# Suppose $\forall x \in X,\sum_{n=1}^\infty|f_n(x)|<\infty$, to prove $\forall F\in X'', \sum_{n=1}^\infty |F(f_n)|\leq C\|F\|$.

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}.

Consider $$\mathcal{E} = \left\{ \sum_{n = 1}^N e_n f_n : N \ge 1 \land \forall n \ge 1, |e_n| \le 1 \right\} \subset X'$$, which we will prove to be bounded.

For all $$x \in X$$, arbitrarily choosing element $$\sum_{n = 1}^N e_n f_n$$ in $$\mathcal{E}$$, $$\left| (\sum_{n = 1}^N e_n f_n) (x) \right| = \left| \sum_{n=1}^N e_n f_n (x) \right| \le \sum_{n=1}^N \left| e_n \right| \left| f_n(x) \right| = \sum_{n = 1}^N \left| f_n(x) \right| < \infty.$$

Thus, $$\sup_{T \in \mathcal{E}} \left| T (x) \right| < \infty, \quad x \in X.$$

As $$X$$ is a Banach space, by uniform boundedness principle, $$\sup_{T \in \mathcal{E}} \left\Vert T \right\Vert < \infty$$, that is, $$\mathcal{E}$$ is bounded. Let $$C = \sup_{T \in \mathcal{E}} \left\Vert T \right\Vert$$.

We could construct $$\{ e_n : e_n = \overline{\mathrm{sgn}[ F(f_n) ]} \}_{n \ge 1}$$. For all $$n \ge 1$$, we have $$|e_n| \le 1$$ and $$e_n F(f_n(x)) = | F(f_n(x)) |$$.

Therefore, for each $$F \in X''$$, considering arbitrary $$N \ge 1$$, $$\sum_{n = 1}^N | F(f_n) | = \sum_{n = 1}^N e_n F(f_n) = F \left( \sum_{n = 1}^N e_n f_n \right) \le \left\Vert F \right\Vert \left\Vert \sum_{n = 1}^N e_nf_n \right\Vert \le C \left\Vert F \right\Vert$$

Let $$N \to \infty$$, we can conclude that $$\sum_{n = 1}^\infty | F(f_n) | \le C \left\Vert F \right\Vert$$.

• The proof generally looks good, but since $f_n$ could be zero, the $e_n$ may not all be well-defined. If, in the definition of $\mathcal{E}$, you require $|e_n| \le 1$ instead of $|e_n| = 1$, then in your construction at the end you would set $e_n = \operatorname{sgn}[F(f_n)]$. Everything else in your proof remains unchanged.
– kobe
Dec 24, 2021 at 15:38
• Thanks for @kobe. I also find that if we consider complex numbers, we need to set $e_n = \overline{\mathrm{sgn}[ F(f_n) ]}$ s.t. $e_n F(f_n) = |F(f_n)|$. Dec 26, 2021 at 8:57

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 $$N\to \infty$$.