TITLE: Suppose $\forall x \in X,\sum_{n=1}^\infty|f_n(x)|<\infty$, to prove $\forall F\in X'', \sum_{n=1}^\infty |F(f_n)|\leq C\|F\|$.
QUESTION [7 upvotes]: Let $X$ be a Banach space, consider $\{ f_n \}_{n \ge 1} \in X'$ s.t.
$$\sum_{n = 1}^\infty | f_n(x) | < \infty, x \in X. \tag{1}\label{cond}$$
Please prove that there exists $C \ge 0$ s.t. for each $F \in X''$, $$\sum_{n = 1}^\infty | F(f_n) | \le C \left\Vert F \right\Vert. \tag{2}\label{goal}$$

I have tried to use the uniform boundedness principle to solve this exercise. From the condition i.e. equation \eqref{cond}, I can prove that
$$\frac{ \sum_{n = 1}^\infty | f_n(x) | }{\Vert x \Vert} < \infty. \tag{3}\label{subcond}$$ And I found that if we could prove $$\sum_{n = 1}^\infty | F(f_n) | < \infty, \tag{4}\label{subgoal}$$ we could prove the required conclusion i.e. equation \eqref{goal}. But I don't know how to deduce equation \eqref{subgoal} from equation \eqref{subcond}.

REPLY [4 votes]: Of course namasikanam's answer is good. Here I give a different answer, which based on well known Goldstine's theorem.
Proof: Consider a linear operator $$T:X\to\ell^1: x\mapsto (f_1(x),f_2(x),...).$$ Your condition ensures that $T$ is well defined. Now we use closed-graph theorem to prove $T$ is bounded. Suppoes $x_n\to x$ in $X$, $Tx_n\to y$ in $\ell^1$. Because each $f_k$ is continuous, $f_k(x_n)\to f_k(x)$. Also note that $\ell^1$ convergence implies $k$-coordinate convergence, so $f_k(x_n)\to y_k$. This shows $Tx=y$. Let $C>0$ be the norm of $T$, i.e: $$\sum_{n}|f_n(x)|\leq C, \text{for $x\in B(X)$}.$$ We claim that $$\sum_n|F(f_n)|\leq C, \text{for $F\in B(X'')$}.$$
Now use Goldstine's theorem to approach $F$ by $B(X)$. For each $N$ and $\epsilon>0$ fixed. By Goldstine's theorem, there exists $x=x(N,\epsilon) \in B(X)$, such that $$|F(f_n)-f_n(x)|< \epsilon, \text{for all $n=1,2,...,N$}.$$ This implies $$\sum_{n=1}^N |F(f_n)|\leq \sum_{n=1}^N |f_n(x)|+N\epsilon \leq C+N\epsilon.$$ The proof is completed by letting $\epsilon\to 0$ and then $N\to \infty$.