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.


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)$.

  • 1
    $\begingroup$ I should never forget that $a \leq b \leq a$ implies $a = b$ $\endgroup$ – user8469759 Feb 8 '17 at 10:54

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