minkowski functional being equivalent to original norm Definition1: A subset $ A$ of a normed space $E$ is absolutely convex if $ \forall x,y \in A, \quad \forall \alpha ,\beta $ s.t $|\alpha |+ |\beta| \leq 1, \quad  $ $ \alpha x +\beta y \in A.$
Definition2: Let $A$ be an absolutely convex subset of a normed space $E$. For $x \in E $ the Minkowski functional of A is defined by $ q_A(x) := \inf \{\lambda \geq 0 : x\in \lambda A  \}. $
Theorem: Let $A \subset E$ be absolutely convex.Then, the minkowski functional of A is equivalent to the original norm of $E$.
What I did so far is the following:
If $x=0$ , then obviously we have $q_A(x)= ||x||$. 
For $0\ne x \in A$, with $||x ||=k$ , since $A$ is absolutely convex, $q_A(x) \leq 1 = || \frac{x}{k}  ||$. Then, playing algebraically, we get $  kq_A(x) \leq ||x ||   $. But , I cannot guarantee the other side, such as $ ||x|| \leq \frac{1}{k} q_A(x) ?$
I have similar arguments for $x \notin A $ but again, I have a problem with one side. How can I proceed from here or was my starting point wrong? (because I could not use the absolute convexity of A)
 A: This is not true; perhaps something is missing from the assumptions (see the end of this post). Let $q$ be any seminorm on $E$; then the set $A=\{x\in E: q(x)\le 1\}$ is absolutely convex since 
$$
q(\alpha x+\beta y)\le q(\alpha x)+q(\beta y)=|\alpha | q(x)+|\beta|q(y)
\le  |\alpha |+|\beta| \le 1
$$
And the Minkowski functional of this $A$ is just $q$ itself:
$$\inf \{\lambda \geq 0 : x\in \lambda A  \} = \inf \{\lambda \geq 0 : q(x)\le  \lambda    \} =q(x)$$
But of course, $q$ need not be equivalent to any specific norm. It does not even have to be a norm itself; consider $q(x)=|x_1|$ on $\ell^2$. Or it can be another, inequivalent, norm, like $q(x)=\sup|x_k|$ on $\ell^2$.  
Remark
Absolute convexity, as defined above, is equivalent to the requirements that $A$: 


*

*is convex

*contains $0$

*is balanced, meaning $\alpha A=A$ whenever $|\alpha|=1$


Indeed, it's clear that absolute convexity implies 1-2-3. Conversely, if 1-2-3 hold and $|\alpha|+|\beta|\le 1$, $x,y \in A$, then let 
$$x'= \frac{\alpha}{|\alpha|}x,\quad  y'= \frac{\beta}{|\beta|}y$$
Since $A$ is balanced, it contains $x',y'$. Then 
$$\alpha x +\beta y = |\alpha| x'+|\beta|y'+(1-|\alpha|-|\beta|)0$$
is a convex combination of three points in $A$, and therefore lies in $A$. 
With additional assumptions: $A$ is open and bounded
Then the claim is true. Let $r=\inf\{\|x\|:x\in A\}$ and $R=\sup\{\|x\|:x\in A\}$. Note that $r>0$ and $R<\infty$. Using the monotonicity of Minkowski functional with respect to set containment (or just its definition), we get $\|x\|/R\le q_A(x)\le \|x\|/r$.
