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 $\|x\|/R\le q_A(x)\le \|x\|/r$.
First of all, I think the key to the trouble of your understanding is the following relation:
$$ \forall a>0\; \frac{x}{a} \in K \iff x \in aK $$
So the definition you wrote for the Minkowski functional and the gauge Lax defines are essentially the same, note that you missed some key esssential properties: In the definition for the Minkowski functional you have the assume that $0 \in K$ is an interior point (we call such sets absorbing), since only then every vector $v \in V$ has some scalar small enough such that $v \in \lambda K$, otherwise the infimum maybe empty. The same thing goes for gauge: Lax explicitly mentions that he assumes the interior point is $0$.
Note that your intuition in 1) is reasoable (yes $p(0)=0$), but we are not only talking about points in $K$, but in the whole of $V$, this is used in the Geometric Version of the Hahn Banach Theorem for example.
Best Answer
Let $E=\mathbb{R}^2$ and $K=\{0\}\cup\mathbb{S}^1\cup\{(2,0),(-2,0),(0,2),(0,-2)\}$. $K$ satisfies all conditions but convexity. Then $\parallel (1,0)\parallel_K=\parallel (0,1)\parallel_K=\tfrac{1}{2}$ and $\parallel (1,1)\parallel_K=\sqrt{2}\nleq\tfrac{1}{2}+\tfrac{1}{2}$, violating the triangle inequality.