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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 (sequentially) complete if every weakly Cauchy sequence in $X$ is weakly convergent. Prove that every reflexive Banach space is weakly (sequentially) complete. Could anyone help me in this proof please?

Dean Miller
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2 Answers2

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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 $\alpha$.

Mark
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  • It is not clear for me why you used $\alpha$ could you clarify please? –  Apr 15 '20 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 '20 at 15:33
  • Could you please tell me why $\alpha $ is a linear functional?or include this in your proof? –  Apr 15 '20 at 16:13
  • Why I need to show that $\alpha $ is a bounded linear functional? –  Apr 15 '20 at 16:15
  • 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? –  Apr 15 '20 at 16:27
  • Why $\alpha$ is in $X^{*}$? –  Apr 15 '20 at 16:29
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    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 '20 at 19:22
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$\{x_n\}$ is weakly Cauchy, i.e. $\forall\varphi\in X^\ast$ $|\varphi(x_n)-\varphi(x_m)|\to0$ for $n,m\to\infty$. So the numerical sequence $\{\varphi(x_n)\}$ is Cauchy and converges to some number, say $a_\varphi$. Since a convergent sequence is always bounded, then $\forall\varphi\in X^\ast$ the 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 (from the uniform boundedness principle). By 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 to the uniqueness of the limit, we obtain $a_\varphi=\varphi(x_0)$, i.e. $x_{n}\rightharpoonup x_0$.

Smiley1000
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thing
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  • I am not understanding the general procedure you are using for the proof …. could you please explain it? –  Apr 15 '20 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 $a$? –  Apr 15 '20 at 16:51
  • Is $a$ an element of $X^{*}$? if so why? –  Apr 15 '20 at 16:52
  • yes I know but we have to prove that it is in $X$ ? am I correct? –  Apr 15 '20 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. – thing Apr 15 '20 at 16:56
  • I do not have a definition for weak boundedness –  Apr 15 '20 at 17:04
  • why the boundedness of ${x_{n}}$ is needed? –  Apr 15 '20 at 17:06
  • If $\forall\varphi$ sequence ${\varphi(x_n)}$ is bounded, then ${x_n}$ is weakly bounded. – thing Apr 15 '20 at 17:06
  • Boundedness of $x_n$ need for applying theorem: every bounded sequence in reflexive space have weakly convergent subsequence. – thing Apr 15 '20 at 17:08
  • I can not use this definition .. I did not studied it. –  Apr 15 '20 at 17:15
  • Can you modify your proof without using it? –  Apr 15 '20 at 17:16
  • I think your continuity argument is flipped as $x_{n} \to x$ implies $Tx_{n} \to Tx$ and not the reverse. Could you explain this in details please? –  Apr 15 '20 at 20:10
  • You're right. Need so: due to $\varphi(x_n/n)\to 0=\varphi(0)$, we have $x_n/n\rightharpoonup0$. But, as is well known, any weakly convergent sequence is bounded, and this is contradiction with $|x_n|>n^2$. – thing Apr 16 '20 at 00:43
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    The fact that weakly convergent sequences are bounded depends on the uniform boundedness principle, so it is better to directly invoke uniform boundedness. – Smiley1000 Feb 16 '24 at 13:23