The following mapping induces a norm on $\mathbb{R}^n$ $$N: \begin{cases} U & \longrightarrow \mathbb{R} \ x & \longmapsto \sum_{i=1}^n i \vert x_i \vert \end{cases} $$
Where $U:= \lbrace x \in \mathbb{R}^n \mid x_i \neq 0 \text{ for } 1 \leq i \leq n \rbrace $

Show that $N$ is differentiable on $U$ but not at $0$

Question

The following mapping induces a norm on $\mathbb{R}^n$ $$N: \begin{cases} U & \longrightarrow \mathbb{R} \ x & \longmapsto \sum_{i=1}^n i \vert x_i \vert \end{cases} $$
Where $U:= \lbrace x \in \mathbb{R}^n \mid x_i \neq 0 \text{ for } 1 \leq i \leq n \rbrace $

Show that $N$ is differentiable on $U$ but not at $0$

anonymous · Answer

Already answered it myself. Thanks