# Metric induced from norm

I was trying to understand the following:

Every norm on $R^n$ is continuous (as a map from $R^n$ to $R$). Proof. We use the maximum metric on $R^n$: $d(x, y) = \max{|x_j − y_j| : j ∈ \{1, . . . ,n\}}$. Let $||\cdot||$ be a norm, and let $C = \Sigma_{j=1}^ne_j$, where $e_1, \dots , e_n$ is the canonical basis of $R^n$.

Then we have $||(x_1,\dots, x_n)|| = ||x_1e_1+...x_ne_n\le|x_1|||e_1||+...|x_n|||e_n||\le \max\{|x_1,...x_n|\}C$ From the triangle inequality, we then get $|||y|| − ||x||| ≤ ||y − x|| ≤ Cd(y, x)$ . So if $d(y, x) < ε/C$, then$|||y|| − ||x||| < \varepsilon$.

My question: The statement claims that every norm is continuous, and nothing about the metric on $R^n$ is specified, since this metric is defined via that norm, how can we take that $d(x, y) = \max{|x_j − y_j| : j ∈ \{1, . . . ,n\}}$? Is it fine to consider arbitrary metric and proceed? Please help.

## 1 Answer

$R^n$ is if not said otherwise assumed to have a distance that derives from a norm.

But all norms are equivalent in finite dimensional vector space (you can search for that in google), what you have in your question is half of the proof

So using the infinite norm to prove this is like taking any other

• why the downvotes ? – Thomas Apr 19 '14 at 12:19
• Looks incomplete. Probably the reason. – DrHAL Apr 18 '16 at 7:11