Isometric isomorphism from $\ell^1$ to $c^*$ I think I have the right idea how to solve this, but I can't find a specific sequence that does what I want it to.
For some $a=(a_n)\in\ell^1$, we define $\varphi_a:c\to\mathbb{F}$ by
$$\varphi_a(x)=a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n,\quad x=(x_n)\in c,$$
where $c$ is the space of convergent sequences and $\|\varphi_a\|=\sup\{|\varphi_a(x)|:\|x\|_\infty=1\}$.
I'm trying to prove that the mapping $\Phi:\ell^1\to c^*$ defined by $\Phi(a)=\varphi_a$ for $a\in\ell^1$ is an isometric isomorphism.
To show $\|\varphi_a\|\leq\|a\|_1$ is easy. I would like to find some $x\in c$ with $\|x\|_\infty=1$ such that $|\varphi_a(x)|\geq\|a\|_1$; this would show equality since it is a lower bound on $\|\varphi_a\|$.
Also, any idea how to show it is surjective, i.e. taking any $x\in c$ and expressing it as some $\varphi_a(x)$?
 A: Isometry
Indeed, the inequality
$$\left| a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n \right|\le \|x\|_{\infty}\|a\|_1$$
is immediate from the triangle inequality.
To attain equality, we want everything inside the absolute value to have the same sign, that is $x_n=\operatorname{sign} a_{n+1}$. A problem is, this doesn't necessarily have a limit. A compromise is to set $x_n=\operatorname{sign} a_{n+1}$ for $n\le N$ and $x_n=\operatorname{sign} a_{1}$ for $n>N$, thus ensuring $\lim_{n\to\infty}x_n=\operatorname{sign} a_{1}$. This allows the terms $a_{n+1}x_n $ with $n>N$ to potentially be negative, so the estimate becomes 
$$
\left| a_1\lim_{n\to\infty}x_n+\sum_{n=1}^\infty a_{n+1}x_n \right|\ge \|x\|_{\infty}\|a\|_1 - 2 \sum_{n=N+1}^\infty |a_{n+1}|
$$
Since $a\in\ell_1$, the subtracted term can be made arbitrarily small.
Surjectivity
The linear combinations of the standard basis $e_n$ and the constant sequence $\sum_{n\in\mathbb{N}} e_n$ are dense in $c$. Therefore, every bounded functional $f$ on $c$ is determined by its values on this set; call them $\beta_n=f(e_n)$ and $\beta_\infty = f(\sum e_n)$. The boundedness of $f$ implies $\beta\in \ell_1$.  But for any such  $\beta $ there is a corresponding $a\in \ell_1$, namely
$$
a= (\beta_\infty,\beta_1,\beta_2,\dots)
$$
