# Help Trying to understand the proof of the Banach-Eberlein-Smulian Theorem

I'm currently reading Ciarlet's Linear Functional Analysis with Applications but I'm slightly unsure if I'm fully understanding the proof. I've writen the proof below along with my thoughts/reasoning in italics. I'm not a mathematician, rather an economist hoping to pick up some more maths along the way. So I apologise if some of this is trivial. If someone could verify if my reasoning is correct or provide clarification I would be very greatful. Many thanks.

The corollary to the Banach Steinhaus theorem is as follows.

Let $X$ be a Banach space, let $Y$ be a normed vector space and let $(A_n)_{n=1}^\infty$ be a family of mappings $A_n \in B(X,Y)$ such that, for each $x \in X$, the sequence $(A_n x)_{n=1}^\infty$ converges in $Y$. Then \begin{equation*} \sup_{n \geq 1} \|A_n\|<\infty \end{equation*} Furthermore, let the mapping $A:X\rightarrow Y$ be defined by $$Ax:=\lim_{n \rightarrow \infty}A_nx$$ for each $x\in X$.

Then $A\in B(X,Y)$ and $\|A\|\leq \liminf_{n \rightarrow \infty} \|A_n\|$

Banach-Eberlein-Smulian Theorem. Any bounded sequence in a reflexive Banach Space contains a weakly convergent subsequence.

Proof

Notation. X': Dual space, X'': Second Dual space. $\mathbb{F}$: field

We assume that $X$ is separable. Since $X$ is reflexive there exists a linear isometry from $X$ onto $X''$ by definition, therefore the space $X''$ is also separable. Since $X''=(X')'$, $X'$ is also separable. Let $x_k'\in X'$, $k\geq 1$ be such that \begin{equation*} X'=\overline{\bigcup_{k=1}^\infty \{x_k'\}} \end{equation*} (From separability)

Let $(x_n)_{n=1}^\infty$ be a bounded sequence of elements $x_n \in X$. Therefore, for each $x'\in X'$, \begin{equation*} |x'(x)|\leq M \|x'\|\hspace{0.5cm}\text{for all}\hspace{0.25cm}n\geq 1, \hspace{0.5cm}M:=\sup\|x_n\|<\infty \end{equation*} The sequence $(x'(x_n))_{n=1}^\infty$ is thus bounded and as it belongs to $\mathbb{F}$ (As $x'$ is a linear functional) it must contain a convergent subsequence (by Bolzano–Weierstrass theorem). In particular, the sequence $(x'_1(x_n))_{n=1}^\infty$ contains a convergent subsequence $(x'_1(x_{\sigma_1(n)}))_{n=1}^\infty$. The sequence $(x'_2(x_{\sigma_1(n)}))_{n=1}^\infty$ is also bounded in $\mathbb{F}$ so contains a subsequence $(x'_2(x_{\sigma_2(n)}))_{n=1}^\infty$ and so on. We consider the diagonal sequence \begin{equation*} (x_{\sigma(n)})_{n=1}^\infty, \hspace{0.5cm}\textit{where}\hspace{0.25cm}\sigma(n):=\sigma_n(n) \hspace{0.25cm}n\geq 1 \end{equation*} We have constructed a subsequence (of a bounded sequence) of $(x_n)_{n=1}^{\infty}$ and thus for each $k\geq 1$, the sequence $(x'_k(x_{\sigma(n)}))_{n=1}^\infty$ converges in $\mathbb{F}$ as $n \rightarrow \infty$. (Why does it follow that a continuous functional applied to a convergent sequence converges. It does if $\;\;f:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function doesn't it?).

