Given a subset $A$ of a vector space $X$, we have the following definitions:
$A$ is said to be convex if for every $t\in [0,1]$ and for every $x,y\in A$, $ty+(1-t)x\in A$.
$A$ is said to be absorbing if for every $x\in X$, there exists a $t>0$ such that $x/t\in A$.
Minkowski functional of $A$, denoted by $m_A$ is defined as $m_A(x):=\inf \{t>0, x/t\in A\}$.
It is to be proven that if $A$ is convex and absorbing, then $m_A$ satisfies the triangle inequality. This can be done as follows: Given an $\epsilon>0$, there exist $s,t>0$ such that $x/s, y/t\in A$ and $s\lt m_A(x)+\epsilon/2, t<m_A(y)+\epsilon/2$. By convexity, $\frac{x+y}{s+t}\in A$, whence $m_A(x+y)\le s+t\lt m_A(x)+m_A(y)+\epsilon$, whence the result follows.
Rudin's proof goes along the following lines: If $t=m_A(x)+\epsilon, s=m_A(y)+\epsilon$ for some $\epsilon>0$, then $x/t$ and $y/s$ are in $A$. Rest of the proof is along the same lines as above.
My question is: Is it true that for any $t>m_A(x), x/t\in A$ because that's what Rudin seems to be using? If it is true, then how can one prove it?
Thanks.
Since $A$ is absorbing, then $0\in A.$ If $t>m_A(x)$ then there exists $t_0$ such that $m_A(x)\le t_0<t$ and $x/t_0\in A.$ By convexity of $A$ is we get $${x\over t}=\left (1-{t_0\over t}\right ) 0+{t_0\over t} {x\over t_0}\in A$$