5
$\begingroup$

$X$ is a normed vector space. Assume $X$ is reflexive, then $X$ must be a Banach space.

I guess we only need to show any Cauchy sequence is convergent in $X$.

$\endgroup$
10
$\begingroup$

Hint: (1) If $X$ is reflexive, $X$ is isomorphic to $X^{**}$. (2) Dual spaces are allways complete.


Regarding (2), we will prove, that $L(X,Y)$ the space of bounded linear operators from $X$ to $Y$ is complete in the operator norm if $Y$ is complete. Then (2) follows, as $X^* = L(X, \mathbb K)$ and $\mathbb K$ is complete. So let $(T_n)$ be an operator norm Cauchy sequence, then $(T_n x)$ is Cauchy for each $x$, as $\def\norm#1{\left\|#1\right\|}$ $$ \norm{T_nx-T_mx} \le \norm{T_n - T_m}\norm x $$ As $Y$ is complete, we may define $T\colon X \to Y$ by $Tx := \lim_n T_n x$. $T$ is linear, as the $T_n$ and the limit is, an bounded since $$ \norm{Tx} \le \sup_n\norm{T_n x} \le \sup_n\norm{T_n}\cdot \norm x $$ and Cauchy sequences are bounded. Now given $\epsilon > 0$, we can find a $N$, such that $$ \norm{T_n - T_m} < \epsilon, \text{ all $n,m \ge N$} $$ giving $$ \norm{T_n x - T_m x} < \epsilon, \text{ all $\norm x \le 1$, $n,m \ge N$} $$ for $m \to \infty$ $$ \norm{T_n x - T x} \le \epsilon, \text{ all $\norm x \le 1$, $n\ge N$} $$ that is $\norm{T_n - T} \le \epsilon$, $n \ge N$. So $T_n \to T$.

$\endgroup$
  • $\begingroup$ Is there a simple proof of (2)? I guess the weak compactness and convexity of the unit ball is enough. $\endgroup$ – Rabee Tourky May 22 '13 at 7:49
  • $\begingroup$ Can you explain why dual spaces are complete? $\endgroup$ – Falang May 22 '13 at 8:11
  • $\begingroup$ So (2) implies the dual space is complete and thus the dual of the dual space is also complete. Thanks! $\endgroup$ – Falang May 22 '13 at 8:52

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.