Next we show that, for each $x' \in X'$, the sequence $(x'(x_{\sigma(n)}))_{n=1}^\infty$ converges in $\mathbb{F}$ as $n \rightarrow \infty$. Let $x' \in X'$ and $\varepsilon>0$ be given. There exists an integer $k=k(x',\varepsilon)\geq 1$ such that $\|x'-x_k'\|\leq \frac{\varepsilon}{4M}$ (by separability of X'). Then for integers $m,n\geq 1$ \begin{align*} |x'(x_{\sigma(m)})-x'(x_{\sigma(n)})|&\leq |x'_k(x_{\sigma(m)})-x'_k(x_{\sigma(n)})|+|(x'-x'_k)(x_{\sigma(m)}-x_{\sigma(n)})| \\ &\leq |x'_k(x_{\sigma(m)})-x'_k(x_{\sigma(n)})|+\frac{\varepsilon}{2} \end{align*} Also using $\|x_{\sigma(m)}-x_{\sigma(n)}\|\leq 2M\varepsilon$ (Separability of X) (Also the inequality on the second line comes from) $|(x'-x'_k)(x_{\sigma(m)}-x_{\sigma(n)})|\leq \|x'-x_k'\|\|x_{\sigma(m)}-x_{\sigma(n)}\|=\frac{\varepsilon}{4M}2M\varepsilon=\frac{\varepsilon}{2}$. Since $(x_k'(x_{\sigma(n)}))_{n=1}^\infty$ is a Cauchy sequence $|x'_k(x_{\sigma(m)})-x'_k(x_{\sigma(n)})|$ can be made as small as we like. In particular we can choose $n_0=n_0(k)=n_0(x',\varepsilon)\geq 1$ large enough that $|x'_k(x_{\sigma(m)})-x'_k(x_{\sigma(n)})|\leq \frac{\varepsilon}{2}$. Therefore \begin{equation*} |x'(x_{\sigma(m)})-x'(x_{\sigma(n)})|\leq \varepsilon \end{equation*} Thus $(x'(x_{\sigma(n)}))_{n=1}^\infty$ is a Cauchy sequence itself and converges in $\mathbb{F}$. Now, let $J:X \rightarrow X''$ denote the linear isometry defined above. We know $J$ is surjective as $X$ is reflexive (by definition). By our previous work, the continuous linear functionals $Jx_{\sigma(n)}\in X''=B(X',\mathbb{F})$, which are thus defined for each $n \geq 1$ by \begin{equation*} Jx_{\sigma(n)}(x')=x'(x_{\sigma(n)})\hspace{0.5cm}\text{for all}\hspace{0.25cm}x' \in X' \end{equation*} have the following property: \begin{equation*} \lim_{n\rightarrow \infty} Jx_{\sigma(n)}(x')\hspace{0.5cm}\text{exists in}\hspace{0.25cm}\mathbb{F}\hspace{0.25cm}\text{as}\hspace{0.25cm}n\rightarrow \infty \hspace{0.25cm}\text{for each}\hspace{0.25cm}x' \in X' \end{equation*} Since the space $X$ is complete we can apply the corollary to the Banch-Steinhaus theorem. This shows that there exists $x'' \in X''=B(X',\mathbb{F})$ such that (From here I really don't follow what's going on) \begin{equation*} Jx_{\sigma(n)}(x')\rightarrow x''(x')\hspace{0.5cm} \text{as}\hspace{0.25cm}n\rightarrow \infty \hspace{0.25cm}\text{for each}\hspace{0.25cm}x' \in X' \end{equation*} But this is the same as \begin{equation*} x'(x_{\sigma(n)})\rightarrow x'(x)\hspace{0.5cm} \text{as}\hspace{0.25cm}n\rightarrow \infty \hspace{0.25cm}\text{for each}\hspace{0.25cm}x' \in X',\hspace{0.25cm}\text{where}\hspace{0.25cm} x:=J^{-1}x'' \end{equation*} Hence the subsequence $(x_{\sigma(n)})_{n=1}^\infty$ weakly converges to $x$ as $n \rightarrow \infty$ (Using the definition of weak convergence in a normed space)

The first item that requires comment is your question about why the sequence $\{x'_k(x_{\sigma(n)}\}$ is convergent for all $k$. Fixing $k$, we have $\{x'_k(x_{\sigma(n)})\}_{n\geq k}$ is a subsequence of $\{x'_k(x_{\sigma_k(n)})\}$, and we know that the latter sequence converges, and hence so does the subsequence. Adding $k-1$ terms to $\{x'_k(x_{\sigma(n)})\}_{n\geq k}$ to obtain the full $\{x'_k(x_{\sigma(n)})\}$ sequence does not alter this convergence.
The second issue is when you try to reason as to why $|(x'-x'_k)(x_{\sigma(m)}-x_{\sigma(n)})|\leq\frac{\varepsilon}{2}$. We don't have $\|x_{\sigma(m)}-x_{\sigma(n)}\|\leq 2M\varepsilon$ as you state, but rather $\|x_{\sigma(m)}-x_{\sigma(n)}\|\leq 2M$ (this comes directly from the triangle inequality). Other than that, it is fine.
Lastly, we look at the final paragraph, where you say you can't follow the proof, and where the corollary to the Banach-Steinhaus theorem is applied. We define $x'':X'\to\mathbb F$ by $$x''(x')=\lim_{n\to\infty}x'(x_{\sigma(n)})=\lim_{n\to\infty}Jx_{\sigma(n)}(x')\qquad (x'\in X').$$ By the corollary, $x''\in X''$, and by reflexivity of $X$ we know $x''=Jx$ for some $x\in X$.
• Many thanks! Sorry the statement $\|x_{\sigma(m)}-x_{\sigma(n)}\|\leq 2M\varepsilon$ was a typo. Not sure how I managed to write $2M\varepsilon$. Also I now realise that the inequality does not come from separability of $X$ but the triangle inequality and boundedness as you say. Thank you for clearing up the third point. I've done a bit more reading and I'm happier with the concept of reflexivity. I'm still not so sure about first point. I'll need to look at it some more. Thanks. – mark Jun 19 '17 at 19:17
• Took me a while to get my head around that $(x_k'(x_{\sigma(n)}))_{n\geq k}$ is a subsequence of $(x_k'(x_{\sigma_k(n)}))$. Now that I'm there I follow the rest of your argument. Cheers! – mark Jun 19 '17 at 21:13