4
$\begingroup$

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., $J'(f) = h(f)$.
|cite|improve this question
$\endgroup$
5
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '13 at 6:00
  • $\begingroup$ 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 '13 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 '13 at 9:15

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.