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If a normed space $X$ is reflexive, show that $X'$ is reflexive.

Suppose $X$ is reflexive. Then by definition the Canonical mapping $J : X \to X''$ defined by $x \mapsto g_x$ where $g_x(f) = f(x)$ is an isomorphism. We want to show that the mapping $J' : X' \to X'''$ defined by $f \mapsto h_f$ where $h_f(g_x) = g_x(f)$ is an isomorphism. It will suffice to show that $J'$ is onto.

I am unsure about what to do after this. Any help would be greatly appreciated.

Some ideas:

  1. Choose $h \in X'''$, then by definition $h : X'' \to \mathbb R$ is a linear bounded functional.
  2. Try to find $f \in X'$, that is $f$ such that $f : X \to \mathbb R$ (a linear bounded functional) such that the cannonical mapping maps $f \mapsto h$, i.e., $J'(f) = h(f)$.
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If $h \in X'''$ is given, define the functional $\tilde h \in X'$ by $\tilde h(f) = h(J(f))$ for all $f \in X$.

For all $g \in X''$, you have $$J'(\tilde h)(g) = g(\tilde h) = \tilde h(J^{-1}(g)) = h(g).$$ This shows that $J'(\tilde h) = h$.

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  • $\begingroup$ How do you know $\tilde h$ is in fact in $X'$. I am understanding the rest, but struggling to show this fact. Maybe I am missing something. $\endgroup$
    – Robert Cardona
    Mar 26, 2013 at 6:00
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    $\begingroup$ My reasoning thus far is that $J(f) \in X''$ and $X \cong X''$ by hypothesis, so $J(f)$ 'is in' $X$ or at least corresponds to an element in $X$. $\endgroup$
    – Robert Cardona
    Mar 26, 2013 at 6:33
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    $\begingroup$ Note that $\tilde h = J\circ h$. Moreover, you have that $J$ maps $X$ to $X''$ and $h$ maps $X''$ to $\mathbb{R}$ (both mappings are linear and bounded). Hence, their composition $h \circ J$ maps $X$ to $\mathbb{R}$ linearly and boundedly. Thus, $\tilde h = h\circ J$ belongs to $X'$. $\endgroup$
    – gerw
    Mar 26, 2013 at 9:15
  • $\begingroup$ @gerw Why does the second equality hold? I.e. that $g(\tilde{h}) = \tilde{h}(J^{-1}(g))$? I don't see how we can get rid of the $J^{-1}$. $\endgroup$
    – Ponky
    Jun 29, 2020 at 8:23
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    $\begingroup$ This is the definition of $J$: We have $\bar h(f) = (Jf)(\bar h)$ for all $f \in X$ and we apply this with $f = J^{-1}(g)$. $\endgroup$
    – gerw
    Jun 29, 2020 at 8:57

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