How to prove $C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$? I want to prove the following theorem (no idea whether it has a name):
Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $\|\cdot\|$ be a norm on $V$. Then, there exist $C_1, C_2 > 0$ such that for all $x \in V$:
$$C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$$
I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $\|x\|_\infty$ and normalized the vector $x$ such that $\|x\|_\infty = 1$. That left me with
$$C_1 \leq \|x\| \leq C_2$$
but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers.
 A: If you are interested in an elementary proof, try $C_1 = \min_j \| e_j \|$ and $C_2 = \sum_j \| e_j \|$.
A: I try to make a self-contained answer to the question (more info can be found on this page). The short answer is: yes you are on the right track!
What you are missing:

*

*$\|x\|$ is a continuous function with respect to the $\|\cdot\|_\infty$ norm. You can write $x= \sum_m x_m e_m$ with $e_m$ a basis with $\|e_m\|=1$ (similarly, $y= \sum_m y_m e_m$). We are interested in $$\|x-y\| = \left\| \sum_m (x_m - y_m) e_m \right\| \leq \sum_m | (x_m - y_m)|  \leq n \| x - y\|_\infty$$ using the triangle inequality. So for every $\epsilon>0$, we can choose a $\delta =\epsilon/n$ such that with $\| x - y\|_\infty < \delta$ it follows that $\|x-y\| < \epsilon$. Therefore, $\|x\|$ is $\infty$-continuous.


*The unit sphere $S= \{x :\|x\|_\infty = 1\}$ is a compact set thereby $\|x\|$ assumes a minimum $C_1$ and a maximum $C_2$ on it.
A: The absolute value of every component is smaller than the absolute value of the max of all components, so the sum of the absolute values of components is smaller than n*max of components. this gives C_2=n
on the other hand, the absolute value of the max of components is smaller than the sum of the absolute values of all components, which is nothing but the first norm. C_1=1. 
