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

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Theorem

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

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:

$(1): \quad$ $\map N x = 0$ if and only if $x = 0$
$(2): \quad$ $\map N {\lambda x} = \cmod \lambda \map N x$ 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:

\(\ds \map N {x + y}\) \(=\) \(\ds \map N y\)
\(\ds \) \(=\) \(\ds \map N x + \map N y\)

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

\(\ds \map N {x + y}\) \(=\) \(\ds \map N x\)
\(\ds \) \(=\) \(\ds \map N x + \map N y\)

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:

\(\ds \map N {\frac x {\map N x} }\) \(=\) \(\ds \frac {\map N x} {\map N x}\) since $N$ is positive homogeneous
\(\ds \) \(=\) \(\ds 1\)

so:

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

Similarly:

\(\ds \map N {\frac y {\map N y} }\) \(=\) \(\ds \frac {\map N y} {\map N y}\) since $N$ is positive homogeneous
\(\ds \) \(=\) \(\ds 1\)

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:

\(\ds \frac {\map N x} {\map N x + \map N y} \paren {\frac x {\map N x} } + \frac {\map N y} {\map N x + \map N y} \paren {\frac y {\map N y} }\) \(=\) \(\ds \frac x {\map N x + \map N y} + \frac y {\map N x + \map N y}\)
\(\ds \) \(=\) \(\ds \frac {x + y} {\map N x + \map N y}\)
\(\ds \) \(\in\) \(\ds B\)

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.

$\blacksquare$


Sources