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If $V \subset H \subset V^*$ is Hilbert triple, and $h \in H$ what's $\langle h, v \rangle_{V^*, V}$?

I know we interpret it to be $(h,v)_H$. But is this correct: $$\langle h, v \rangle_{V^*, V} := \langle Rh, v\rangle_{V^*, V} = Rh(v) = (h,v)_H$$ where $R:H \to H^*$ is the Riesz map and the last equality follows by Riesz representation.

I am still a bit confused about the relation between $V$ and its dual and what we identify and what we don't.

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1 Answer 1

up vote 2 down vote accepted

There are three maps in play here:

  1. the embedding $\iota : V \to H$,
  2. its transpose, the embedding $\iota^T : H^\ast \to V^\ast$,
  3. and the Riesz (anti)isomorphism $R : H \to H^\ast$.

In particular, the inclusion $H \to V^\ast$ is actually given by the composition $\iota^T \circ R$. Thus, $$ \left\langle h,v \right\rangle_{V^\ast,V} := \left\langle (\iota^T \circ R)(h), v\right\rangle_{V^\ast,V} = \left\langle R(h),v \right\rangle_{H^\ast,H} = \left(h,v\right)_H,$$ as you indeed computed.

EDIT: Note that since $V$ will be a proper subspace of $H$, endowed with a topology (typically, that of a nuclear Fréchet space) finer than the subspace topology inherited from $H$, you'll really have that $V \subsetneq H \subsetneq V^\ast$.

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thanks good answer. –  george.s Feb 10 '13 at 12:04

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