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 !



Weak*-null sequences in dual spaces
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
$$a=\lim\_{m}\lim\_{n}f\_{m}(x\_{k\_{n}})=\lim\_{m}f\_{m}(y\_{i\_{0}})=0.$$


