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.
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 $l^p(A)$.