1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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. $x_{n}\rightharpoonup x_0$.

$\endgroup$
13
  • $\begingroup$ I am not understanding the general procedure you are using for the proof …. could you please explain it? $\endgroup$ – user591668 Apr 15 '20 at 16:48
  • $\begingroup$ Can you take a look at the other proof …. seems your ideas are different … where is the proof of boundedness and linearity of your $a$? $\endgroup$ – user591668 Apr 15 '20 at 16:51
  • $\begingroup$ Is $a$ an element of $X^{*}$? if so why? $\endgroup$ – user591668 Apr 15 '20 at 16:52
  • $\begingroup$ yes I know but we have to prove that it is in $X$ ? am I correct? $\endgroup$ – user591668 Apr 15 '20 at 16:54
  • $\begingroup$ No, we need to prove that $\varphi(x_n)\to\varphi(x_0)$ for all $\varphi$. It's definition of weakly convergence. $\endgroup$ – thing Apr 15 '20 at 16:56
4
$\begingroup$

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 $\alpha$.

$\endgroup$
7
  • $\begingroup$ It is not clear for me why you used $\alpha$ could you clarify please? $\endgroup$ – user591668 Apr 15 '20 at 15:17
  • $\begingroup$ 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. $\endgroup$ – Mark Apr 15 '20 at 15:33
  • $\begingroup$ Could you please tell me why $\alpha $ is a linear functional?or include this in your proof? $\endgroup$ – user591668 Apr 15 '20 at 16:13
  • $\begingroup$ Why I need to show that $\alpha $ is a bounded linear functional? $\endgroup$ – user591668 Apr 15 '20 at 16:15
  • $\begingroup$ I think I understood now why you need it bounded linear functional as the Cauchy sequence should be weakly convergent to an element in $X.$ But $X$ is reflexive which means that it is isometric isomomophic to $X^{**}$ so it is a space of bounded linear functionals .. am I correct? $\endgroup$ – user591668 Apr 15 '20 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy