Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $V \subsetneq H \subsetneq V^\ast$.

share|cite|improve this answer
thanks good answer. – george.s Feb 10 '13 at 12:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.