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}.$
How do you prove the $p$-norm is not a norm in $\mathbb R^n$ when 0?
12
$\begingroup$
real-analysis
norm
examples-counterexamples
lp-spaces
2 Answers
23
Consider $(1,0,0,\ldots, 0)+(0,1,0,\ldots,0) = (1,1,0,\ldots,0)$.
6
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.