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Let $V$ and $H$ be Banach spaces and suppose that $$V\subset H.$$ Which topology (either from $V$ or $H$) is finer? If topology induced by $H$ is finer, then is topology in $H^*$ finer than $V^*$? Can we ommit the assumption that $V$ and $H$ are Banach spaces and obtain the same relations?

I think that we should distinguish two cases. The first one is that $V$ has topology inherited from $H$, which means that it consists of all subsets from $V$ that has nonempty intersection with open sets from $H$. Then, topology in $V$ is at least as big as topology from $H$. Hence, topology in $H^*$ is finer than topology in $V^*$.

What about the second case - we have two topologies induced by different norms? Can we somehow compare those topologies? For instance $V=H^1(\mathbb{R})$ and $H=L^2(\mathbb{R})$.

zorro47
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  • Could you provide some context and tell us what you have done? – Kavi Rama Murthy Aug 02 '18 at 10:19
  • I'm trying to understand $"H^\subset V^$ and I found https://math.stackexchange.com/questions/655718/why-is-this-inclusion-of-dual-of-banach-spaces-wrong?rq=1 But, I don't know much about topology. – zorro47 Aug 02 '18 at 10:24
  • In the post you have quoted it is shown that we cannot consider $H^{}$ as a subspace of $V^{}$ in general. So where is the question of comparing topologies on $H^{*}$? – Kavi Rama Murthy Aug 02 '18 at 10:27
  • Exactly, but if $V$ is dense in $H$ than the restriction map is injective and hence, we may in some sense think of "$H^=V^$". But still, we write $H^\subset V^$. According to the post, the inclusion may be in common with topologies. – zorro47 Aug 02 '18 at 10:31

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