In $\mathbb{R}^{n}$ all norms are equivalent

Question

While trying to prove the Theorem mentioned in the Title, I got stuck in the inequality shown below.

I think that the proof uses the $\epsilon$ and $\delta$ definition of continuity but I am not able to understand the steps in between the inequality.

Thank you in advance for your help!

Answer 1

The first inequality is the triangle inequality (subadditivity):

$$ \rho(x) = \rho(x - y + y) \leq \rho(x - y) + \rho(y). $$

You can similarly show

$$ \rho(y) \leq \rho(x - y) + \rho(x) $$

establishing the absolute values.

For the second inequality, we write

$$ x = \sum\xi_i e_i, \quad y = \sum\eta_i e_i $$

so

$$ x - y = \sum (\xi_i - \eta_i) e_i $$

and once again use the triangle inequality, along with homogeneity of norms (i.e. $\rho(cx) = |c|\rho(x)$ for a scalar $c$ and vector $x$).

You might think of the proof as using the limit definition of continuity (if $y \to x$, then the rightmost expression becomes very small, so the leftmost expression has to get smaller as well, showing $\rho(y) \to \rho(x)$), but of course this is equivalent to the $\epsilon - \delta$ definition, so you could easily (and correctly) think of it that way too.