# Prove that every reflexive Banach space is weakly complete.

Here is the question:

A sequence $$(x_{n})$$ in a normed linear space $$X$$ is weakly Cauchy if $$(Tx_{n})$$ is a Cauchy sequence for every $$T \in X^*.$$ The space $$X$$ is weakly complete if every weakly Cauchy sequence in $$X$$ is weakly convergent. Prove that every reflexive Banach space is weakly complete. Could anyone help me in this proof please?

Let $$\{x_n\}$$ is weakly Cauchy, i.e. $$\forall\varphi\in X^\ast$$ $$|\varphi(x_n)-\varphi(x_m)|\to0$$ by $$n,m\to\infty$$. So the numerical sequence $$\{\varphi(x_n)\}$$ is Caushy and it's converges to some number, say $$a$$. Since the convergent sequence is always bounded, then $$\forall\varphi\in X^\ast$$ a sequence $$\{\varphi(x_n)\}$$ is bounded. This means that $$\{x_n\}$$ is weakly-bounded in $$X$$. It's easy to see that boundedness $$\{x_n\}$$ follows from here (if $$\{x_n\}$$ is unbounded then we can assume that $$\|x_n\|>n^2$$. But $$\forall\varphi\in X^\ast$$, due the weak-boundedness $$\varphi(x_n/n)\to0$$ -- a contradiction). By the reflexivity, this means that for a sequence $$\{x_n\}$$ one can find a weakly convergent subsequence $$\{x_{n_k}\}$$, i.e. $$x_{n_k}\rightharpoonup x_0\in X$$ or $$\forall\varphi\in X^\ast$$ $$\varphi(x_{n_k})\to\varphi(x_0)$$. Due of the uniqueness of the limit, we obtain $$a=\varphi(x_0)$$, i.e. $$x_{n}\rightharpoonup x_0$$.

• I am not understanding the general procedure you are using for the proof …. could you please explain it?
– user591668
Apr 15, 2020 at 16:48
• Can you take a look at the other proof …. seems your ideas are different … where is the proof of boundedness and linearity of your $a$?
– user591668
Apr 15, 2020 at 16:51
• Is $a$ an element of $X^{*}$? if so why?
– user591668
Apr 15, 2020 at 16:52
• yes I know but we have to prove that it is in $X$ ? am I correct?
– user591668
Apr 15, 2020 at 16:54
• No, we need to prove that $\varphi(x_n)\to\varphi(x_0)$ for all $\varphi$. It's definition of weakly convergence. Apr 15, 2020 at 16:56

Suppose $$(x_n)$$ is a weakly Cauchy sequence and let $$f\in X^*$$. Then $$f(x_n)$$ is a Cauchy sequence and by completeness of $$\mathbb{C}$$ it converges to some element $$\alpha(f)\in\mathbb{C}$$. It is easy to see that $$\alpha$$ is a linear functional $$X^*\to\mathbb{C}$$. It is a bit more tricky so show this functional is bounded. For that we identify the elements of $$X$$ with elements in $$X^{**}$$. So for each $$f\in X^*$$ we have $$x_n(f)=f(x_n)\to\alpha(f)$$. Hence for each $$f\in X^*$$ the sequence $$x_n(f)$$ is bounded by some constant $$C(f)$$. But now by the uniform boundedness principle the sequence $$(x_n)$$ itself must be bounded in $$X^{**}$$. Since $$||x_n||_{X^{**}}=||x_n||$$ we conclude that the sequence $$(x_n)$$ is bounded in $$X$$ by some $$M>0$$. Hence for each $$f\in X^*$$:

$$|f(x_n)|\leq M\times||f||$$

By passing to the limit we get $$|\alpha(f)|\leq M||f||$$. So $$\alpha$$ is indeed bounded, hence in $$X^{**}$$. But since $$X$$ is reflexive we conclude that $$\alpha\in X$$. And by definition for each $$f\in X^*$$ we have:

$$f(x_n)\to \alpha(f)=f(\alpha)$$

So indeed $$x_n$$ converges weakly to $$\alpha$$.

• It is not clear for me why you used $\alpha$ could you clarify please?
– user591668
Apr 15, 2020 at 15:17
• For each $f$ the sequence $f(x_n)$ converges to a complex number, I call the limit $\alpha(f)$. This gives us a function $\alpha: X^*\to\mathbb{C}$. As I showed this is a bounded linear functional, i.e an element in $X^{**}$. But since $X$ is reflexive this implies $\alpha\in X$. We are using the isometry between $X$ and $X^{**}$ all the time.
– Mark
Apr 15, 2020 at 15:33
• Could you please tell me why $\alpha$ is a linear functional?or include this in your proof?
– user591668
Apr 15, 2020 at 16:13
• Why I need to show that $\alpha$ is a bounded linear functional?
– user591668
Apr 15, 2020 at 16:15
• Yes, $X$ is isomorphic to $X^{**}$, the space of bounded linear functionals on $X^*$. That's why I showed that $\alpha\in X^{**}$. As for why $\alpha$ is linear-take $f,g\in X^*$. Then $\lim_{n\to\infty} (f+g)(x_n)=\alpha(f+g)$. On the other hand, $(f+g)(x_n)=f(x_n)+g(x_n)\to \alpha(f)+\alpha(g)$ by arithmetic of limits. So $\alpha(f+g)=\alpha(f)+\alpha(g)$. Similarly you can show that $\alpha(cf)=c\alpha(f)$ for $c\in\mathbb{C}$.
– Mark
Apr 15, 2020 at 19:22