A paradox on Hilbert spaces and their duals I am making some elementary mistakes here. Could you please help me point out the problems? Thank you very much!
Suppose on some space $H$ we have two inner products, which make $H$ after completion two real Hilbert spaces. Suppose that these two inner products are comparable:
$$
(f,f)_1 \le (f,f)_2,\quad \forall f\in H. 
$$
Denote these two Hilbert spaces by $H_1$ and $H_2$. It is clear that
$$
H_2 \subseteq H_1.
$$
Let $H_1'$ and $H_2'$ be the dual spaces of $H_1$ and $H_2$, respectively. They are also Hilbert spaces. Noticing that the fact that
$$
||u||_{H_i'} = \sup_{(x,x)_i \le 1} |u(x)|,\qquad i=1,2,
$$
we have that 
$$
H_1' \subseteq H_2'\:.
$$
Being Hilbert spaces, $H_i \cong H_i'$ (i.e., $H_i$ is isomorphic to $H_i'$). Since the Hilbert spaces are real, we can identify $H_i'$ by $H_i$. Then the above two inclusions imply that
$$
H_1 = H_2 ,
$$
which cannot be true in general. What is wrong in my arguments? Thank you very much for your great help! :-)
Anand
 A: A key question here is "What precisely does $H_2\subseteq H_1$ mean"? 
Let's begin with the identity map on the
pre-Hilbert space $H$: $$\text{id}:(H,(\cdot,\cdot)_2)\to (H,(\cdot,\cdot)_1).$$
where we assume that, for all $h\in H$, we have  $(h,h)_1\leq (h,h)_2$.
This map lifts uniquely to the completions 
$$i: H_2\to H_1.$$
It is often overlooked, however, that the resulting map $i$ may not be injective.
The map $i$ is injective if and only if the 
bilinear form $((h,h)_2, H)$ is closable on $H_1$.
Let's suppose  that $i$ is injective so that we can really think of
elements of $H_2$ as also belonging to $H_1$. 
We should consider $H_2\subseteq H_1$ as shorthand for $i:H_2\to H_1$.
Now $i$ is a linear map, and the dual map gives  $i^\prime:H_1^\prime\to H_2^\prime$.
This gives a precise meaning to $H_1^\prime\subseteq H_2^\prime$. 
Combining this with  Riesz representations on $H_1$ and $H_2$, we can write
$$H_1\overset{j}{\rightarrow} H_1^\prime \overset{i^\prime}{\rightarrow} H_2^\prime
\overset{k}{\rightarrow} H_2.$$
I think that this is what you mean by $H_1\subseteq H_2$.    
In short, all of the paradoxes disappear when you keep careful
track of the mappings involved. 
