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According to my lecture notes (we're using Folland' Real Analysis textbook), if $X$ is a normed vector space, then $L(X,Y) = \left\lbrace \text{all bounded linear operators T} : X \rightarrow Y \right\rbrace$. We specified that the dual space $X^*$ is $L(X,\mathbb{R})$ or $L(X,\mathbb{C})$.

However, I don't understand what the dual of $X^*$ means (denoted $X^{**}$). There's no explanation for it in the book, just that it's the "dual of $X^*$", but since the dual space is already mapping to $\mathbb{R}$ or $\mathbb{C}$, I'm not sure what the dual of $X^*$ is mapping.

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  • $\begingroup$ $X^{**}=\mathcal{L}(X^*, \mathbb{F})$, that is $X^{**}$ is the set of all bounded linear operators whose domain is $X^*$. Hence an element of $X^{**}$ is a function $\Lambda$ that is evaluated on functionals $\varphi \in X^*$ and that takes it values in the field: $\Lambda(\varphi) \in \mathbb{F}$ $\endgroup$ – Alonso Delfín Feb 24 '16 at 23:49
  • $\begingroup$ think to $X^{**}$ as an extension of $X$ : take some $x \in X$, clearly for every $y \in X^*$ (so $y$ being a bounded linear function $X \to \mathbb{R}$), then $y(x)$ is a real number, and the function $x^{**} : y \to y(x)$ is a $X^* \to \mathbb{R}$ function which is linear in $y$, and bounded. thus $x^{**} \in X^{**}$. proving that $X^{**}$ is a little more than that, is the second part. $\endgroup$ – reuns Feb 25 '16 at 0:18
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The dual space of $X^*$ is the set of linear applications $f:X^*\rightarrow R$ that is $L(L(X,R),R)$.

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    $\begingroup$ So you're applying a bounded linear operator to a bounded linear operator which maps to $R$, to produce a real number? $\endgroup$ – Brenton Feb 24 '16 at 23:51
  • $\begingroup$ $X^*$ is viewed here as a normed space $\endgroup$ – Tsemo Aristide Feb 24 '16 at 23:53
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    $\begingroup$ @Brenton : yes, $X^*$ is a normed space, and you have defined the dual of a normed space, so you have naturally defined the dual of $X^*$ $\endgroup$ – Tryss Feb 25 '16 at 0:01

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