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Given a normed vector space $X$. Let $\mathcal{T}$ represent the topology on $X$ induced by the norm.


  • $A$:={topologies that can make the norm continuous},
  • $B$:={topologies that can make the addition and scalar multiplication continuous},
  • $C$:={topologies that can make every element in the continuous (wrt $\mathcal{T}$?) dual space $X^{**}$ continuous}. ( The coarsest in $C$ is called the weak topology on $X$?)

Is it true that

  1. neither of $A$ and $B$ is a subset of the other;
  2. $A$ can be ordered by finer/coarser, while $B$ cannot be (because the codomain of the norm is $\mathbb{R}$ with a known topology, while the domains and codomains of addition and of scalar multiplication all depend on $X$?);
  3. $\mathcal{T}$ is the coarsest topology in $A\cap B$;
  4. $C \equiv A \cap B$.
  5. Same questions as in 1, 2, and 3, if "norm" is replaced with "inner product" in the above description.
  6. Are there other types of topologies on a normed/inner product space often encountered, besides those mentioned above?

Thanks and regards!

share|cite|improve this question
The third assertion is true. The norm topology consists exactly of the pre-images of the norm on open subsets of $\mathbb{K}$. Thus if $\mathcal{T}' \in A \cap B \subseteq A$, the continuity of the norm wrt $\mathcal{T}'$ gives $\mathcal{T} \subseteq \mathcal{T}'$. – user18063 Jan 6 '12 at 4:41
About the first question, I don't think a general answer can be given. If $X$ is finite dimensional, and if $\mathcal{T}' \in B$ then all norms are continuous in $(X,\mathcal{T}')$. This means $B \subseteq A$. On the other hand, the norm induced on $\ell^p$ via an isomorphism with $\ell^q$, where $q>p$, will be discontinuous with respect to the usual topology. So for $(\ell^p, ||\cdot||_q)$, there is a topology in $B$ (the usual topology on $\ell^p$) that is not in $A$, i.e. $B \nsubseteq A$. – user18063 Jan 6 '12 at 8:46
@user18063: I edited your comments to fix the issue introduced by the OP's edit. – Willie Wong Jan 6 '12 at 9:39

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