Let $\|\cdot\|_0$, a norm on $\mathbb{R}^n$. Show that the function $\|\cdot\|_0$ is $1$-lipschitz and hence, continuous.
Meaning, I need to prove that: $$\big|\|x\|_0-\|y\|_0\big| \le \|x-y\|$$
Now, I'm familiar with the reverse triangle inequality proof which will results: $$\big|\|x\|_0-\|y\|_0\big| \le \|x-y\|_0$$
but please notice that the $\|x-y\|$ should be the euclidean norm.
Best Answer
It is meant to be 1-Lipschitz with respect to $\| \cdot \|_0$. Otherwise the claim would be false, consider for example the norm $\| \cdot \|_0 := 2 \| \cdot \|$ for $y = 0$ and an arbitrary $x \ne 0$.