \to 0$ in $L^2$, does $v_n \to 0$ in $L^8$?
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functional-analysis
integration
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0Are you talking about a bounded domain in $\mathbb R^n$ with Lebesgue measure? – 2012-12-09
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1@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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2I presume you are speaking of sequences $(v_n)$? – 2012-12-09
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0@maximumtag That would be some good information to add to your original post... – 2012-12-09
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3$L^2$ 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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0@DavidMitra You're right, I was very careless. Sorry all. I edited the post. – 2012-12-09
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0@JonasMeyer Thanks. Is it ever true with Lebesgue measure? – 2012-12-09
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0@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