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I need to prove that if a vector space $V$ has two norms $\| \space \|$ and $\| \space \|'$ such that with respect to both norms it becomes a Banach space, then the topologies generated by these norms are either equivalent or incomparable.

I don't understand two things- Firstly, by topologies generated by these norms do we mean that the values obtained from these norm functions create an open set? Secondly, how do I prove the equivalence or incomparable condition?

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By topology generated by $||.||$, we mean that a neighbourhood basis of $0$ is given by $\{ x : ||x|| < r\}$. That is, every open set containing zero, contains some set of that kind i.e. for some small enough $r$.

Suppose that $||.||$ and $||.||'$ are comparable. Then suppose that $d ||x||' > ||x||$. For all $x$.

Note that the map from $(V,||.||') \to (V,||.||)$ given by the identity is a bounded linear map. Further this map is surjective. Hence, by the open mapping theorem it is an open map. Further, because the inverse exists (the identity), by the bounded inverse theorem the inverse is also a bounded map.

Therefore, the inverse map from $(V,||.||) \to (V,||.||')$ is also bounded, which by definition means that $||x||' \leq c ||x||$ for all $x$ for some constant $c$.

Hence, the norms $||.||$ and $||.||'$ are equivalent, if they are comparable.

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  • $\begingroup$ Isn't there any difference between the equivalence of two norms and equivalence of topologies generated by the norms? $\endgroup$ – user371842 Sep 24 '16 at 11:06
  • $\begingroup$ No, there is no difference. To see this, note that by equivalence of norms, every open set of one kind is contained, and contains an open set of the other kind. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '16 at 3:57

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