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I want to show that $\ell^1( \mathbb N )$ enjoys the Schur property. More precisely, I have to prove the following

Theorem. Let $X = \ell^1 (\mathbb N )$, $\{ x^{(n)} \} \subset X$, $x \in X$. The following statements are equivalent:

  1. $x^{(n)} \overset{n \to \infty}{\rightharpoonup} x$ ;
  2. $x^{(n)} \overset{n \to \infty}{\to} x$.

Now, $(2) \implies (1)$ is trivial. For the $(1) \implies (2)$ side, I tried something like: let $f \in \ell^1(\mathbb N)^\prime$ be an arbitrary linear functional. Then, by definition,

\begin{equation} x^{(n)} \overset{n \to \infty}{\rightharpoonup} x \iff \langle f, x^{(n)} \rangle \to \langle f, x \rangle \implies \lvert \langle f, x^{(n)} - x \rangle \rvert \to 0 \quad (n \to \infty) . \end{equation}

Hence

\begin{equation} \lvert \langle f, x^{(n)} - x \rangle \rvert \leq \lVert f \rVert \lVert x^{(n)} - x \rVert. \end{equation}

(This reasoning is actually applied for the other side; I wonder if something useful could be extracted in a similar way.)

Now, by the Hahn-Banach theorem, we can choose $f$ such that $\lVert f \rVert = 1$ and $\langle f, x^{(n)} - x \rangle = \lVert x^{(n)} - x \rVert$.

Since the left hand side of the last equation tends to zero for every $f \in \ell^1(\mathbb N)^\prime$ by definition of weak convergence, and in particular for the $f$ given by H-B theorem, it seems that the result would hold for every Banach space. This is clearly absurd, but I can't understand where the error is. Notice that it is obvious that the proof relies on the particular norm of $\ell^1$ and my original idea was to exploit it in same manner from the point where I'm stucked.

I suspect there is no way to complete the preceding attempt and achieve a correct solution of the problem. (as one should expect in most cases he tries to copy the other part in a proof like that.) Nevertheless, I wanna detect the mistakes for many reasons. (proving the theorem, understanding the usage of the Hahn-Banach theorem, better myself, avoid similar mistakes in different situations, etc...) So I ask for a "error-checking" and for a proof of the theorem. (or references as well.) Thank you!

user91126
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2 Answers2

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As Giusepped pointed out in the comments, the problem is that $f$ is not a priori the same for each $n$.

Here are the ideas for a proof of the Schur property.

Community
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Davide Giraudo
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Suppose $\{x_n\}\subset\ell_1$ converges weakly to $x$. Then, $\{x_n\}$ is bounded on $\ell_1$ by the Banach-Steinhaus theorem.

Also, for any $A\subset\mathbb{N}$, $\langle x_n\mathbb{1}_A\rangle\xrightarrow{n\rightarrow\infty}\langle x,\mathbb{1}_A\rangle$. Thus, the measures $\mu_n(dm)=x_n(m)\,\kappa(dm)$ on $(\mathbb{N},2^{\mathbb{N}})$, where $\kappa$ is the counting measure, converge set wise to $\mu(dm)=x(m)\,\kappa(dm)$.

Consider the measure $\nu(dm)=2^{-m}\kappa(m)$. Clearly $\mu_n\ll\kappa\ll \nu$ By Vitaly-Hanh-Saks theorem (see Yosida's book for a reference), for any $\varepsilon>0$, there is $\delta>0$ such that $$\mu_n(A),\mu(A)<\varepsilon\qquad\text{whenever}\quad\nu(A)<\delta\tag{1}\label{one}$$ This implies that $\{x_n,x\}$ are uniform integrable with respect to $(\mathbb{N},2^{\mathbb{N}},\kappa)$. For any $m\in\mathbb{N}$, setting $A=\mathbb{1}_{\{m\}}$, we have that $x_n(m)\xrightarrow{n\rightarrow\infty}x(m)$. that is $x_n$ converges to $x$ pointwise. This shows the $x_n$ converges to $x$ in $L_1(\kappa)$, that is $x_n$ converges to $x$ in $\ell_1$.

The last assertion is contained in the following Theorem

Theorem: Suppose that $\mu$ is $\sigma$--finite and let $f_n\in L_1(\Omega,\mathscr{F},\mu)$, $n\in\mathbb{N}$. The following statements are equivalent.

  • There is $f\in L_1$ to which $f_n$ converges in $L_1$.
  • $f_n$ is a Cauchy sequence in $L_1$.
  • $\{f_n\}$ is uniformly integrable and there is a measurable function $f$ to which $f_n$ converges in measure.

(Klenke's Probability Theory: He treats the $\sigma$ finite case, which I quoted)

Here is a version of the Vitali-Hanh-Saks Theorem

Theorem: Let $(\Omega,\mathscr{F})$ be a measurable space, and let $\mu_n$ be a sequence of finite signed (or complex) measures on $\mathscr{F}$ converging to $\mu$ setwise, that is, $\mu_B:=\lim_n\mu_n(B)$ exists in $\mathbb{R}$ (or $\mathbb{C}$) for each $B\in\mathscr{F}$. If $\mu_n\ll \nu$ for some $\sigma$--finite measure $\nu$ in $\mathscr{F}$, then $\mu$ is a finite signed (or complex) measure and $\mu\ll \nu$. Moreover, $\{\mu_n,\mu\}$ is uniformly continuous with respect to $\nu$ (i.e. $\eqref{one}$)

(By changing $\nu$ to some equivalent probability measure on $(\Omega,\mathscr{F})$, we may assume without loss of generality that $\nu$ is a probability measure)

Mittens
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