I heard of this problem that caught my attention and I am curious now thus I would appreciate if I could have a hint or a solution.
Let $(x_n)$ a sequence in a normed space $X$ such that $\sum_{n=1}^{\infty} {|f(x_n)|}<+\infty$ for every $f\in X^*$. Prove that there exists $M>0$ such that $$\sum_{n=1}^{\infty} {|f(x_n)|} \leqslant M \|f\|$$ for every $f\in X^*$.
I have strong suspicions that the following fact is somehow used.
If $X,Y$ are Banach and $F:X\rightarrow Y$ a linear, bounded and onto operator then there exists $M>0$ such that for every $y\in Y$ there exists a $x\in X$ with $F(x)=y$ and $\|x\| \leqslant M \|y\|$.
So, my thought would be to define an operator $T:K \rightarrow X^*$ from the subspace of $l^1$ , $K=\{ (f(x_n))_n , f\in X^* \}$ with $T((f(x_n))_n)=f$ and then the result would follow from the above fact if I prove it's bounded. However, I don't think that this operator is well defined so the reasoning doesn't work. And I am not sure even if $K$ is a closed subset of $l^1$ to have the Banach space requirement so this reasoning fails. Am I even thinking in the right direction?
Thanks!
