According to Wikipedia, the $L_p$-norm is not subadditive when $p\in(0,1)$. How can I show that the map $n_p(f)=(\int_0^1|f(x)|^p~\mathrm{d}x)^{2p}$ is not subadditive for $f\in C[0,1]$ for $0 ?
$L_p$ norm not subadditive for 0 when endowed on $C[0,1]$
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functional-analysis
measure-theory
banach-spaces
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3Is the power of the integral purposely $2p$ or should it be $\frac{1}{p}$? – 2012-09-11
1 Answers
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You can show this by just posing an counterexample,for instance,take $f(x)=x$and $g(x)=1-x$. For $p=\frac{1}{2}$,then one has $\left(\int_{0}^{1}|f+g|^{\frac{1}{2}}dx\right)^2=1>\frac{8}{9}=\left(\int_{0}^{1}|f|^{\frac{1}{2}}dx\right)^2+\left(\int_{0}^{1}|g|^{\frac{1}{2}}dx\right)^2$ Another result for $L^p$,$0 is that it has no non-zero continuous linear functional, so people rarely use them in analysis.
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0@Norbert: this is simplier,thanks – 2012-09-12