A set $C\subseteq X$ is convex if for any $x,y\in C$ and any $0\leq t \leq 1,$ we have $tx+(1-t)y\in C.$
A set $C\subseteq X$ is absorbing if for any $x\in X,$ there exists $t>0$ such that $tx\in C.$
Let $X$ be a vector space over $\mathbb{R}$ and let $C$ be a convex and absorbing subset of $X.$ Define $\rho:X\to\mathbb{R}$ by $$\rho(x) = \inf\{t>0: \frac{x}{t}\in C \}.$$ Show that
(a) $\rho$ is well-defined and that $\rho(x)\geq 0$ for all $x\in X.$
(b) $\rho(ax) = a\rho(x)$ for any $x\in X$ and any $a\geq 0.$
(c) $\rho(x+y) \leq \rho(x)+\rho(y)$ for any $x,y\in X.$
Moreover, $$\{x\in X:\rho(x)<1 \} \subseteq C \subseteq \{ x\in X:\rho(x)\leq 1\}.$$
My attempt:
part (a): Fix $x\in X.$ Clearly $\{t>0:\frac{x}{t}\in C\}$ is bounded below by $0.$ Since $C$ is absorbing, there exists $t>0$ such that $$tx = \frac{x}{\frac{1}{t}} \in C.$$ Therefore, the set $\{t>0:\frac{x}{t}\in C \}\neq \emptyset$ and hence $\rho$ is well-defined.
Fix $\varepsilon>0.$ Then there exists $t>0$ such that $\frac{x}{t}\in C$ and $$\rho(x) +\varepsilon>t>0.$$ Since $\varepsilon >0$ is arbitrary, therefore $\rho(x)\geq 0.$
part (b): Fix $x\in X$ Clearly $\rho(ax)=a\rho(x)$ holds if $a=0$ as $\rho(0)= 0.$ Assume that $\alpha>0.$ Then $$\rho(ax) = \inf\{a\frac{t}{a} : \frac{ax}{t} = \frac{x}{\frac{t}{a}} \in C\} = a\inf\{\frac{t}{a}: \frac{x}{\frac{t}{a}}\in C \} = a\rho(x).$$
part (c): Fix $\varepsilon>0.$ Then there exist $t_x$ and $t_y$ such that $\frac{x}{t_x} \in C, \frac{y}{t_y}\in C,$ $$t_x<\rho(x)+\frac{\varepsilon}{2} \text{ and } t_y < \rho(y) + \frac{\varepsilon}{2}.$$ It follows that $$t_x+t_y < \rho(x) + \rho(y) + \varepsilon.$$ By convexity of $C,$ $$\frac{x+y}{t_x+t_y} = (\frac{t_x}{t_x+t_y})\frac{x}{t_x} + (\frac{t_y}{t_x+t_y})\frac{y}{t_y} \in C.$$ Therefore, $$\rho(x+y) \leq t_x+t_y < \rho(x)+\rho(y)+\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, hence $$\rho(x+y)\leq \rho(x)+\rho(y).$$
'Moreover' part: Let $x\in \{x\in X:\rho(x)<1\}.$ Let $\varepsilon = 1 - \rho(x)>0.$ Then there exists $t>0$ such that $\frac{x}{t}\in C$ and $$1 = \rho(x)+\varepsilon > t.$$ Note that If $C$ is absorbing, then $0\in C.$ By convexity of $C,$ $$x = t(\frac{x}{t}) + (1-t)0\in C.$$
Finally, let $x\in C.$ Then $\frac{x}{1} = x \in C.$ Therefore, $\rho(x)\leq 1.$
Are my proofs correct?