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i know that we can identify $H$ with his $H^*$(dual) then if we have an other sub-space $V$ of$H$ then we will have $V\subset H=H^*\subset V^*$ can we identify here $V$ with $V^*$ then we will have V=H sure we can not but why ?

kamerove
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  • Why do you think that $V \subset H$ entails $H^{\ast} \subset V^{\ast}$? That's quite wrong. – Daniel Fischer Nov 15 '15 at 15:29
  • Probably relevant. – Daniel Fischer Nov 15 '15 at 15:32
  • thank you very much ,,, i think i confused with the notion of "densly embedded" in this case will be true normally , i wanna ask you about this notion : for example : let $V$ be a banach space , $H$ a hilbert space , $V$ is densly embedded in $H$ means that $\overline V$=$H$ and $V \subset H$ ? – kamerove Nov 15 '15 at 15:56
  • and what do you mean by the adjoint of the inclusion plaise ? – kamerove Nov 15 '15 at 15:59
  • And the topology on $V$ must be (strictly) finer than the subspace topology induced by $H$ to get something interesting. Then you have a continuous inclusion $H^{\ast} \hookrightarrow V^{\ast}$, which is the adjoint/dual/transpose [choose your favourite term, these are different words for the same thing] of the inclusion $V \hookrightarrow H$. – Daniel Fischer Nov 15 '15 at 16:02
  • thanks Daniel Fischer for unswering i am really confused about this and i have an exam soon, well : what we did is $V\subset H=H^* \subset V^* $ – kamerove Nov 15 '15 at 16:15
  • and this by identification $H$ and $H^$ in this case if for example $V$ is a hilbert space then why i can not identify $V$ and $V^$ (here every thing is densly embedded ) – kamerove Nov 15 '15 at 16:18
  • Well, we don't have $H = H^{\ast}$. We have an isometric isomorphism (or an anti-isomorphism for complex Hilbert spaces) via the Riesz map of $H$. Composing the various maps, we get a continuous inclusion $V \hookrightarrow V^{\ast}$, but that is a different map from the Riesz map of $V$. Unless $V = H$, it is not surjective. – Daniel Fischer Nov 15 '15 at 16:23
  • aah ok i think you mean from $V \hookrightarrow V^{\ast}$ is not an isomety ? for that we don't have V=H ? am i right ? but for $H=H^*$ this what we wrote in our cours, the prof. told us that when we identify two spaces by an isometry we write the equality – kamerove Nov 15 '15 at 17:04

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