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On comparing norms in a linear space

  1. Jan 28, 2010 #1
    a norm [tex] ||\cdot||_{1} [/tex] is said to be stronger than [tex] ||\cdot||_{2} [/tex] if there exists some constant k such that

    [tex]||\cdot||_{1} \geq k||\cdot||_{2} [/tex]

    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. jcsd
  3. Jan 29, 2010 #2
    The point is in the topologies they generate: suppose you have an open set [tex]O[/tex], relative to [tex]||\cdot||_{1}[/tex]; this means that, for any [tex] a \in O [/tex], there is an [tex]r > 0[/tex], 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 [tex]||\cdot||_{1} \geq k||\cdot||_{2}[/tex], the set [tex]O[/tex] must remain open if you switch to [tex]||\cdot||_{1} \geq k||\cdot||_{2}[/tex]. Therefore, the topology generated by [tex]||\cdot||_{1}[/tex] is stronger (has at least as many open sets) as the one generated by [tex]||\cdot||_{2}[/tex].

    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.
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