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The weak topology on normed linear space $X$ can be defined as being induced by semi-norms $\|\cdot\|_{x'}$, $x'\in X'$ with $\|x\|_{x'}=|x'(x)|$. Similarly the weak* topology is induced by $\|\cdot\|_x$, $x\in X$ with $\|x'\|_x=|x'(x)|$.

My question: If $A'$ (resp $A$) is a norm dense subset of $X'$ resp ($X$), are the topologies generated by semi-norms $$\{\|\cdot\|_{x'} \mid x' \in A'\}\qquad\qquad\{\|\cdot\|_x \mid x \in A\}$$

the same as the weak (resp weak*) topology?

It is straightforward to show that these topologies agree with the weak (resp weak*) topologies on bounded subsets of $X'$ (resp $X$), but I think it is not true on all of $X'$ (resp $X$).

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For a subspaces $A'$ of $X'$ the topology on $X$ given by the seminorms $x\mapsto |a'(x)|$ with $a'\in A'$ is somtimes denoted $\sigma(X,A')$. A standard result is that the dual of $(X,\sigma(X,A'))$ is equal to $A'$. In particular, $\sigma(X,A')$ is strictly coarser than $\sigma(X,X')$ if $A'\neq X'$. The situation becomes more symmetric for general dual pairs. (I don't have references at hand, but google will help.)

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  • $\begingroup$ Thanks for the hints, if I have time I might write up the proofs tomorrow. A related question: you were assuming that $A'$ is a vector subspace, this is not really what I was considering, but $\sigma(X,A')=\sigma(X,\text{span}(A'))$ true. $\endgroup$ – s.harp May 28 '16 at 18:51

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