Problem: Let $U:= \lbrace x \in \mathbb{R}^n \mid x_i \neq 0 \text{ for } 1 \leq i \leq n \rbrace $ and show that the Norm given by $$ N: \begin{cases} U & \longrightarrow \mathbb{R} \\ x & \longmapsto \displaystyle \sum_{i=1}^n i\vert x_i \vert \end{cases} $$ is differentiable on $U$ but $N$ is not differentiable at $0$
My Approach P1: I first tackled the easier part (in my opinion). I know that if $N$ is differentiable at $0$ then all directional derivatives at that point exist, thus I considered to use the contra positive of that statement.
Let $h$ be any direction and then consider the following $$\frac{\partial N}{\partial x_i}(0)= \lim_{h \to 0}\frac{N(0+h)-N(0)}{h}=\lim_{h \to 0} \frac{\sum_{i=1}^n i |h|-0}{h}= \lim_{h \to 0}\frac{n|h|}{h} $$ Which shows that the limit does not exists, it depends on the direction of $h$ and therefore the directional derivative does not exist at $0$. Can you comment please (in comments or in your answer) if this approach was correct?
P2: Now to show that $N$ is differentiable on $U$ I am a bit stuck, should I try to compute the Jacobi-Matrix? (I kind of dislike doing that because it feels very artificial to me) Or can I try to compute $$\lim_{ \epsilon \to 0} \frac{N(x+h \epsilon)-N(x)}{\epsilon}= \lim_{\epsilon \to 0} \frac{\sum i|x_i + \epsilon h|-\sum i |x_i|}{\epsilon} \leq n |h| $$ Which shows that the limit is bound by a constant independent of the actual direction (just by the magnitude). I believe I miss some accurate reasoning here in this step.
I often try showing that a function is differentiable by using the actual definition of differentiability in $\mathbb{R}^n$ which forces me to find a linear mapping, but I don't succeed with that.
To prove differentiability on $U$, pick a point $x\in U$. Let $\sigma_i = x_i/|x_i|$ for each $i$; this is either $1$ or $-1$, i.e., the sign of $x_i$. Note that if $h\in\mathbb R$ is sufficiently small, then the coordinates of $x+h$ have the same signs as the coordinates of $x$. More precisely, let $r=\min |x_i|$. If $h$ satisfies $\max |h_i| <r $, then $\operatorname{sign}(x_i+h_i)=\operatorname{sign} x_i$ for all $i$. All this long-winded setup is to show that in a neighborhood of $x$, the norm $N$ is a linear function, namely $$N(x+h) = \sum_{i} i \sigma_i (x_i+h_i)$$ Clearly, the linear function is differentiable.
As for non-differentiability at $0$: the non-existence of the partial $\partial N/\partial x_1$, shown in your post, settles the issue.