Homogeneity on Minkowski functional

Question

Let $ X $ be a vector space over a field $ \mathbb{K} = \mathbb{R}, \mathbb{C} $ and $ A \subseteq X $ convex and absorbing.

Then $ \forall x, y \in X $, $ \forall t > 0 $

1. $ \mu_A (x + y) \leq \mu_A (x) + \mu_A (y) $
2. $ \mu_A (tx) = t \mu_A (x) $

In Rudin's Functional Analysis, theorem 1.35, it states that $ \mu_A $ is a seminorm if $ A $ is also balanced, so that one has to prove $ \forall \alpha \in \mathbb{C} $ $ \mu_A (\alpha x) = |\alpha| \mu_A (x) $; according to the book, it follows from 1. and 2., still I have no clue why.

I suppose I may start with

$ \mu_A (\alpha x) = \mu_A ((a+ib)x) \leq \mu_A (ax) + \mu_A (ibx) $

but I cannot go much further.

Can anyone provide (a sketch of) a complete proof?

Answer 1

Assume $\alpha \neq 0$. $\{t>0: \frac 1 t {\alpha x} \in A\} \subseteq \{t>0: \frac 1 t {|\alpha| x} \in A\} $ because $y \in A$ implies $\frac {|\alpha|} {\alpha} y \in A$. Taking infimum we get $\mu_A(\alpha x) \geq \mu_A ({|\alpha| x})$. To get the reverse inequality first change $x$ to $\frac 1 {\alpha} x$ and the change $\alpha$ to $\frac 1 {\alpha}$.

Answer 2

Combine the following:

$\tag1\alpha=|\alpha|e^{is}\ \text{for some}\ 0\le s< 2\pi.$

$\tag2 \{t>0: \frac 1 t {\alpha x} \in A\} = \{t>0: \frac 1 t {|\alpha| (e^{is}x)} \in A\}=\{t>0: \frac 1 t {|\alpha| x} \in A\}.$

$\tag3 \mu_A(|\alpha|x)\le \frac 1 t {|\alpha| x}\Rightarrow \frac{1}{|\alpha|}\mu_A(|\alpha|x)\le \mu(x)\Rightarrow \mu_A(|\alpha|x)\le |\alpha|\mu_A(x).$

$\tag 4|\alpha|\mu_A(x)\le |\alpha|\frac{1}{t}x=\frac{1}{t}(|\alpha|x)\Rightarrow |\alpha|\mu_A(x)\le \mu_A(|\alpha|x).$