# A property of the Minkowski functional

Let $E$ be a normed vector space let also $C$ be a convex subset of $E$ such that $0 \in E$ defining.

$$p(x) = \inf \left\{ \alpha : \alpha^{-1}x \in C \right\}$$

I want to prove that for each $\beta > 0$ I have $p(\beta x) = \beta p(x)$, apparently this think should sound obvious but it is not to me... So I want to prove it rigorously...

My attempt is

$$p(\beta x) = \inf \left\{ \alpha : \alpha^{-1} \beta x \in C \right\} = \inf \left\{ \alpha \beta^{-1} \beta : \alpha^{-1} \beta x \in C \right\}$$

Defining $\gamma = \alpha \beta^{-1}$ end up with

$$p(\beta x) = \inf \left\{ \gamma \beta : \gamma^{-1} x \in C \right\}$$

But I got stuck here... all the proofs I've seen they kind of factor the $\beta$ from the set, I'm missing the very reason that allow to do that.

• – Surb Feb 8 '17 at 10:47

I write $a,b$ instead aof $\alpha, \beta$.

For $b>0$ we have $p(bx) \le a^{-1}bx$ for all $a>0$, hence $\frac{p(bx)}{b} \le a^{-1}x$ for all $a>0$.

This gives $\frac{p(bx)}{b} \le p(x)$ or

(1) $p(bx) \le bp(x)$.

From $p(x) \le a^{-1}x$ for all $a>0$ we get $bp(x) \le a^{-1}bx$ for all $a>0$, thus

(2) $bp(x) \le p(bx)$.

• I should never forget that $a \leq b \leq a$ implies $a = b$ – user8469759 Feb 8 '17 at 10:54