This is the problem which I'm struggling with:
"Let $V$ be a finite vector space, let $p$ be a norm on $V$, with $p$ being differentiable on $V$\ $\{0\}$. Show that: For every $x\in V$ exists a $Y\in$ Lin$(V,\mathbb{R})$, such that $|Y(z)|\leq p(z)$ $\forall z\in V$ and $Y(x)=p(x)$."
My first idea was to maybe define $Y(v) = Dp(x)(v)$, since $Dp(x)$ is linear and $p$ is defined as being differentiable. (For the case $x=0$, you simply take $Y=0$, since $p(0)=0$). I tried it with $p$ being the euclidean norm as a quick test and it worked, so I'm almost quite sure that this the right solution, but I don't really know how to prove it. For example, if I try $Y(x)$, I get: $$Y(x) = \sum_{i=1}^n \partial_i p(x)x_i$$ And this somehow has to be equal to $p(x)$. What can I try here? I feel like there are some properties of differentiable norms which can be used here. Does someone have some advice for me?
Edit: I think I found a very simple way: $V$ is isomorph to the $\mathbb{R}^n$, so every norm is equivalent to each other, so I can simple use the euclidean norm, right?
Edit 2: No, it's not possible to use another norm which is equivalent, also, $V$ was never specified to be a $\mathbb{R}$ vector space, the solution is much much easier: Let $Y(v) = Dp(x)(v)$, now use the definition of the derivative in a direction $v$: $$Y(v) = \lim_{t\to 0} \frac{p(x+vt) - p(x)}{t}$$
With this definition it's much easier to prove that $Y$ satisfies the given conditions, it's actually a very easy problem.