On comparing norms in a linear space

1. Jan 28, 2010

Somefantastik

a norm $$||\cdot||_{1}$$ is said to be stronger than $$||\cdot||_{2}$$ if there exists some constant k such that

$$||\cdot||_{1} \geq k||\cdot||_{2}$$

Can someone explain the deeper meaning of this? I know that in general, a norm with smaller value will produce a larger unit ball. Is that the extent of the meaning?

2. Jan 29, 2010

JSuarez

The point is in the topologies they generate: suppose you have an open set $$O$$, relative to $$||\cdot||_{1}$$; this means that, for any $$a \in O$$, there is an $$r > 0$$, such that the ball:
$$B\left(a,r\right)=\left\{x:[tex]||x-a||_{1}<r\right\}\subseteq O$$
Now, notice that, because of the inequality $$||\cdot||_{1} \geq k||\cdot||_{2}$$, the set $$O$$ must remain open if you switch to $$||\cdot||_{1} \geq k||\cdot||_{2}$$. Therefore, the topology generated by $$||\cdot||_{1}$$ is stronger (has at least as many open sets) as the one generated by $$||\cdot||_{2}$$.

The real interesting fact is when you have equivalent norms, that is, when:
$$k_2||\cdot||_{2} \geq ||\cdot||_{1} \geq k_1||\cdot||_{2}$$

This implies that the topologies they generate are equal, and it's a nontrivial fact that, in finite-dimensional spaces, all norms are equivalent.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook