On comparing norms in a linear space

[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?

 

[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:
[tex] B\left(a,r\right)=\left\{x:$$||x-a||_{1}<r\right\}\subseteq O$$<br /> 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 $$||\cdot||_{2}$$.<br /> <br /> The real interesting fact is when you have <b>equivalent</b> norms, that is, when:<br /> $$k_2||\cdot||_{2} \geq ||\cdot||_{1} \geq k_1||\cdot||_{2}$$<br /> <br /> This implies that the topologies they generate are equal, and it's a nontrivial fact that, in finite-dimensional spaces, <b>all</b> norms are equivalent.[/tex]