# Show reflexive normed vector space is a Banach space

$X$ 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 $$T_n \to T$$.