Characterization of Unit Open Balls of Norms of Euclidean Space

== Theorem ==
Let $\struct {\R^n, \norm \cdot}$ be the Euclidean $n$-space. 
Let $K \subseteq \R^n$ be a non-empty open set of $\struct {\R^n, \norm \cdot}$.
Then there exists a norm $\norm \cdot_* : \R^n \to \R$ such that: 
:$\set {x \in \R^n : \norm x_* < 1} = K$
:
:$(1): \quad$ $K$ is bounded in $\struct {\R^n, \norm \cdot}$
:$(2): \quad$ $K$ is convex
:$(3): \quad$ $K$ is symmetric.
That is: 
:an open set $K \subseteq \R^n$ is the unit ball of some norm on $\R^n$ $K$ is bounded, (with respect to the Euclidean norm), convex and symmetric.
== Proof ==
=== Necessary Condition ===
==== Proof of $(1)$ ====
Suppose that $\norm \cdot_* : \R^n \to \R$ is a norm on $\R^n$ with: 
:$\set {x \in \R^n : \norm x_* < 1} = K$
From Norms on Finite-Dimensional Real Vector Space are Equivalent, there exists $M > 0$ such that: 
:$\norm x \le M \norm x_*$
for each $x \in \R^n$.
So, for each $x \in K$, we have: 
:$\norm x < M$ 
So $K$ is bounded.
==== Proof of $(2)$ ====
From Open Ball is Convex Set, we have that $K$ is convex.
==== Proof of $(3)$ ====
Let $x \in K$. 
Then we have: 
:$\norm x_* < 1$ 
We have: 
:$\norm {-x}_* = \size {-1} \norm x = \norm x$
so we also have: 
:$\norm {-x}_* < 1$
So:
:$-x \in K$.
So $K$ is symmetric.
=== Sufficient Condition ===
Suppose that $K \subseteq \R^n$ satisfies:
:$(1): \quad$ $K$ is bounded in $\struct {\R^n, \norm \cdot}$
:$(2): \quad$ $K$ is convex
:$(3): \quad$ $K$ is symmetric.
We first show that $0 \in K$.
Since $K$ is not empty, we can pick $x \in K$. 
Since $K$ is symmetric, we have $-x \in K$. 
Then since $K$ is convex, we have: 
:$\ds \frac 1 2 x + \frac 1 2 \paren {-x} = 0 \in K$
So we can consider the Minkowski functional for $K$, $p_K$.
We show that $p_K$ gives a norm. 
From Minkowski Functional of Open Convex Set in Normed Vector Space is Bounded, there exists a real number $c > 0$ such that: 
:$\map {p_K} x \ge c \norm x$ for all $x \in X$. 
From Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional, we have that: 
:$\map {p_K} {x + y} \le \map {p_K} x + \map {p_K} y$ for each $x, y \in \R^n$
and:
:$\map {p_K} {\lambda x} = \lambda \map {p_K} x = \size \lambda \map {p_K} x$ for each $\lambda \in \R_{\ge 0}$ and $x \in \R^n$. 
So and half of of the norm axioms are satisfied. 
It remains to show that: 
:$\map {p_K} x = 0$ $x = 0$
and:
:$\map {p_K} {\lambda x} = \size \lambda \map {p_K} x$ for each $\lambda \in \R_{< 0}$ and $x \in \R^n$. 
Since $K$ is bounded, there exists $R > 0$ such that whenever $x \in K$, we have: 
:$\norm x < R$ 
Then, for each $x \in \R^n \setminus \set 0$, we have: 
:$\ds \frac {2 R} {\norm x} x \not \in K$
So:
:$\ds \frac {\norm x} {2 R} \not \in \set {t > 0 : t^{-1} x \in K}$
From , we have: 
:$\openint {\map {p_C} x} \infty \subseteq \set {t > 0 : t^{-1} x \in K}$
So, we must have: 
:$\ds \frac {\norm x} {2 R} \le \map {p_C} x$ 
for $x \in \R^n \setminus \set 0$.
This clearly holds for $x = 0$ also.
Then, if $\map {p_K} x = 0$, we have: 
:$\norm x \le 0$
So from positive definiteness, we have: 
:$x = 0$
So:
:$\map {p_K} x = 0$
verifying .
Now it just remains to show that: 
:$\map {p_K} {\lambda x} = \size \lambda \map {p_K} x$ for each $\lambda \in \R_{< 0}$ and $x \in \R^n$.
Let $\lambda < 0$ and $x \in \R^n$.
Since $K$ is symmetric, for $t > 0$ we have $t^{-1} x \in K$ $-t^{-1} x \in K$.
That is, $\paren {-\lambda t}^{-1} \paren {\lambda x} \in K$. 
Since $\lambda > 0$, this gives:
:$t^{-1} x \in K$ $\paren {\size \lambda t}^{-1} \paren {\lambda x} \in K$ 
So: 
:$t \in \set {t > 0 : t^{-1} x \in K}$ $\size \lambda t \in \set {t > 0 : t^{-1}\paren {\lambda x} \in K}$
So we have: 
:$\set {t > 0 : t^{-1} \paren {\lambda x} \in K} = \size \lambda \set {t > 0 : t^{-1} x \in K}$
and so:
:$\map {p_K} {\lambda x} = \size \lambda \map {p_K} x$
from Multiple of Infimum.
So $p_K$ satisfies . 
So $p_K$ is a norm on vector space.
So, defining $\norm \cdot_\ast : \R^n \to \R$ by: 
:$\norm x_\ast = \map {p_K} x$
for each $x \in \R^n$ gives a norm.
Further, from Minkowski Functional of Open Convex Set in Normed Vector Space recovers Set, we have: 
:$\set {x \in \R^n : \norm x_\ast < 1} = K$
So $\norm \cdot_\ast$ is the required norm.
== Sources ==
* $3.1$: Norms
Category:Normed Vector Spaces