Let $(X,S,\mu )$ a measure speace, f measurable function, $E=\left\{ { p\ge 1 }|{ \int { { |f| }^{ p }d\mu } <\infty } \right\} $ and $\varphi (p)=\int { { |f| }^{ p } } d\mu $ then $\varphi (p)$ is continuous at the interior of E?
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Yes, it is. You can prove that $E$ is an interval, and that $\log\varphi$ is convex in the interior of $E$. – Rigel May 10 '17 at 17:23
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https://math.stackexchange.com/questions/1880656/log-convexity-of-the-p-norms-of-a-fixed-function – Jonas Dahlbæk May 10 '17 at 21:09
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I've already proved that $log(\varphi)$ is convex, but how can i use that for prove that $\varphi$ is continuous ? – Parcos May 11 '17 at 01:36
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@PacoFrancisco, convex in an open interval $\implies$ continuous. See https://math.stackexchange.com/questions/258511/proof-of-every-convex-function-is-continuous. – Martín-Blas Pérez Pinilla May 11 '17 at 15:10