Weak*-null sequences in dual spaces

Question

Let $(x_{n})_{n}$ be a sequence in a Banach space $X$. Assume that the set $\{x_{n}:n=1,2,\cdots\}$ is finite. Let $(f_{m})_{m}$ be a weak*-null sequence in $X^{*}$ satisfying the following conditions:

(1) the limit $a_{m}:=\lim\limits_{n}\langle f_{m},x_{n}\rangle$ exists for each $m$;

(2) the limit $a:=\lim\limits_{m}a_{m}$ exists.

Question. $a=0$ ?

Thank you !

Answer 1

Let's write $S=\{x_n:n\in\mathbb{N}\}$. WLOG we can assume that $\{n\in\mathbb{N}:x_n=y\}$ is infinite for each $y\in S$ (otherwise, we can comfortably cut off the finitely many terms that doesn't repeat infinitely often in the sequence, and work with the rest).

Since $S$ is finite, then so is $S_f=\{f(x_n):n\in\mathbb{N}\}\subseteq\mathbb{C}$ for each $f\in X^*$. Clearly, $$S_f=\{L\} \Leftrightarrow \lim_{n\to\infty} f(x_n) = L.$$ In this case, $L=f(x_1)$. Now let $(f_m)$ be given as above. $$ a = \lim_{m\to\infty}\lim_{n\to\infty} f_m(x_n) = \lim_{m\to\infty} f_m(x_1) = 0.$$

Answer 2

I think what is intended is that $x_n$ is bounded. With this interpretation, the answer is no. In $L^1$, let $x_n=n\chi_{[-1/n,1/n]}$ and $f_m=\chi_{[-1/m,1/m]}$.

Answer 3

Let's write $\{x_{n}:n=1,2,\cdots\}=\{y_{1},y_{2},\cdots,y_{N}\}$. For each $i=1,2,\cdots,N$, we set $A_{i}=\{n:x_{n}=y_{i}\}$. Then $\cup_{i=1}^{N}A_{i}=\mathbb{N}$. Then there exists $1\leq i_{0}\leq N$ so that $A_{i_{0}}$ is infinite. Write $A_{i_{0}}=\{k_{n}:n=1,2,\cdots\}$. Hence $x_{k_{n}}=y_{i_{0}}$ for all $n$. So we get $$a=\lim_{m}\lim_{n}f_{m}(x_{k_{n}})=\lim_{m}f_{m}(y_{i_{0}})=0.$$