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, $K$ 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 will show that:


 * $\set {t > 0 : t^{-1} x \in K}$

is non-empty for each $x \in \R^n$.

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:


 * $\set {t > 0 : t^{-1} x \in K} = \openint 0 \infty \ne \O$ if $x = 0$.

Now take $x \in \R^n \setminus \set 0$.

Since $0 \in K$ and $K$ is open, there exists $\delta > 0$ such that whenever:


 * $\norm x < \delta$

we have $x \in K$.

So we have:


 * $\ds \norm {\frac \delta {2 \norm x} x} = \frac \delta 2 < \delta$

so:


 * $\ds \frac \delta {2 \norm x} x \in K$

In particular:


 * $\ds \frac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in K}$

so:


 * $\set {t > 0 : t^{-1} x \in K}$ is non-empty.

Clearly, it is also bounded below for each $x \in \R^n$.

So:


 * $\inf \set {t > 0 : t^{-1} x \in K}$

exists by the Continuum Property.

Now, we define $N : \R^n \to \R$ by:


 * $\map N x = \inf \set {t > 0 : t^{-1} x \in K}$

for each $x \in \R^n$.

Since:


 * $0 \in \set {t > 0 : t^{-1} x \in K}$

we have:


 * $\map N x \ge 0$

for each $x \in \R^n$ by the definition of infimum.

We will first show, in view of Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm, that:


 * $\map N x = 0$ $x = 0$

and:


 * $\map N {\lambda x} = \size \lambda \map N x$ for each $\lambda \in \R$ and $x \in \R^n$.

We establish a useful bound on $N$.

We already have:


 * $\ds \frac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in K}$

where $\delta$ is chosen such that $x \in K$ whenever $\norm x < \delta$.

So we have:


 * $\ds \map N x \le \frac {2 \norm x} \delta$

from the definition of infimum.

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, if $t > 0$ has $t^{-1} x \in K$, we must have:


 * $\ds t^{-1} < \frac {2 R} {\norm x}$

That is:


 * $\ds t > \frac {\norm x} {2 R}$

So we obtain the bound:


 * $\ds \frac {\norm x} {2 R} \le \map N x$

for $x \in \R^n \setminus \set 0$.

This clearly holds for $x = 0$ also.

So we have:


 * $\ds \frac {\norm x} {2 R} \le \map N x \le \frac {2 \norm x} \delta$

for each $x \in \R^n$.

Then, if $x \in \R^n$ has $\map N x = 0$, we immediately obtain:


 * $\ds \frac {\norm x} {2 R} = 0$

so:


 * $x = 0$

Conversely, we have:


 * $0 \le \map N 0 \le 0$

so:


 * $\map N 0 = 0$

We now verify that:


 * $\map N {\lambda x} = \size \lambda \map N x$ for each $\lambda \in \R$ and $x \in \R^n$.

If $\lambda = 0$, this is obvious since $\map N 0 = 0$.

Now take $x \in \R^n$ and $\lambda \in \R \setminus \set 0$.

Suppose that $\lambda > 0$.

Then for $t > 0$ we have $t^{-1} x \in K$ $\paren {\lambda t}^{-1} \paren {\lambda x} \in K$.

We therefore have:


 * $t \in \set {t > 0 : t^{-1} x \in K}$ $\lambda t \in \set {t > 0 : t^{-1} \paren {\lambda x} \in K}$.

So we have:


 * $\set {t > 0 : t^{-1} \paren {\lambda x} \in K} = \lambda \set {t > 0 : t^{-1} x \in K}$

From Multiple of Infimum, we then obtain:


 * $\map N {\lambda x} = \lambda \map N x = \size \lambda \map N x$

Now suppose that $\lambda < 0$.

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 we again have:


 * $\set {t > 0 : t^{-1} \paren {\lambda x} \in K} = \size \lambda \set {t > 0 : t^{-1} x \in K}$

and so:


 * $\map N {\lambda x} = \size \lambda \map N x$

from Multiple of Infimum.

In order to apply Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm, we show that:


 * $\set {x \in \R^n : \map N x \le 1}$ is convex.

We finally verify that indeed:


 * $\set {x \in \R^n : \map N x < 1} = K$

We first show that:


 * $K \subseteq \set {x \in \R^n : \map N x < 1}$

Since $K$ is open set, there exists $\epsilon$ such that whenever $y \in \R^n$ has:


 * $\ds \norm {y - x} \le \epsilon$

We have $y \in K$.

From the Reverse Triangle Inequality, this gives that whenever $y \in \R^n$ has:


 * $\paren {1 - \epsilon} \norm x \le \norm y \le \paren {1 + \epsilon} \norm x$

we have $y \in K$.

In particular:


 * $\ds \paren {1 + \frac \epsilon 2} x \in K$

So:


 * $\ds \frac 2 {2 + \epsilon} \in \set {t > 0 : t^{-1} x \in K}$

Since:


 * $\ds \frac 2 {2 + \epsilon} < 1$

We obtain:


 * $\map N x < 1$

So we get:


 * $K \subseteq \set {x \in \R^n : \map N x < 1}$

We now show that:


 * $\set {x \in \R^n : \map N x < 1} \subseteq K$

Let:


 * $x \in \set {x \in \R^n : \map N x < 1}$

Then there exists $t < 1$ such that:


 * $t^{-1} x \in K$

Again, since $K$ is convex, we have:


 * $t \times \paren {t^{-1} x} + \paren {1 - t} \times 0 \in K$

so:


 * $x \in K$

So we have:


 * $\set {x \in \R^n : \map N x < 1} \subseteq K$

hence:


 * $\set {x \in \R^n : \map N x < 1} = K$