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.
 A: 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.
A: Sketch of one direction:
Observe that $\|\alpha x\|=\inf\{\lambda>0:\alpha x/\lambda\in B\}$.  Since this is an infimum, by the definition of an infimum, for all $\varepsilon>0$, there exists a $\lambda_\varepsilon$ such that 


*

*$\frac{\alpha x}{\lambda_\varepsilon}\in B$

*$\|\alpha x\|\leq \lambda_\varepsilon<\|\alpha x\|+\varepsilon$.
Consider $\lambda'_\varepsilon:=\frac{\lambda_\varepsilon}{|\alpha|}$.  This lambda satisfies $\frac{x}{\lambda'_\varepsilon}\in B$ since $\frac{x}{\lambda'_\varepsilon}=\frac{|\alpha| x}{\lambda_\varepsilon}$, which we know is in $B$ since $\frac{\alpha x}{\lambda_\varepsilon}\in B$ and $B$ is symmetric about the origin.  Therefore, we know that $\lambda'_\varepsilon$ is one of the elements of $\{\lambda>0:x/\lambda\in B\}$.  Hence, for the infimum, $\|x\|\leq\lambda'_\varepsilon$.
From the inequalities above, we know that 
$$
\frac{1}{|\alpha|}\|\alpha x\|\leq \lambda'_\varepsilon<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}.
$$
Combining inequalities, we know that
$$
\|x\|\leq\lambda'_\varepsilon<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}.
$$
In other words,
$$
\|x\|<\frac{1}{|\alpha|}\|\alpha x\|+\frac{\varepsilon}{|\alpha|}.
$$
Since $\varepsilon$ was arbitrary, we can let it be as small as possible, and, in the limit, we get
$$
\|x\|\leq \frac{1}{|\alpha|}\|\alpha x\|.
$$
This gives the proof of one side, for the other direction, mimic this proof, but start with $\|x\|$.  You can, alternatively, replace $\alpha x$ by $x$ and $\alpha$ by $\frac{1}{\alpha}$ to use this proof as a lemma.
