TITLE: Isometrical isomorphism $T : \ell _1\to c^*$
QUESTION [0 upvotes]: $c^*$ is the dual space of $c$, space of convergent sequences
Prove $T : \ell _1\to c^*$ is isometrical isomorphism when:
$$(Ta)(x) = \alpha _0\xi + \sum_{j=1}^\infty \alpha _j\xi _j,\qquad x = (\xi _j)_{j=1}^\infty\in c,\ \  \xi _j\xrightarrow[j\to\infty]{}\xi\quad a = (\alpha _j)_{j=0}^\infty\in\ell _1 $$
I have already verified that $T\in\mathcal{L}(\ell _1,c^*)$ and $\|Ta\|\leq \|a\|$. Remains to show that $\|Ta\|\geq \|a\|$.
The tough spot is surjectivity. Let $f\in c^*$, need to find an $a$ s.t $(Ta)(x) = f(x)$ for all $x\in c$. If $x\in c$ we can express it as $x = \sum_{j=1}^\infty \xi _je_j$, where $e_j = \{\underbrace{0,0,\ldots ,1}_{j},0,\ldots\}$. Then $f(x) = \sum_{j=1}^\infty f(e_j)\xi _j$. Ok, since $f$ is bounded we can take $a = (0, f(e_1),f(e_2),\ldots)\in\ell _1$ and
$$(Ta)(x) = 0 + \sum_{j=1}^\infty f(e_j)\xi _j = f(x) \quad\forall x\in c$$
..but something seems out of place. The definition of $T$ would be redundant if the above were really true. Have I cheated somewhere?  
For $\|a\|\leq \|Ta\|$ we need
$$\|a\| = \sum_{j=0}^\infty |\alpha _j|\overset{?}\leq\sup\limits_{x\in S_c}|(Ta)(x)| = \|Ta\|$$
$S_c$ is the unit sphere in $c$ 
but it doesn't seem all that obvious.
Hints on former inequality, please and also need clarification on surjectivity and the rather strange result.

REPLY [4 votes]: If $a = (a_0,a_1,...)$, $x=(x_1,....)$, then
$(Ta)(x) = a_0 \lim_k x_k +\sum_k a_k x_k$. If $\|x\|_\infty \le 1$, we
have $|(Ta)(x)| \le |a_0|+ \sum_k |a_k| = \|a\|_1$ and so $\|Ta\|_\infty \le \|a\|_1$. If we take $x$ with $x_k = \operatorname{sgn} a_k$, then $\|x\|_\infty = 1$ and $|(Ta)(x) | = \|a\|_1$ and so $\|Ta\|_\infty = \|a\|_1$.
Note: The above is slightly incorrect as the $x_k$ may have no limit. Choose
$\epsilon>0$ and select $n$ such that $\sum_{k>n} |a_k| < \epsilon$. Define $x_k = \operatorname{sgn} a_0$ for $k > n$ instead. This shows that for any $\epsilon>0$ that there is some $x$ such that $|(Ta)(x)| \ge \|a\|_1 -\epsilon$,
and hence $\|Ta\|_\infty = \|a\|_1$.
If $x \in c$, let $Lx = \lim_k x_k$, let $e_k$ be the $k$th unit vector and $e=(1,1,...)$.
Given $y \in c^*$ and $x \in c$, note that $x-(Lx)e \in c_0$ and
$x= (Lx)e+x-(Lx)e $, of course.
Then $y(x) = y(x-(Lx)e) + (Lx)y(e)$. Since $y$ is a linear functional on ${c_0}$,
there is some $\{y_k\}_k \in l_1$ such that for any $x \in C$ we have $y(x-(Lx)e) = \sum_k y_k (x_k -Lx)$.
Then $y(x) = (Lx)y(e) + \sum_k y_k (x_k -Lx) = (y(e)+\sum_k y_k)Lx + \sum_k y_k x_k$.
In particular, if we choose $a_0 = y(e)+\sum_k y_k$, and $a_k = y_k$ for $k >0$,
we have $Ta = y$, hence $T$ is surjective.
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