The following argument is wrong, because the conclusion is impossible. Could someone please help me identify the problem(s)?

Consider the space $l^2(\mathbb{R})$ of square summable sequences. This is a Hilbert space with inner product given by $\langle (a_1, a_2, \dots) , (b_1, b_2, \dots ) \rangle = \sum_i a_ib_i.$

We can define a new inner product $\langle, \rangle^1$ by setting $\langle (a), (b) \rangle^1 = \sum (1/2)^ia_i b_i,$ for $(a), (b)$ sequences in $l^2(\mathbb{R}).$

Hence, by the Riesz representation theorem, any linear functional on $l^2(\mathbb{R})$ can be obtained by the map $(u) \mapsto \langle (v), (u) \rangle^1$ for $(u), (v) \in l^2(\mathbb{R}).$

In particular, if we define a linear functional via the usual inner product by fixing $(a) \in l^2(\mathbb{R})$ and defining $v \mapsto \langle (a), (v) \rangle,$ then there is an element $(a)^1 \in l^2(\mathbb{R})$ such that $\langle (a), \cdot \rangle$ and $\langle (a)^1, \cdot \rangle^1$ define the same functional.

But this can't be right, as it would imply that $(a_1, a_2, \dots) = (a_1^1, 2 a_2^1, 2^2 a_3^1, \dots),$ and the last sequence needn't be in $l^2(\mathbb{R}).$

  • $\begingroup$ Edited the question. $\endgroup$ – user142700 Nov 13 '14 at 17:47
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    $\begingroup$ what makes you believe that $\ell^2$ is a Hilbert space wrt to $\langle . ,. \rangle^1$? Your example sequence even shows it is not. $\endgroup$ – Thomas Nov 13 '14 at 17:50
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    $\begingroup$ Hilbert spaces are complete. $\endgroup$ – Thomas Nov 13 '14 at 17:54
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    $\begingroup$ You see, the Riesz theorem is an existence theorem. Existence is based on completeness. $\endgroup$ – Thomas Nov 13 '14 at 17:55
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    $\begingroup$ @MichaelHardy: I think that is a very unusual notation. At least I frequently use $\ell^2(I)$ for the $L^2$ space on $I$ with counting measure, where the range ($\Bbb{R}$ or $\Bbb{C}$) of the functions is implicitly understood. I have also seen a lot of other people use that notation. Hence, $\ell^2(\Bbb{R})$ would be the set of all $(c_i)_{i \in \Bbb{R}}$ that are square summable. $\endgroup$ – PhoemueX Nov 13 '14 at 17:58

With $\ell_2$ denoting the original Hilbert space, the space $\left( \ell_2, \langle .,. \rangle^1\right)$ is not complete. This is actually an easy consequence of the example you constructed yourself, which shows that the completion of $\ell_2$ with respect to $\langle .,. \rangle^1$ contains $\ell_2 $ as a proper subspace.

The Riesz representation theorem, however, relies on the hypothesis of completeness, is therefore not applicable to $\left( \ell_2, \langle .,. \rangle^1\right)$.


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