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Let $\Omega$ $\subseteq$ $\Bbb R^{n}$ be bounded and $u$ be a measurable function on $\Omega$ Such that $|u|^{p} \in {L^{1}_{\mathrm{loc}}(\Omega)}$ for some $p \in \Bbb R$. Then how to show the following?

$$\lim_{p\rightarrow0} \left(\frac{1}{|\Omega|}\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac{1}{p}} = \exp \left(\frac{1}{|\Omega|}\int_{\Omega}\log|u|\,\mathrm{d}x\right)$$

I attempted as follows:

$$\lim_{p\rightarrow0} \left(\frac{1}{|\Omega|}\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac1p} = \lim_{p\rightarrow0} \frac{1}{|\Omega|^\frac{1}{p}}\left(\int_{\Omega}|u|^{p}\,\mathrm{d}x\right)^{\frac1p} = \exp \lim_{p\rightarrow0} \frac{1}{|\Omega|^\frac1p} \log\left(\int_{\Omega} |u|^{p}\,\mathrm{d}x\right)^{\frac1p}= \exp\left(\frac{1}{|\Omega|}\int_{\Omega}\log|u|\,\mathrm{d}x\right)$$ However;I think some steps especially the last two are not satisfactory for me.If you see that you're well come!

Jean-Claude Arbaut
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user176456
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1 Answers1

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Assuming that limits and integrals can be interchanged you can use l 'Hopitâl: \begin{eqnarray*} f(p)^{1/p} &=&\exp [\frac{1}{p}\ln f(p)] \\ \lim_{p\downarrow 0}f(p)^{1/p} &=&\exp [\lim_{p\downarrow 0}\frac{1}{p}\ln f(p)]=\exp [\lim_{p\downarrow 0}\partial _{p}f(p)] \end{eqnarray*} \begin{eqnarray*} f(p) &=&\frac{1}{\Omega }\int_{\Omega }dx|u|^{p} \\ \partial _{p}f(p) &=&\frac{1}{\Omega }\int_{\Omega }dx\partial _{p}\exp [p\ln |u|]=\frac{1}{\Omega }\int_{\Omega }dx\{\ln |u|\}\exp [p\ln |u|] \\ &=&\frac{1}{\Omega }\int_{\Omega }dx\{\ln |u|\}|u|^{p} \\ \lim_{p\downarrow 0}(f(p)^{1/p} &=&\lim_{p\downarrow 0}(\exp [\ln f(p)])^{1/p}=\lim_{p\downarrow 0}(\exp [\frac{1}{p}\ln f(p)]) \\ \lim_{p\downarrow 0}\frac{1}{p}\ln f(p) &=&\lim_{p\downarrow 0}\partial _{p}\ln f(p)=\lim_{p\downarrow 0}\frac{1}{f(p)}\partial _{p}f(p)=\lim_{p\downarrow 0}\partial _{p}f(p) \\ &=&\frac{1}{\Omega }\int_{\Omega }dx\ln |u| \end{eqnarray*}

Urgje
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