Paradox or Error: On the inclusion of dense subspaces into Hilbert spaces 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?
 A: The error lies in "As $i$ is a bounded injection, $i^{\ast}$ is a bounded surjection". In fact, the range of $i$ is closed in $H$ if and only if the range of $i^{\ast}$ is closed in $U^{\ast}$ by the closed range theorem.
However, in the present situation, we need not use this rather difficult result (which is true for all Banach spaces) but it's more an exercise in the yoga of orthogonal complements.

Added Later
As I failed to mention (and Nate was kind enough to point out in his answer), the image of $i^{\ast}$ is dense of course. This is implicit in my yoga exercise above.
To see why $i^{\ast}$ need not be surjective, let me treat the concrete situation of the question by exhibiting an explicit example of a functional on $H^1$ which is not in the image of $i^{\ast}$.
Recall the following facts: The Sobolev space $H^1[0,1]$ is dense in $L^2[0,1]$ (since it contains the smooth functions) and consists of continuous functions. In fact, there is an inclusion  $H^1[0,1] \to C[0,1]$ which is compact (hence continuous) by one of the standard compactness theorems (whose exact name I tend to forget), when the former is equipped with its Sobolev norm and the latter is equipped with the sup norm.
Therefore it makes sense to define the evaluation functional $\phi(f) = f(p)$ for $f \in H^1$ and $p \in [0,1]$ and $\phi$ is continuous with respect to the Sobolev norm.
On the other hand, $\phi$ is not continuous with respect to the $L^2$-norm, as can be seen by choosing an appropriate sequence of smooth bump functions such that $\|f_n\|_2 \to 0$ and $\phi(f_n) = f_n (p) \to \infty$.
This means that $\phi$ cannot be in the image of $i^{\ast}$ (which is simply the restriction of functionals on $L^2$ to $H^1$ via $i$), and hence $i^{\ast}$ fails to be surjective.
A: I think it's worth mentioning that although, as Theo points out, $i^*$ need not be surjective, it is close: under these assumptions, $i^*$ has dense image.  Proving this is a nice short exercise.
