Related to the dual space of $c$ In the analysis class, the instructor defined the following map between $\ell^1$ and $c^*$:
$$T:\ell^1\to c^*,\quad g\mapsto\phi_g$$
where $\phi_g$ is defined as
$$\phi_g:c\to\mathbb{C},\quad \phi_g(f)=g(1)\lim_{n\to\infty}f(n)+\sum_{n=1}^\infty g(n+1)f(n)$$
By showing that $T$ is an isometric isomorphism, we can conclude that $c^*=\ell^1$. 
The proof of isometry can be seen here in the body of question. However, I am not quite sure about the onto part. My instructor proved the same way as in the answer of the linked question: for each $\phi\in c^*$, define
$$g(n)=\begin{cases}\phi(\mathbf{1}),&n=1\\
\phi(e_{n-1}),&n>1\end{cases}$$
where $\mathbf{1}$ is the sequence with all terms equal to 1, and $e_n$'s are the canonical coordinates. We can show that $g\in\ell^1$. Then my instructor said it is obvious that $\phi=\phi_g$. However, this does not seem quite convincing to me. I did the following: we have
$$\phi(f)=\sum_{n=1}^\infty f(n)\phi(e_n)=\sum_{n=1}^\infty f(n)g(n+1).$$
which is missing a term $g(1)\lim_{n\to\infty}f(n)$.
Where are my mistakes and what is the right way to show this?
 A: The mistake is that you wrote
$$\phi(f) = \sum_{n=1}^\infty f(n)\phi(e_n).$$
That would hold if we had
$$f = \sum_{n=1}^\infty f(n)\cdot e_n\tag{$\ast$}$$
in the space $c$, but the series on the right hand side of $(\ast)$ converges only if $\lim\limits_{n\to\infty} f(n) = 0$. For we have
$$\left\lVert f - \sum_{n=1}^N f(n)\cdot e_n\right\rVert = \sup_{n > N} \lvert f(n)\rvert \geqslant \left\lvert \lim_{n\to\infty} f(n)\right\rvert.$$
The system $\{ e_n : n\in \mathbb{N}\setminus \{0\}\}$ is a Schauder basis of the closed subspace $c_0$ of null-sequences, not of $c$.
The correct representation of $f\in c$ is
$$f = \lambda\cdot \mathbf{1} + \sum_{n=1}^\infty \bigl(f(n)-\lambda\bigr) \cdot e_n,\tag{1}$$
where $\lambda = \lim\limits_{n\to\infty} f(n)$.
Using the representation $(1)$, we obtain
$$\phi(f) = \lambda\phi(\mathbf{1}) + \sum_{n=1}^\infty \bigl(f(n)-\lambda\bigr)\phi(e_n) = \lambda\cdot \left(g(1) - \sum_{n=2}^\infty g(n)\right) + \sum_{n=1}^\infty f(n)\cdot g(n+1),$$
which however in general is different from $\phi_g$. One must choose a slightly different $g$ to get $\phi_g = \phi$.
