$\mathbf x_n\xrightarrow{w^*}\bf x$ in $c_0^*\cong\ell^1$ iff $\mathbf x_n$ converges componentwise to $\bf x$ and $\sup_n\|\mathbf x_n\|_1<\infty$ 
Let $\mathbf{x}_n, \mathbf{x} \in \ell^1. $ Suppose that $\mathbf{x}_n \xrightarrow{w^*} \mathbf{x}.$ This means that for all $\mathbf{y} \in c_0$,  $\sum_{i=1}^{\infty} x_{ni}y_i \rightarrow \sum_{i=1}^{\infty} x_{i}y_i$ as $n \rightarrow \infty$. Define $f_{\mathbf{x}}: c_0 \rightarrow \mathbb{R}$ by $f_{\mathbf{x}}(\mathbf{y}) = \sum_{i=1}^{\infty}x_iy_i.$ Then $x_{ni} = f_{\mathbf{x}_n}(\mathbf{e}_i) = \sum_{i=1}^{\infty}x_{ni}e_i \rightarrow \sum_{i=1}^{\infty}x_{i}e_i= f_{\mathbf{x}}(\mathbf{e}_i) = x_i. $ Does this imply that weak convergence implies componentwise convergence. Since $\mathbf{x}_n \in \ell_1,$ does that not already mean that $\sup ||\mathbf{x}_n||_1 < \infty $ (why is that included as a condition). Does this also mean that $f_{\mathbf{x}_n}$ convnerges pointwise to $f_\mathbf{x}$? How would I do the reverse implication?
 A: $(\Longrightarrow)$: Your proof of weak* convergence implying componentwise convergence is correct. However, $\mathbf{x}_n\in\ell^1$ doesn't imply $\sup_n\|\mathbf{x}_n\|_{\ell^1}<\infty$. For example, if we take $\mathbf{x}_n=n\mathbf{e}_1\in\ell^1$, then $\sup_n\|\mathbf{x}_n\|_{\ell^1}= \sup_n n=\infty$. Here is a proof of $\sup_n\|\mathbf{x}_n\|_{\ell^1}<\infty$ if we assume $\mathbf{x}_n \xrightarrow{w^*} \mathbf{x}.$ We need to use the uniform boundedness principle. Since $\mathbf{x}_n \xrightarrow{w^*} \mathbf{x}$, we have
$$f_{\mathbf{x}_n}(\mathbf{y})\to f_{\mathbf{x}}(\mathbf{y}),\qquad \text{for all }\mathbf{y}\in c_0,$$
where $f_{\mathbf{x}}$ is defined in OP; in particular, we have
$$\sup_n|f_{\mathbf{x}_n}(\mathbf{y})|<\infty,\qquad \text{for all }\mathbf{y}\in c_0.$$
Now, uniform boundedness principle implies that $\sup_n\|f_{\mathbf{x}_n}\|_{c_0^*}<\infty$, and thus by $c_0^*\cong\ell^1$, we have
$$\sup_n\|\mathbf{x}_n\|_{\ell^1}<\infty.$$
$(\Longleftarrow)$: Given $\mathbf{x}_n=(x^{(n)}_1, x^{(n)}_2, \cdots), \mathbf{x}=(x_1, x_2,\cdots)\in\ell^1$ such that $\mathbf{x}_n$ converges componentwisely to $\mathbf x$ and $\sup_n\|\mathbf{x}_n\|_{\ell^1}<\infty,$ we show that $\mathbf{x}_n \xrightarrow{w^*} \mathbf{x}.$ Fix any $\mathbf{y}=(y_1, y_2, \cdots)\in c_0$, then we need to prove
$$\lim_{n\to\infty}\sum_{i=1}^\infty x^{(n)}_i y_i=\sum_{i=1}^\infty x_iy_i.$$
It suffices to show
$$\lim_{n\to\infty}\sum_{i=1}^\infty \left|x^{(n)}_i-x_i\right| |y_i|=0.\tag{$*$}$$
Now we prove $(*)$. Denote $M:=\sup_n\|\mathbf{x}_n\|_{\ell^1}+\|\mathbf{x}\|_{\ell^1}<\infty$.
Fix any small $\epsilon>0$. Since $y_i\to0$, we can find $i_0>0$ such that $|y_i|<\frac\epsilon{2M}$ for all $i>i_0$. Since $\mathbf{x}_n$ converges componentwisely to $\mathbf x$, we have $x^{(n)}_i\to x_i$ as $n\to\infty$ for all $1\leq i\leq i_0$, hence
$$\lim_{n\to\infty}\sum_{i=1}^{i_0} \left|x^{(n)}_i-x_i\right| |y_i|=0.$$
So we can find $N>0$ such that
$$\sum_{i=1}^{i_0} \left|x^{(n)}_i-x_i\right| |y_i|<\frac\epsilon2,\qquad \forall n>N.$$
Now, for all $n>N$, we have
\begin{align*}
\sum_{i=1}^\infty \left|x^{(n)}_i-x_i\right| |y_i|&=\sum_{i=1}^{i_0} \left|x^{(n)}_i-x_i\right| |y_i|+\sum_{i=i_0+1}^\infty \left|x^{(n)}_i-x_i\right| |y_i|\\
&<\frac\epsilon2+\frac\epsilon{2M}\sum_{i=i_0+1}^\infty \left(\left|x^{(n)}_i\right|+\left|x_i\right|\right)\\
&\leq \frac\epsilon2+\frac\epsilon{2M} \left(\|\mathbf{x}_n\|_{\ell^1}+\|\mathbf{x}\|_{\ell^1}\right)\\
&\leq \frac\epsilon2+\frac\epsilon{2M}\cdot M=\epsilon.
\end{align*}
This completes the proof of $(*)$ and thus completes the proof of $(\Longleftarrow)$.
Remark. Although the proof of $(\Longleftarrow)$ part seems much longer than the proof of $(\Longrightarrow)$ part, we still consider the proof of $(\Longrightarrow)$ part harder than the reverse implication. Because the uniform boundedness principle is a very non-trivial theorem, which is considered one of the cornerstones of functional analysis.
