the following observations are very simple, but I suppose they contain an error, which I haven't been able to find it so far. Maybe somebody can help how to fix it:
Let $H$ be a Hilbert space, $U$ be a dense subspace. Assume we can equip $U$ with another Hilbert space norm by itself. We denote this space by $\mathcal U$, to avoid misunderstandings.
We then have the linear inclusion $ i : \mathcal U \rightarrow H$ with image dense in $H$. Assume furthermore, $i$ is bounded.
Let us now inspect the dual arrow, which acts on the topological dual spaces: $i^\ast : H^\ast \rightarrow \mathcal U^\ast, \;\; w( \cdot) \mapsto w( i \cdot )$
As $i$ is a bounded injection, $i^\ast$ is now a bounded surjection. What is the kernel of $i^\ast$? We have
$\ker i^\ast = \{ w \in H^\ast : w(x) = 0, x \in \operatorname{Im} i \}$
But then $\ker i^\ast = (\operatorname{Im} i)^\perp = U^\perp = H^\perp = \{0\}$, so $i^\ast$ is injective. Hence it is an isomorphism.
Strange: But if we dualize $i^\ast$ again, we then see $i$ is an isomorphism, too.
Furthermore, as $H$ and $\mathcal U$ are Hilbert spaces, we are given (isometric) isomorphisms $H \simeq H^\ast$, $\mathcal U \simeq \mathcal U^\ast$. We can compose these morphisms.
Strange: We obtain $\mathcal U \simeq H$, where the injection is in fact an isomorphism.
So this eventually means, if we equip any dense subspace of a Hilbert space with a stronger topology, then the injection $i$ as above is an isomorphism.
This seems paradoxical, and I suppose there is an error in the above. For example, this implies the injection $H^1(\mathbb R) \rightarrow L^2(\mathbb R)$ is an isomorphism.
So, can anybody please either: (a) Point out where I have been wrong (b) Point out how to interpret this paradox?