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! :-)


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    $\begingroup$ When you identify $H_1$ with $H_1'$, it's with respect to the first inner product, whereas the identification between $H_2$ and $H_2'$ is with respect to the second inner product. We cannot do these identifications "in the same time". An example is $H=\ell^2$, and $V:=\{u\in H,\sum_n nu_n^2<\infty\}$ with the inner product $((u,v))=\sum_n nu_nv_n$. $\endgroup$ – Davide Giraudo Jan 28 '12 at 15:56
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    $\begingroup$ $H_1'\not \subseteq H_2'$. They are functionals on different spaces. $\endgroup$ – azarel Jan 28 '12 at 16:03

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.

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    $\begingroup$ Just a comment... the composition $ki'j$ is just the Hilbert space adjoint $i^*$. Then $i^*i$ is a positive operator which "implements" $(.|.)_1$ as $(i^*i(h)|h')_2 = (i(h)|i(h'))_1 = (h|h')_1$ for $h,h'\in H$. So we'd only have a "paradox" if $i^*i$ were the identity-- but we now see that this happens only when $(.|.)_1 = (.|.)_2$... $\endgroup$ – Matthew Daws Jan 28 '12 at 20:57
  • $\begingroup$ Thank Professors Byron Schmuland and Matthew Daws for your nice answer and nice comment. I am clear now. :-) $\endgroup$ – Anand Jan 29 '12 at 12:14

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