What is $\|\cdot \|_p$ for $0<p<1$?

Question

1) Is $$\|(x,y) \|_p=(|x|^p+|y|^p)^{1/p}$$ a norm for $0<p<1$ ? I can prove that $\|\lambda (x,y)\|_p=|\lambda |\|x\|_p$ and $\|x\|_p=0\iff x=0$. But I have difficulties to prove the triangle inequality.

2) Is this a type of non-convex norm ? Or non convex distance (if we set $d(x,y)=\|(x,y)\|_p$) ?

Answer 1

If $p\in(0,1)$, the set $\{(x,y)\in\mathbb{R}^2: |x|^p+|y|^p=1\}$ is asteroid-shaped and convexity is clearly violated:

^{In this diagram I picked $p=\frac{1}{2}$.}

In particular the midpoint between $(0,1)$ and $(1,0)$ has a $p$-norm which is always $>1$.
Similarly $$ 1=\left\|\left(0,\tfrac{1}{2}\right)\right\|_p+\left\|\left(\tfrac{1}{2},0\right)\right\|_p \color{red}{<}\left\|\left(\tfrac{1}{2},\tfrac{1}{2}\right)\right\|_p = 2^{\frac{1}{p}-1}$$ hence such "norms" do not induce distances.

Answer 2

The function you describe fails to be a norm for $0 < p < 1$. In particular, you can verify that the triangle inequality will always fail since $$ \|(1,1)\|_p > \|(1,0)\|_p + \|(0,1)\|_p $$