Is the square of a norm differentiable?

Question

Let $(E,||.||)$ be a normed vector space.
Is $f:E\rightarrow \mathbb{R},\ f(x)=||x||^2$ differentiable ?
I know that if $||.||$ is the norm associated to an inner product then $f$ is indeed differentiable, but what if $||.||$ doesn't come from an inner product ?
Thank you in advance for your help.

Answer 1

Let $f(x) = (|x_1|+|x_2|)^2$ and note that $f((t,1)) = t^2 + 2|t| +1$, which is not differentiable at $t=0$.

Hence $x \mapsto \|x \|_1^2$ is not differentiable.

Answer 2

In other words, your question is, Do there exist norms that are not differentiable?

How about $E = \mathbb{R}^2$, and let $|| \cdot ||$ be, say, the max norm: $$ || \, (x_{1}, x_{2}) \,|| = \left( \, \sqrt[3]{|x_{1}|} + \sqrt[3]{|x_{2}|} \, \right)^3? $$ I haven't explored this in detail (hence the community wiki), but it seems that the square of $|| \, (x_{1}, x_{2}) \,||$ will have cube roots of $|x_{k}|$.

Answer 3

Not necessarily. Take the $L^\infty$ norm in the plane $$ F(x,y)=\max\{|x|,|y|\} $$ If you square it, you get a non-differentiable function along the lines $x=\pm y$.