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 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 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 is Minkowski 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 $(\text N 3)$ and half of $(\text N 2)$ 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 $(\text N 2)$.

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 $(\text N 2)$.

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 recovers Set, we have:


 * $\set {x \in \R^n : \norm x_\ast < 1} = K$

So $\norm \cdot_\ast$ is the required norm.