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I want to prove that, in $l_1$, if $x_n\to x$ weakly then $x_n\to x$.

i tried to use proprieties of $l_1$. If $x_n\to x$ weakly, then for all $f\in {\ell^1}'$, $f(x_n)\to f(x)$. Considering $T\in {\ell^1}'$:

\begin{align}T&:\ell^1\to \mathbb{K} \\ x&\to T(x)=\sum_{n=1}^{\infty}x_n \end{align}, then if $x_n=\{\xi_{j}^{n}\}$ and $x={\xi_j}$:

Computing $|T(x_n)-T(x)|=|\sum_{j=1}^{\infty}\xi_j^n-\sum_{j=1}^{\infty}\xi_j|\geq |\|x_n\|_1-\|x\|_1|\to 0$ then $\|x_n\|_1\to \|x\|_1$. For the other hand, trying \begin{align}G&:\ell^1\to \mathbb{K} \\ x&\to G(x)=x_n\end{align}, n fixed. I computed to $|G(x_n)-G(x)|$ to get some inequality to $\|x_n-x\|_1$. Someone can help?

Luiza Camile
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  • my attempet is getting $G\in \ell_1'$ that $ |x_n-x|_1 \leq |G(x_n)-G(x)|$ ... – Luiza Camile Jun 29 '20 at 20:35
  • I on't know if there is a trivial proof for this. I do have a proof that uses a big gun, namely, Vitali-Hahn-Saks theorem. It is not complicated but it s theorem that is not usually taught course of integration or functional analysis. – Mittens Jun 29 '20 at 22:25
  • Here another proof along the lines of your initial approach: https://math.stackexchange.com/questions/2208898/schurs-theorem-in-ell1-weak-convergence-of-x-n-is-the-same-as-convergenc – Mittens Jun 29 '20 at 22:35

1 Answers1

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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 (a.k.a. uniform boundedness principle).

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 $\mu_n\ll\kappa\ll \nu$. It follows from Vitaly-Hanh-Saks theorem (see Yosida's book for a reference) that 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 $x_n(m)\xrightarrow{n\rightarrow\infty}x(m)$; in other words, $x_n$ 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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