Real Analysis, Folland Problem 6.1.21 if and only if condition for weak convergence in $l^p(A)$ 
6.1.21 - If $1 < p < \infty$, $f_n\rightarrow f$ weakly in $l^p(A)$ iff $\sup_{n}\|f_n\|_{p} < \infty$ and $f_n\rightarrow f$ pointwise.

Attempted proof - Suppose $f_n\to f$ weakly in $l^p(A)$. If $a\in A$ then $\chi_{a}\in l^{p/(p-1)} = l^{q}$ , so for each $a\in A$ 
$$\sum_{a\in A}f_n(a)\chi_{a} =f_n(a)$$
and similarly
$$\sum_{a\in A} f(a) \chi_{a} = f(a)$$
So we see that $f_n\to f $ pointwise.
Now consider $T_n: l^q(A) \to \mathbb{C}$ that takes any $g$ into $\int f_n g$. We see that $\|T_n\| = \|f_n\|_{p}$ is linearly continuous and $$\int f_n g \to \int fg$$
I am pretty confused with proving this, any suggestions is greatly appreciated.
 A: 
6.1.21 - If $1 < p < \infty$, $f_n\rightarrow f$ weakly in $l^p(A)$ iff $\sup_{n}\|f_n\|_{p} < \infty$ and $f_n\rightarrow f$ pointwise.

Proof 
($\Rightarrow$) Suppose $f_n\to f$ weakly in $l^p(A)$. If $a\in A$ then $\chi_{\{a\}}\in l^{p/(p-1)}(A) = l^{q}(A)$ , so for each $a\in A$ 
$$\sum_{a\in A}f_n(a)\chi_{\{a\}} =f_n(a)$$
and similarly
$$\sum_{a\in A} f(a) \chi_{\{a\}} = f(a)$$
So we see that $f_n\to f $ pointwise.
Now consider $T_n: l^q(A) \to \mathbb{C}$ that takes any $g$ into $\int f_n g$. We see that $\|T_n\| = \|f_n\|_{p}$ is linearly continuous. On the other hand, given any $g\in l^q(A)$, he have that  $$T_n(g)=\int f_n g \to \int fg$$
So $\{T_n(g)\}_n$ is a convergent sequence of complex number and so it is a bounded. It means  $\sup_n|T_n(g)|<\infty$. So we have that, for all $g\in l^q(A)$, $\sup_n|T_n(g)|<\infty$. Then we can apply the Uniform Boundedness Principle (theorem 5.13 in Folland's book), and conclude that  $\sup_n\|T_n\|<\infty$. Since, for all $n$, $\|T_n\| = \|f_n\|_{p}$, we have proved that 
 $\sup_n\|f_n\|_{p}<\infty$.
($\Leftarrow$) Suppose $\sup_{n}\|f_n\|_{p}=M < \infty$ and $f_n\rightarrow f$ pointwise. 
Let us first prove that $f \in l^p(A)$ and $\|f\|_p\leq M$.
For each $n$, we have $$\sum_{x\in A}|f_n(x)|^p < \infty$$
So there is a countable subset $U_n$ of $A$ such that, for any $x \in A \setminus U_n$, we have $f_n(x)=0$. Let $U = \bigcup_n U_n$. Then $U$ is countable and for any  $x \in A \setminus U$,  we have $f_n(x)=0$ for all $n$. Since  $f_n\rightarrow f$ pointwise, we also have, for any  $x \in A \setminus U$,  that $f(x)=0$. 
We can enumerate the elements of $U$ and write $U=\{u_j : j\in \mathbb{N} \}$. So we have, for all $n$, 
$$\sum_{j}|f_n(u_j)|^p = \sum_{x\in A}|f_n(x)|^p \leq M^p $$
and 
$$\sum_{j}|f(u_j)|^p = \sum_{x\in A}|f(x)|^p $$
For any $Q \in \mathbb{N}$, we have 
$$\sum_{j=0}^Q|f_n(u_j)|^p \leq\sum_{j}|f_n(u_j)|^p = \sum_{x\in A}|f_n(x)|^p \leq M^p $$
Since $f_n\rightarrow f$ pointwise, we can easily see that 
$$\sum_{j=0}^Q|f_n(u_j)|^p \to \sum_{j=0}^Q|f(u_j)|^p \quad \textrm{ as } n \to \infty $$
So we have that, for any $Q \in \mathbb{N}$, 
$$\sum_{j=0}^Q|f(u_j)|^p\leq M^p $$
So we have that 
$$ \sum_{x\in A}|f(x)|^p = \sum_{j}|f(u_j)|^p \leq M^p $$
So we have proved that $f \in l^p(A)$ and $\|f\|_p\leq M$.
Now, given any $g\in l^q(A)$, we have that 
$$\sum_{x\in A}|g(x)|^q < \infty$$
So there is a countable subset $S$ of $A$ such that, for any $x \in A \setminus S$, we have $g(x)=0$. We can enumerate the elements of $S$ and write $S=\{x_k : k\in \mathbb{N} \}$. So we have 
$$\sum_{k}|g(x_k)|^q = \sum_{x\in A}|g(x)|^q < \infty$$ 
So, given $\epsilon >0$, there is $K$ such that 
$$\sum_{k>K}|g(x_k)|^q <\left (\frac{\epsilon}{4 M}\right )^q$$ 
So we have 
\begin{align*}
\left |\sum_{x \in A}(f_n(x) -f(x))g(x)   \right | & \leq \sum_{x \in A}|f_n(x)-f(x)|\,|g(x)| = \sum_{k}|f_n(x_k)-f(x_k)|\,|g(x_k)| = \\
& = \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+  \sum_{k>K}|f_n(x_k)-f(x_k)| \,|g(x_k)| \leq \\
& \leq \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ \\
& \phantom{\sum_{k=0}^K|f_n(x_k)iii} +\left (\sum_{k>K}|f_n(x_k)-f(x_k)|^p \right )^{1/p}  \left (\sum_{k>K} |g(x_k)|^q \right )^{1/q} \leq \\
& \leq \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ \left (\sum_{x \in A}|f_n(x)-f(x)|^p \right )^{1/p}  \frac{\epsilon}{4 M} = \\
& = \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ \|f_n -f \|_p  \frac{\epsilon}{4 M} \leq \\
& \leq \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ (\|f_n\|_p + \| f \|_p)  \frac{\epsilon}{4 M} \leq \\
& \leq \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ 2M \frac{\epsilon}{4 M} = \\
& = \sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)|+ \frac{\epsilon}{2} \\
\end{align*}
Since for each $k\in \{0, \ldots ,K\}$, $|f_n(x_k)-f(x_k)| \to 0$, as $n \to \infty$, we have that 
$$\sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)| \to 0 \quad \textrm{ as } n \to \infty$$
So there is $N$ such that, for all $n>N$, 
$$\sum_{k=0}^K|f_n(x_k)-f(x_k)|\,|g(x_k)| <\frac{\epsilon}{2}$$
So we have that, for all $n>N$,
$$ \left |\sum_{x \in A}(f_n(x) -f(x))g(x)   \right | < \frac{\epsilon}{2} + \frac{\epsilon}{2} =\epsilon $$
So $(f_n -f) \rightarrow 0$ weakly in $l^p(A)$, which means $f_n\rightarrow f$ weakly in $l^p(A)$. 
