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I am quite confused at the notation used for $J$ in the following question:

For a normed space $\Omega$, let $\Omega^{**}$ denote the dual space of the dual space. Let $J: \Omega \rightarrow \Omega^{**}$ be defined by $\langle x^*, J(x)\rangle = \langle x, x^*\rangle = x^*(x)$. Show that $J$ is a linear isometry.

Where, for $x^* \in \Omega$, $\langle x, x^*\rangle = x^*(x)$.

I am confused on how to decipher the notation for $J$. I am normally use to things like $J(x) = $ fill in the blank. Could someone please explain this notation?

Martin Argerami
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9301293
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  • If $x\in \Omega$, then $J(x)\in \Omega^{*}$ so it is a functional on $\Omega^$, and thus you have to plug in an element $x^* \in \Omega^$. The definition should be $J(x)(x^):=...$, but you can also denote this pairing as $<x^*, J(x) >$. – Dimitris May 08 '16 at 21:20
  • If you write the function application as $f(y)$, the definition becomes $(J(x))(x^{\ast}) = x^{\ast}(x)$. Or $J(x) \colon x^{\ast} \mapsto x^{\ast}(x)$. Verbally, $J(x)$ is evaluation at $x$. – Daniel Fischer May 08 '16 at 21:20
  • @DanielFischer So, this is just the evaluation map, as in linear algebra? – 9301293 May 08 '16 at 21:23
  • Yes. Only in linear algebra one doesn't consider continuity, so this is the restriction of the evaluation map to continuous linear forms. And it needs to be shown (though it's easy) that $J(x) \colon \Omega^{\ast} \to \mathbb{C}$ is continuous. – Daniel Fischer May 08 '16 at 21:28
  • Hint: Riesz representation theorem for linear functionals. I'm typing a full answer below if you'd like to just read the proof. – D_S May 09 '16 at 02:39
  • @D_S Well, I answered the question. But, no one has answered it officially. – 9301293 May 09 '16 at 02:40
  • Ah okay, I thought you had still had a question about surjectivity – D_S May 09 '16 at 02:41
  • @D_S I did, but then I figured it out. This question does need to be answered, though. Answer, so I can give points! – 9301293 May 09 '16 at 02:43
  • Haha, thank you for the offer :) – D_S May 09 '16 at 02:50

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