Let $X$ be a normed space and $U\subseteq X$ convex, now define $p_{U}:X\to [0,\infty]$ where $p_{U}(x):=\inf\{\lambda>0: \frac{x}{\lambda}\in U\}$ and $0\in \operatorname{int}(U)$. Show that:
$i)U$ is absorbing, that is $\forall x \in X, \exists r > 0, \forall \alpha \in \mathbb C: \vert \alpha \vert < r \Rightarrow \alpha x \in U$
$ii)$ Show that $p_{U} < \infty$
My ideas:
$i)$ let $x \in X$ arbitrary and since $0\in \operatorname{int}(U)$ we know that there exists $r_{0}>0$ such that $B_{r_{0}}(0)\subseteq U$ we thus choose $r_{x}:=\frac{r_{0}}{\vert \vert x\vert \vert}$ and hence any $\vert\alpha \vert < \frac{r_{0}}{\vert \vert x\vert \vert}\Rightarrow\vert\vert\alpha x-0\vert\vert<r_{0}\Rightarrow \alpha x \in U$
$ii)$ I have been told that I should get to the bound $p_{U}(x)\leq \frac{\vert \vert x \vert \vert}{\epsilon}$ but I do not see how I can do this. Any help?
You know that $\frac{r_0}{2 \|x\|} x \in U$. This means that if you take $\lambda = \frac{2 \|x\|}{r_0}$ then $\frac{x}{\lambda} \in U$ and hence $p_U(x) \leq \lambda$. This implies the bound you want with $\epsilon = \frac{r_0}{2}$.