# The theorem about the equivalence of the norms

The theorem about the equivalence of the norms says that every two norms in $\Bbb R^n$ are equivalent. It can be written as $c_1||x|| \le ||x||_2 \le c_2 ||x||$. I struggle with imaging it and don't know how to prove it (or I don't understand that proof).

• Does it mean that it doesn't matter which norm I would use for my given vector space if I want to measure something there or that If I am using one type of the norm it is easy to tranfer to another? The Lipschitz continuity says if I have the linear mapping between two metric spaces such that $f: (M, d_m) \to (N, d_n)$ than we have some constant $K \gt 0$ for every $x,y \in M$ such that $d_N(f(x), f(y)) \le Kd_M(x-y)$. So transfering between the norms then means the linear transformation (homomorphism).

• I have alson read that $f^{-1}$ exists and that the mapping works for the finite (bounded) vector spaces, the first condition I understand, the second I don't.

• Why I should compare all the norms with euclidean norms?

For example, you might have started with the Euclidean norm on the unit disc and then realised that you're in a space that looks pretty hyperbolic, so you might find that switching to the norm generated by the Poincaré metric ($d_P(z_1, z_2) = \tanh^{-1}\left| \frac{z_1-z_2}{1-z_1\bar{z_2}}\right|$) simplifies many of your calculations.