Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$ - why? I have trouble understanding the following proof:

Theorem: Every symmetric convex body in $\mathbb{R}^n$ is the closed unit ball of a norm on $\mathbb{R}^n$
Proof: Given a symmetric convex body $K$ in $\mathbb{R}^n$ we define its (so-called) Minkowski functional $$p_K(x) = \inf\{t>0, x\in tK\}$$
It is clear that $\{p_K(x)\leq 1\} = K$, so it remains to check that $p_K$ is a norm (exercise).


*

*In the definition of $p_K(x)$, there is $``x\in tK"$ - what's $tK$? The convex body scaled in all directions by a factor of $t\in\mathbb{R}$? Also what does $p_K(x)$ really mean intuitively?

*Why is $\{p_K(x)\leq 1\} = K$?

*To see that $p_K$ is a norm:(i) Consider $p_K(x) = 0$. This means $\inf\{t>0, x\in tK\} = 0$. I need to show that $x=0$. This is a little weird, innit? Perhaps the definition of $p_K(x)$ should have been $p_K(x) = \inf\{t\geq0, x\in tK\}$? 
(ii) To show $p(ax) = |a|p(x), a\in\mathbb{R}$: we must show $\inf\{t>0, ax\in tK\} = |a|\inf\{t>0, x\in tK\}$. Why does this hold? 
(iii) Lastly we also need to show the triangle-inequality, i.e. $p(x) + p(y) \ge p(x+y), x,y\in\mathbb{R}^n$.

I'm not necessarily interested in a detailed solution (though that would help), some hints or pointers to get me started would suffice too!
 A: Since there is already an answer on your questions, I will try to explain the intuition behind the minkowski functional.
You can think  of the minkowski functional $ ρ_Α$  (geometrically) as the fraction $ \frac{d(x,0)}{d_x'}$ where $d(x,0)$ is the distance from x to $0$ and $d_x'$ is the distance from $0$, to the "farthest" point of $A$, in the direction of $x$.
See here for some visual examples.
In the first example: $A \subset \mathbb R^n$ is a ball of radius $R$ centered at $0$. It turns out that  $ ρ_A(x) = \frac{||x||}{R}$. Notice that     $ d(x,0)=||x||$ and $ d_x' =R$.
In the second example: $A$ is a half plane in $\mathbb R^2$. The distance from $0$ to the farthest point of $A$ in the direction of $x$ is actually infinity (since $A$ is unbounded in that direction) and $ρ_A(x) =0 = \frac{(OM)}{\infty} $. On the other hand, the  distance from $0$ to the farthest point in $A$ in the direction of $y$ is $N'$. So, $ρ_A(y) = \frac{(ON)}{(ON')}= 1/μ $.
A: *

*In the definition of $p_K(x)$, $tK = \{tk ; k\in K\}$.

*Why is $\{p_K(x)\leq 1\} = K$?

First, if $x\in K$ then $x\in 1K$ so $\inf \{t>0 ; x\in tK\} \leqslant 1$, so $p_K(x) \leqslant 1$.
If $x\in \{p_K(x)\leq 1\}$, then there exists $t\leqslant 1$ such that $x\in tK$. So there exists $k\in K$ such that $x = tk = tk + (1-t)0$ and since $K$ is convex, we have that $x\in K$ ($0$ is in $K$ since $K$ is symmetric).


*To see that $p_K$ is a norm:(i) $p_K(x) = 0$ means $\inf\{t>0, x\in tK\} = 0$, which means there exists a sequence $(t_n)$ of positive numbers such that $t_n \rightarrow 0$ and for all $n$, $x\in t_n K$. So for $n$, there exists $k_n \in K$ such that $x = t_n k_n$. As $K$ is compact (finite-dimension), we can extract $k_{\varphi (n)}$ which converges to $k$. Since $t_{\varphi (n)} \rightarrow 0$, we have $x = \lim t_n k = 0$.
(ii) To show $p(ax) = |a|p(x), a\in\mathbb{R}$: as you said, we must show $\inf\{t>0, ax\in tK\} = |a|\inf\{t>0, x\in tK\}$. It's true because $x\in tK$ also means $ax \in (at)K$. 
(iii) For the triangle-inequality, remark that if $x = t_x k_x$ and $y= t_y k_y$, then after some work, we get $x+y = (t_x + t_y)k$ for some $k$. So $p_k (x+y) \leqslant t_x + t_y$. Then since this is true for all $t_x$ and $t_y$ such that $x\in t_x K$ and $y\in t_y K$, we can take the inf of the equation and we get the result.

A: *

*$p_K(x)$ is measuring, roughly speaking, the smallest dilation $t$ for which $x$ sits on the boundary of the convex body. It might help to consider the case where $K=\overline{B(0,1)}$, the closed Euclidean unit ball. We look for the greatest lower bound of $t$ for which $x\in tK$--but this $t$ is precisely the $t$ for which $x$ lies on the sphere of radius $t$. But that radius tells us the distance of $x$ to the origin, so it's in this way scalings of $K$ can give us a notion of norm.


*If $x\in 1\cdot K$, then $p_K(x)\leq 1$ since $p_K(x)$ is the infimum of $t$ for which $x\in tK$. Thus $K\subset\{x: p_K(x)\leq 1\}$. The reverse inclusion is a little trickier. Suppose $0<p_K(x)\leq 1$. There is a minimizing sequence $t_n\to p_K(x)$ such that $x\in t_nK$ for each $n$. But then $x/t_n\in K$, and since $K$ is closed, we conclude $x/p_K(x)\in K$. Thus $x\in p_K(x)K$, which is a subset of $K$ since $p_K(x)\leq 1$ and $K$ is convex. If $p_K(x)=0$, then it is not hard to see $x=0$ (if one assumed $x\neq 0$ and $x\in (1/n)K$ for all positive integers $n$, then we get $nx\in K$ for all positive integers $n$--this will contradict boundedness of $K$).


*(i) I already mentioned this case. You'll want to use boundedness of $K$.
(ii) Just show that, for $a\neq 0$, $\{t>0: ax\in tK\}=\{t>0: x\in (t/|a|)K\} = \{|a|s>0: x\in sK\}$. The infimum of the last set is $|a|p_K(x)$, so the result will follow.
(iii) If $x\in t_1K$ and $y\in t_2K$, then show that $(x+y)\in (t_1+t_2)K$. It follows $p_K(x+y)\leq t_1+t_2$. Now take infimums over $t_1,t_2$.
