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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}.

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2 Answers 2

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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 $\sum_{n = 1}^\infty | F(f_n) | \le C \left\Vert F \right\Vert$.

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    $\begingroup$ 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. $\endgroup$
    – kobe
    Commented Dec 24, 2021 at 15:38
  • $\begingroup$ 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. $e_n F(f_n) = |F(f_n)|$. $\endgroup$ Commented Dec 26, 2021 at 8:57
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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 $N\to \infty$.

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