Closed Unit Ball of X Being Weakly Comp. Implies X is a Banach Space

# Closed Unit Ball of X Being Weakly Compact Implies X is a Banach Space

 Proposition: Let $X$ be a normed space and let $B_X = \{ x \in X : \| x \| \leq 1 \}$ denote the closed unit-ball of $X$. If $B_X$ is weakly compact then $X$ is a Banach space.
• Proof: Let $(x_n) \subset X$ be a Cauchy sequence, and let $\epsilon > 0$ be given. Since $(x_n)$ is Cauchy, it is bounded, and thus and so there exists a $k > 0$ such that $(x_n) \subset k B_X$ where:
(1)
\begin{align} kB_X = \{ x \in X : \| x \| \leq k \} \end{align}
• Since $B_X$ is weakly compact and since $X$ equipped with the weak topology is a locally convex topological vector space, (so that in particular, addition and scalar multiplication are continuous) we have that $kB_X$ is weakly compact. So by definition of compactness, the sequence $(x_n) \subset k B_X$ has a weak accumulation point $x \in k B_X$.
• We will now show that $(x_n)$ norm converges to $x$ to complete our proof.
• Since $(x_n)$ is a Cauchy sequence there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:
(2)
\begin{align} \| x_m - x_n \| < \frac{\epsilon}{2} \end{align}
• Moreover, since $(x_n)$ weakly converges to $x$, we have that for every $f \in X^*$ that $\displaystyle{\lim_{n \to \infty} f(x_n) = f(x)}$. In particular, if $f \in X^*$ is such that $\| f \| = 1$ then there exists an $m \in \mathbb{N}$ with $m \geq N$ such that:
(3)
\begin{align} |f(x_m) - f(x)| < \frac{\epsilon}{2} \end{align}
• So for each $f \in X^*$ with $\| f \| \leq 1$, if $n \geq N$ then:
(4)
\begin{align} | f(x_n) - f(x) | &= | f(x_n) - f(x_m) + f(x_m) - f(x) | \\ & \leq | f(x_n) - f(x_m) | + | f(x_m) - f(x) | \\ & \leq \| f \| \| x_n - x_m \| + \| f(x_n) - f(x) \| \\ & \leq \| x_m - x_n \| + \| f(x_n) - f(x) \| \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & < \epsilon \end{align}
• Since this holds for all $f \in X^*$ with $\| f \| = 1$ we see that if $n \geq N$ then:
(5)
\begin{align} \quad \| x_n - x \| = \| \hat{x_m} - \hat{x} \|_{X^{**}} = \sup_{\| f \| = 1} |(\hat{x_m} - \hat{x})(f)| = \sup_{\| f \| = 1} |f(x_m) - f(x)| < \epsilon \end{align}
• So $(x_n)$ norm converges to $x$. Thus every Cauchy sequence in $X$ norm converges in $X$, so $X$ is a Banach space. $\blacksquare$