8
$\begingroup$

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., $J'(f) = h(f)$.
$\endgroup$

1 Answer 1

9
$\begingroup$

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 $J'(\tilde h) = h$.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Mar 26, 2013 at 6:00
  • 1
    $\begingroup$ 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 $X$. $\endgroup$ Mar 26, 2013 at 6:33
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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