How can we characterize weak convergence in $(c, \Vert \, \Vert _{\infty})$? Let me recall $c = \{ (x_h)_{h \in \mathbb{N}} \subset \mathbb{R} \, | \, \lim_{h \to \infty} x_h = k < \infty \}$ the space of convergent sequences equipped with $\Vert \, \Vert_{\infty}$.
What are sufficient and necessary conditions on a sequence $(x^{(n)})_{n \in \mathbb{N}} \subset c$ to say $ x^{(n)} \rightharpoonup x \in c$?
I found something like this
$$
x^{(n)} \rightharpoonup x \iff 
\begin{cases}
\sup_n \Vert x^{(n)} \Vert_{\infty} < \infty  & (1)\\
\lim_{n \to \infty} x^{(n)}_h = x_h  & (2)\\
\lim_{n \to \infty} \lim_{h \to \infty}x^{(n)}_h =\lim_{h \to \infty} x_h & (3)
\end{cases}
$$
However it looks to me not such an immediate proof
Partial Proof:
$\Rightarrow$: If $x^{(n)}$ weakly converges to $x$ then we know gratis $\sup_n \Vert x^{(n)} \Vert_{\infty} < \infty$ $(1)$. Moreover,  $x^{(n)} \rightharpoonup x \iff \phi ( x^{(n)}) \to \phi (x)$ for every $\phi \in c^*$. The projections $\pi_h (x) = x_h$ lie in $c^*$ and this fact leads to us to $\lim_{n \to \infty} x^{(n)}_h = x_h$ $(2)$. About the condition $(3)$: it is a consequence of $(2)$ if we can exchange the limits. However, $x^{(n)}$ is dominated and it converges pointwise. Then, using Dominated convergence theorem adapted to sequences, we obtain $(3)$. Is the proof of $(3)$ correct?
$\Leftarrow$: If we call $k^{(n)} = \lim_{h \to \infty} x^{(n)}_h$, and $k =\lim_{h \to \infty} x_h$ then (3) tells us $k^{(n)} \to k$. Hence $(x^{(n)}_h - k^{(n)}) \to (x_h-k)$ for every $h$ because of (2). Moreover, $\sup_n \Vert x^{(n)} - k^{(n)} \Vert_{\infty} \leq \sup_n(\Vert x^{(n)} \Vert + \Vert k^{(n)} \Vert ) < \infty$ because of (1).
Now, using the characterization of weak convergence in $c_0$ (indeed $x^{(n)}_h - k^{(n)}$ and  $x_h-k$ are in $c_0$), we discover
$$
(x^{(n)} - k^{(n)}) \rightharpoonup (x-k)
$$
Finally, $x^{(n)}  \rightharpoonup x$.
Does this implication hold?
 A: Crucial for us is the description of the dual space $c'$. Namely, we have that $c' \cong \ell^1$ via the isomorphism $\ell^1 \to c'$ given by
$$(\alpha_k)_{k=0}^\infty \mapsto f, \quad f(x_k)_{k=1}^\infty := \alpha_0\left(\lim_{k\to\infty} x_k\right) + \sum_{k=1}^\infty \alpha_kx_k.$$
Now, assume that $x_n \rightharpoonup x$ in $c$.

*

*By the uniform boundedness principle we get that $\sup_{n\in\Bbb{N}} \|x_n\|_\infty < +\infty$.


*For every $k \in \Bbb{N}$ for the projection we have $\pi_k \in c'$ so $$x_n(k) = \pi_k(x_n) \xrightarrow{n\to\infty} \pi_k(x) = x(k).$$


*For the limit functional $L(x_n)_n := \lim_{n\to\infty} x_n$ we have $L \in c'$ so
$$\lim_{k\to\infty} x_n(k) = L(x_n) \xrightarrow{n\to\infty} L(x) = \lim_{k\to\infty} x(k).$$
Conversely, suppose that $(x_n)_{n=1}^\infty$ is a bounded sequence in $c$ so that $(1)-(3)$ holds for some $x \in c$. Pick $f \in c'$ and we claim that $f(x_n) \to f(x)$. There exists some $(\alpha_k)_{k=0}^\infty \in \ell^1$ such that $f$ is of the above form.
The functions $g_n, g : \Bbb{N}_0 \to \Bbb{C}$ for $n \in \Bbb{N}$ given by $$g_n(k) = \begin{cases}\alpha_0 \left(\lim_{j\to\infty} x_n(j)\right), &\text{ if $k=0$},\\ \alpha_k x_n(k) &\text{ if $k>1$}. \end{cases}, \qquad g(k) = \begin{cases}\alpha_0 \left(\lim_{j\to\infty} x(j)\right), &\text{ if $k=0$},\\ \alpha_k x(k) &\text{ if $k>1$}. \end{cases}$$
are all dominated by the function $k \mapsto \alpha_k\left(\sup_{n\in\Bbb{N}} \|x_n\|_\infty\right)$ which is summable by $(1)$. Moreover, by $(2)$ and $(3)$,  we have $g_n \to g$ pointwise so by the Lebesgue Dominated convergence theorem we get
\begin{align*}
\lim_{n\to\infty} f(x_n) &= \lim_{n\to\infty}\left(\alpha_0 \left(\lim_{j\to\infty} x_n(j)\right)+ \sum_{k=1}^\infty \alpha_kx_n(k)\right) \\
&= \lim_{n\to\infty}\sum_{k=0}^\infty g_n(k)\\
&= \sum_{k=0}^\infty \lim_{n\to\infty} g_n(k)\\
&= \sum_{k=0}^\infty g(k)\\
&= \lim_{n\to\infty}\left(\alpha_0 \left(\lim_{j\to\infty} x(j)\right)+ \sum_{k=1}^\infty \alpha_kx(k)\right)\\
&= f(x).
\end{align*}
Since $f\in c'$ was arbitrary, we conclude $x_n \rightharpoonup x$ in $c$.
