Norm and Dual Norm So, I was studying the norm and its dual in the book Convexity and Optimization in Finite Dimensions I and I came across a sort of theorem and had problem understanding it. I would appreciate if somebody helped me understand it:  

For every compact convex set $K\subset \mathbb{R}^{n}$ of dimension $n$, which contains the origin in its interior, there is precisely one norm $p$ such that $K=B_p$, where $B_p=\{x\in \mathbb{R}^{n}:p(x) \le  1\}$ is a norm body and $p$ is a norm.
Proof:
  Put 
  $$p(x)=\inf \left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace .$$
  Clearly, $p(x)\le 1$ if and only if $x\in K$. Thus, $K=B_p$, provided $p$ is a norm. To prove this, we first show that $p$ is finite on $\mathbb{R}^{n}$. Assume that $p(x)=\infty$.
  Then, $\frac{1}{\rho}x\notin K$ for all $\rho>0$, which contradicts the fact that $0$ lies in the proper interior of $K$. Trivially, $p(x)\ge 0$ and $p(\lambda x)=\lambda p(x)$ for $\lambda\ge 0$. If $p(x)=0$, then $\{\lambda x:\lambda \ge 0\}\subseteq K$. Therefore, $x=0$, as $K$ is bounded.
  Finally,  $$\frac{p(x_1+x_2)}{p(x_1)+p(x_2)}=p\left( \frac{p(x_1)}{p(x_1)+p(x_2)}.\frac{x_1}{p(x_1)}+\frac{p(x_2)}{p(x_1)+p(x_2)}.\frac{x_2}{p(x_2)}\right)\le 1,$$
  since the argument of $p$ in the middle term above belongs to $K$ by convexity. 

Here are what I have problem understanding:  


*

*Clearly, $p(x)\le 1$ if and only if $x\in K$. Like how?  

*Assume that $p(x)=\infty$.
Then, $\frac{1}{\rho}x\notin K$ for all $\rho>0$. Again, how?  

*Trivially, $p(x)\ge 0$. Is it by definition of $p(x)$?    

*If $p(x)=0$, then $\{\lambda x:\lambda \ge 0\}\subseteq K$. Therefore, $x=0$, as $K$ is bounded. Didn't understand the whole thing.  

*The argument of $p$. What is argument of $p$?(embarrassing, I know.)
 A: 
Clearly, $p(x)\le 1$ if and only if $x\in K$. Like how?

If $x \in K$, then $\frac{1}{1}x \in K$, so the infimal value of $\rho$ such that $\frac{1}{\rho} x \in K$ must be less than or equal to $1$.
Inversely, suppose $x \notin K$. By the convexity of $K$, and the fact $0 \in K$, we also have that $\lambda x \notin K$ for $\lambda \ge 1$, as we can write
$$x = \frac{1}{\lambda}(\lambda x) + \left(1 - \frac{1}{\lambda}\right)0,$$
which makes $x$ a convex combination of $\lambda x$ and $0$.
That is, if $x \notin K$ and hence $1 \notin \left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace$, then $1$ is a lower bound for $\left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace$; any $\rho < 1$ in this set would make $\frac{1}{\rho} > 1$, so $\frac{1}{\rho} x \notin K$, hence $\rho \notin \left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace$. This implies that $p(x) \ge 1$.
Why can't $p(x) = 1$? Then there would be $\rho_n \in \left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace$ such that $\rho_n \to 1$, and so $\frac{1}{\rho_n} x \in K \to \frac{1}{1}x = x$. But $K$ is closed, so $x \in K$, against assumption. Thus, indeed, $p(x) > 1$.

Assume that $p(x)=\infty$.
  Then, $\frac{1}{\rho}x\notin K$ for all $\rho>0$. Again, how?

Well, that's really the only way an infimum can be positive infinity (by definition): if it's taken of the empty set. The only way $p(x) = 0$ is if $\left\lbrace \rho\ge 0:\frac{1}{\rho}x\in K \right\rbrace = \emptyset$, which is immediately equivalent to the condition given.

Trivially, $p(x)\ge 0$. Is it by definition of $p(x)$?

Yes. The definition of $p(x)$ is the infimum of the set $\left\lbrace \color{red}{\rho\ge 0}:\frac{1}{\rho}x\in K \right\rbrace$, which consists of only positive numbers (just realised, it probably should be $\rho > 0$!). As such, by its definition, $0$ is a lower bound for the set, and the greatest lower bound must be at least $0$.

If $p(x)=0$, then $\{\lambda x:\lambda \ge 0\}\subseteq K$. Therefore, $x=0$, as $K$ is bounded. Didn't understand the whole thing.

If $p(x) = 0$, then this means that there are positive $\rho \to 0$ such that $\frac{1}{\rho_n} x \in K$. Since $\frac{1}{\rho_n} \to \infty$, if $x \neq 0$, this  would mean that $\{\frac{1}{\rho_n}x\}$ is an unbounded subset of $K$, which would contradict $K$ being compact. But, if $x = 0$, of course, there is no issue; $\frac{1}{\rho_n} x = 0$ for all $n$.

The argument of $p$. What is argument of $p$?(embarrassing, I know.)

The argument of a function is the value being substituted in. In this case, I believe it's referring to the quantity
$$\frac{p(x_1)}{p(x_1)+p(x_2)}.\frac{x_1}{p(x_1)}+\frac{p(x_2)}{p(x_1)+p(x_2)}.\frac{x_2}{p(x_2)}.$$
A: For most of your questions the answers come from the definition of $p(x)$, which is
$$p(x) = \inf \left\{\rho \geq 0 \mid \frac{1}{\rho}x \in K \right\} $$
Hence,


*

*if $x\in K$ then $\rho = 1$ is valid in the definition of $p(x)$.  However, if $\rho < 1$ we get that $1/\rho > 1$ and then $(1/\rho) x \not\in K$ since $K$ is convex.  So the infimum must indeed be $1$.

*if $p(x)=\infty$ then $\rho \rightarrow \infty$ in the definition of $p(x)$.  But as $\rho$ approaches $\infty$ we have that $1/\rho \rightarrow 0$, so all points are being excluded from $K$. (Think of it as the set of points that can lie in $K$ must shrink as $\rho$ increases.)

*Since $\rho \geq 0$ and $p(x)$ is the infimum of possible $\rho$-values, $p(x) \geq 0$ as well.  Yes, it's by the definition.

*This is like point $2$ but in the other direction: if $p(x)=0$ then $\rho \rightarrow 0$ so all positive multiples of $x$ are included in $K$ (e.g. when $\rho =1/2$ we have $2x\in K$ etc.).  We can state that mathematically as $\{ \lambda x \mid \lambda \geq 0\}$.  This set is unbounded (because $\lambda$ can be as large as you like), but $K$ is bounded, so the only way to bound this set is by having $x=0$.

*The argument of $p(x)$ is $x$ -- it's whatever is in the parentheses.  As a further example, for $p(x_1 +x_2)$ the argument is $x_1+x_2$.
