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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

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    Is the power of the integral purposely $2p$ or should it be $\frac{1}{p}$?2012-09-11

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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

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    But neither $f$ nor $g$ is continuous.2012-09-11
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    @ShanLinHuang I've slightly edited your answer because we deal with continuous functions2012-09-12
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    @Norbert: this is simplier,thanks2012-09-12