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:
<br /> B\left(a,r\right)=\left\{x:||x-a||_{1}&amp;lt;r\right\}\subseteq O&lt;br /&gt;<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 /> &lt;br /&gt; k_2||\cdot||_{2} \geq ||\cdot||_{1} \geq k_1||\cdot||_{2} &lt;br /&gt;<br /> <br /> This implies that the topologies they generate are equal, and it&#039;s a nontrivial fact that, in finite-dimensional spaces, <b>all</b> norms are equivalent.