$\left\lVert x \right\rVert_p$ = ($\sum_{i=1}^{n} |x_i|^p$)$^{1/p}$ and triangle inequality

Question

How can I prove that for $0 < p < 1$ the defined function

$\left\lVert x \right\rVert_p$ = ($\sum_{i=1}^{n} |x_i|^p$)$^{1/p}$ for $n \geq 2$

does not fulfill the triangle inequality anymore?

I know that in one dimension, the norm is simply the absolute value and the triangle inequality holds. So here I could take two unit vectors.

For example, $\vec{a}$ = $(0, 1, 0)$ and $\vec{b}$ = $(2/3, 2/3, 1/3)$

But I don't really know how I can put the unit vectors into the sum above to show, that

$\left\lVert a + b \right\rVert$ $\leq$ $\left\lVert a\right\rVert$ + $\left\lVert b \right\rVert$ does not hold.

I have read this question yesterday night here, but it seems to be deleted, that's why I ask

Answer 1

You cannot add vectors in different dimensions. Your $a$ is in $\mathbb R^{2}$ and $b$ is in $\mathbb R^{3}$ so $a+b$ has no meaning. What you are asked to show is that if you fix $n \geq 2$ then we can have vectors $a$ and $b$ in $\mathbb R^{3}$ such that $\|a+b\| >\|a\|+\|b\|$. This is very easy: take the vectors $(1,0,0,...,0)$ and $(0,1,0,...,0)$ and verify that $\|a+b\|=2^{1/p}$, $\|a\|+\|b\|=2$. Since $p<1$ we have $2^{1/p} >2$.

Answer 2

Well, actually, this function is concave when you restrict your study to positive vectors (i.e., vectors with positive coordinates). If $\nabla^2 f(x)$ denotes the Hessian matrix of your function $f(x)=\left(\sum_{i=1}^n x_i^p\right)^{1/p}$ then, for any vector $\vec{u}$ you can show (after a lot of calculations) that $\vec{u}^{T}\nabla^2 f(x)\vec{u}\leq 0,$ hence the Hessian of $-f$ is positive semi-definite, so, by a known theorem, $-f$ convex, so $f$ is concave.