$\left(\|u^p\|_{L^\infty}\right)^{1/p}=\|u\|_{L^\infty}$

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Quick question on that equality. Is this true $$\left(\|u^p\|_{L^\infty}\right)^{1/p}=\|u\|_{L^\infty}?$$

I would expect so (as the $L^\infty$ norm is an infimum or a supremum and inf/sup of product is product of inf/sup) but I don't know if I'm right.

Thank you!

asked 2017-02-09

user id: user

2

Yes: $$\inf\{t\geqslant0\mid\mu(\{x\mid |u(x)|\geqslant t\}=0\}=\inf\{t\geqslant0\mid\mu(\{x\mid |u(x)|^p\geqslant t\}=0\}^{1/p}$$ – 2017-02-09
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What should $u^p$ mean if $u$ can possibly take negative values? I would agree if it was $(\||u|^p\|_{L^\infty})^{1/p}$. – 2017-02-09

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