Positive Definite and Positive Homogeneous Map with Convex Closed Unit Ball is Norm
Jump to navigation
Jump to search
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
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $3.1$: Norms