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I see that it fails to satisfy the triangle inequality by example but I don't see how to prove this is the case for all $0 < p < 1$. The definition I am using for $p$-norm is $$ \|A\|_p= \left(\sum_{k=1}^{n} |x_k|^p\right)^{1/p}.$$

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2 Answers 2

up vote 11 down vote accepted

Consider $(1,0,0,\ldots, 0)+(0,1,0,\ldots,0) = (1,1,0,\ldots,0)$.

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Yes, a short counter-example.

assume that $n=2$,take vectors $(1,0)$ and $(0,1)$.

you'll find it doesn't satisfy triangle inequality.

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