I want to prove that a reflexive normed space $X$ is a Banach space. By the definition of the reflexive space, the evaluation map $J:X\to X''$ is a bijection. All I need is to prove that the evaluation map is bicontinous (isomorphism) to transfer the Cauchy sequence properties from the Banach space $X''$ to $X$. But I don't know how to prove that $J$ is an isomorphism. To apply the "Banach isomorphism theorem", $X$ has to be already a Banach Space.
[Math] Reflexive normed spaces are Banach
banach-spacesfunctional-analysisgeneral-topology
Related Solutions
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 $T_n \to T$.
Reflexive spaces interest mathematicians because they have a lot of nice properties:
Unit ball is weakly compact, so you can exploit compactness to prove exitence of fixed points, convergent subsequences and etc.
Reflexive spaces are characterized by the property that weak and weak* topology coincide. You can forget about weak topology and work with much well understood weak* topology.
Every functional on a reflexive space attains its norm. Simply speaking you always have a vector that tells you almost everything about your functional. More on this matter you can find here.
Reflexivity is a three space property. You can pass to quotients and subspaces of reflexive spaces and get a reflexive space again.
After equivalent renorming all reflexive spaces are strictly convex. In some sense unit ball of a reflexive spaces is round.
Reflexive spaces have Radon-Nykodim property. This allows you to develop a rich theory for vector valued integration and vector valued measures for reflexive spaces.
Reflexivity is a rare property and this helps one to distinguish Banach spaces. For example there is no infinite dimensional reflexive $C^*$-algebras, so $c_0$, $l_\infty$ are not reflexive. Their non-commutative counterparts $\mathcal{K}(H)$ and $\mathcal{B}(H)$ are not reflexive either.
Shauder bases in a reflexive space are very nice and sweet, they are shrinking and boundedly complete.. Beware! There are hereditarily indecomposable (and a fortiori without any basis) reflexive Banach spaces. See this paper.
To find more on reflexive spaces use search on this site, or mathoverflow, or any book on Banach geometry with keyword 'reflexive'.
Best Answer
As pointed out in comments, the main step is to prove $\|Jx\|_{X^{**}}=\|x\|_X$ for all $x\in X$. This, together with surjectivity of $J$, imply that $X$ is isometric to $X^{**}$ and therefore inherits completeness from the latter.
For every unit norm functional $\phi$ we have $|\phi(x)|\le \|x\|$. By taking supremum over such $\phi$, we get $\|Jx\| \le \|x\|$. On the other hand, by the Hahn-Banach theorem (which does not require completeness) there is a unit norm functional $\phi\in X^*$ such that $\phi(x)=\|x\|$. Therefore, $\|Jx\|\ge |\phi(x)| = \|x\|$.