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 $\|x\|/R\le q_A(x)\le \|x\|/r$.
Best Answer
Recall that a point $z$ lies in the convex hull of $A$, $z\in C(A)$, iff we can write,
$$z = \sum\limits_{a\in A}z_a a$$
with $\sum\limits_{z\in A}z_A = 1$ and $z_a\geq 0$ for all $a\in A$.
Then the set $\lambda C(A)$ is the set of all points $\lambda z$, $z\in C(A)$. Since $z\in C(A)$, we can write,
$$\lambda z = \lambda \sum\limits_{a\in A} z_a a$$
and since $\sum\limits_{a\in A}z_a=1\implies \lambda\sum\limits_{a\in A}z_a=\lambda$ we are done.