Differential of the norm in $\mathbb{R}^n$

Question

We consider the normed vector space $(\mathbb{R}^n,\Vert\cdot\Vert)$. Is the map $\Vert\cdot\Vert$ differentiable even if it is not induced by a scalar product?

Answer 1

Sometimes yes, sometimes no. For example, consider the $p$-norms: $$ \|(x_1,\dots,x_n)\|_p = \left(\sum_{k=1}^n |x_k|^p\right)^{1/p} . $$ This is differentiable (except at the origin) if $p>1$, but not if $p=1$.

Answer 2

A norm on $\mathbb{R}^n$ is never differentiable at the origin. Let $x_0\in \mathbb{R}^n\setminus \left\{0\right\}$. Then $$\lim_{t\to 0}\frac{\|(0+tx_0)\|-\|0\|}{t} =\lim_{t\to 0}\frac{\|t\|}{t}\|x_0\|$$ By the norm axiom $x_0\neq 0 \Rightarrow \|x_0\|\neq 0$, and so the limit does not exist, as $$\lim_{t\to 0^+}\frac{\|t\|}{t}=1\neq -1=\lim_{t\to 0^-}\frac{\|t\|}{t} $$

On the other hand, the triangle inequality $$\left|\|x\|-\|y\|\right|\leq \|x-y\| $$ tells us that any norm in $\mathbb{R}^n$ is Lipschitz with Lipschitz constant $L=1$. Thus by Rademacher's theorem, it is differentiable almost everywhere in $\mathbb{R}^n$.

Answer 3

A simple example of a norm on $\Bbb R^2$ which is non-differentiable at points other than the origin is given by $$||(x,y)||=\max(|x|,|y|).$$

If $h>0$ then $||(1+h,1)||=1+h$, while if $-2<h<0$ then $||(1+h,1)||=1$. Hence the function $h\mapsto||(1+h,1)||$ is not differentiable at $h=0$, so $||.||$ is not differentiable at $(1,1)$.