Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm

Theorem
Let $\mathbb F$ be a subfield of $\mathbb F$.

Let $X$ be a vector space over $\mathbb F$.

Let $N : X \to \R_{\ge 0}$ be a positive definite and positive homogeneous function.

That is:


 * $(1): \quad$ $\map N x = 0$ $x = 0$
 * $(2): \quad$ $\map N {\lambda x} = \cmod \lambda \map N x$ for each $x \in X$ and $\lambda \in \mathbb F$.

Suppose further that:


 * $B = \set {x \in X : \map N x \le 1}$ is convex.

Then:


 * $N$ is a norm on $X$.

Proof
Note that $N$ satisfies axioms $\text N 1$ and $\text N 2$.

So we only need to verify $\text N 3$.

That is, we want to show that:


 * $\map N {x + y} \le \map N x + \map N y$

for each $x, y \in X$.

Fix $x, y \in X$.

If $\map N x = 0$, then $x = 0$ and we have:

Similarly, if $\map N y = 0$, then $y = 0$ and:

In both cases we clearly have:


 * $\map N {x + y} \le \map N x + \map N y$

So suppose that $\map N x > 0$ and $\map N y > 0$.

We then have:

so:


 * $\dfrac x {\map N x} \in B$

Similarly:

so:


 * $\dfrac y {\map N y} \in B$

Now note that we can write:


 * $\ds 1 = \frac {\map N x} {\map N x + \map N y} + \frac {\map N y} {\map N x + \map N y}$

Since $B$ is convex, we have:


 * $\ds \frac {\map N x} {\map N x + \map N y} \paren {\frac x {\map N x} } + \frac {\map N y} {\map N x + \map N y} \paren {\frac y {\map N y} } \in B$

That is:

That is:


 * $\ds \map N {\frac {x + y} {\map N x + \map N y} } \le 1$

Using positive homogeneity, we then obtain:


 * $\ds \frac {\map N {x + y} } {\map N x + \map N y} \le 1$

so:


 * $\map N {x + y} \le \map N x + \map N y$

as was required.