$X$ 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 $X$.


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 $T_n \to T$.

  • $\begingroup$ 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

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