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We all know the Riesz representation theorem, so we can represent a bounded linear function on a Hilbert space $H$ with an inner product on $H$ and vice-versa. My question is, given an inner product in $H^*$, say $(a,b)_{H^*}$, can I write it as $$(a,b)_{H^*} = \langle a, f \rangle_{H^*, H}$$ where $f \in H$? This is the RRT applied to the Hilbert space $H^*$ with its dual $H$. I think it works but I never saw it so I should get it clarified.

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How the inner product is given? –  Berci Feb 20 '13 at 15:09
@Berci I guess you pull back the inner product to one on $H$ by using Riesz map. –  michael_faber Feb 20 '13 at 15:40
In that case, it holds. –  Berci Feb 20 '13 at 15:43

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Well, for an arbitrary inner product on $X:=H^*$ it is not going to work, since then $X^*$ need not be isomorphic to $H$.

On the other hand, the Riesz representation gives a linear isomorphism $H\to H^*$, and if the inner product is defined via this isomorphism, i.e. if $$(\langle x,-\rangle,\langle y,-\rangle)_{H^*}=\langle x,y\rangle_{H} $$ for all $x,y\in H, \ f\in H^*$, then your claim is valid:

Let $a,b\in H^*$ then $a=\langle x,-\rangle$ and $b=\langle y,-\rangle$ for some $y\in H$ by Riesz representation, and $a(y)=\langle x,y\rangle$ so we have $$(a,b)_{H^*} = (a,\langle y,-\rangle)=a(y)=\langle a,y\rangle_{H^*,H}\ .$$

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Are you sure it doesn't work for arbitrary inner products? Since $H$ and $H*$ are Hilbert, they're reflexive so $H = H^{**}$ –  michael_faber Feb 20 '13 at 16:01
1. How do you know $H^*$ will be Hilbert space with that arbitrary inner product? In general, it can have nothing common with $H$ or $H^*$, just the underlying vector space structure.. Probably if the generated norm is equivalent to the operator norm in $H^*$.. –  Berci Feb 20 '13 at 16:08
Ok. How about if $H^*$ is given an IP that makes it Hilbert, not necessarily the Riesz map though? –  michael_faber Feb 20 '13 at 18:00
By pulling back the IP to $H$ (via the Riesz map), we get another IP on $H$, so you could start considering what connection is needed between $2$ inner products on the same vector space $H$ such that their duals are the same. Maybe, if they generate equivalent norms it's enough.. –  Berci Feb 21 '13 at 0:32

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