Set defined by infimum How can we prove:
$$|\alpha|*||x|| = ||\alpha x|| $$ 
With the norm defined as:
$$ ||x|| := \inf\left\{ \lambda > 0 \mid  x/\lambda\in B  \right\} $$
Where B is convexe, open, symmetric and bounded and $ 0\in B$.
I thought about to proof this by contradiction, but I never encountered a situation where I wanted to pull out a constant from a definition of a set.

I would suggest letting $m = |\alpha| \mathrm{inf} \{ \lambda > 0 \mid x/\lambda \in B \}$ and verifying that $m$ satisfies the requirements to be the infimum of the set $\{ \lambda > 0 \mid \alpha x / \lambda \in B \}$. That is, prove that


*

*If $\lambda > 0$ satisfies $\alpha x/ \lambda \in B$, then $m \le \lambda$; and

*If $m'$ is another value such that $m' \le \lambda$ for all $\lambda > 0$ such that $\alpha x/\lambda \in B$, then $m' \le m$.


You'll use the fact that $m$ is $|\alpha|$ times an infimum of another set in your proof.
