Prob. 6, Sec. 2.10 in Erwin Kreyszig's functional analysis. Definition (Dual space $X'$). Let $X$ be a normed space. Then the set of all bounded linear functionals on $X$ constitutes a normed space with norm defined by
$$
\left \| f \right \|= \sup_{\substack{x\in X\\x \neq 0}} \frac{\left | f(x) \right |}{\left \| x \right \|} = \sup_{\substack{x\in X\\ \left \| x \right \|=1}} \left | f(x) \right |.
$$
which is called the dual space of $X$ and is denoted by $X'$.
Now comes my question
if $X$ is the space of ordered n-tuples of real numbers and $\left \| x \right \| =\max_{j}\left |\xi_j  \right |$, where $x=\left ( \xi _{1},\cdots ,\xi _{n} \right )$, what is the corresponding norm on the dual space $X'$
my attempt:
Since the normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded (or Shortly $X'=X^*$ where $X^*$ is algebraic dual space of $X$) as stated in Theorem 2.7-8 in Erwin Kreyszig's book.
If $\{e_1,\cdots,e_n\}$ is a basis for $X$, then each $x\in X$ is written as
$x=\sum_{k=1}^{n} \xi_k e_k$
and because $f\in X'$ is linear
\begin{equation}
f(x)=f\left(\sum_{k=1}^{n} \xi_k e_k\right) = \sum_{k=1}^{n} \xi_k \underbrace{f(e_k)}_{=\gamma_k} = \sum_{k=1}^{n} \xi_k \gamma_k
\end{equation}
Next,
\begin{equation}
\left | f(x) \right |=\left | \sum \xi_k \gamma_k \right |\leq \sum \left | \xi_k \gamma_k \right | \leq \left \| x \right \| \sum \left | \gamma_k \right |
\end{equation}
and because $f\in X'$ is bounded, 
$\left \| f \right \| = \sup_{\left \| x \right \|\neq 0} \frac{\left | f(x) \right |}{\left \| x \right \|}\leq \frac{\left \| x \right \|\sum \gamma _{k}}{\left \| x \right \|}=\sum \gamma _{k}$
hence 
$\left \| f \right \| = \sum \gamma _{k}$ which is the norm on $l^1$
Is the proof correct and complete?
 A: After the fruitful discussion, the full proof goes as follows.
Since the normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded (or Shortly $X'=X^*$ where $X^*$ is algebraic dual space of $X$) as stated in Theorem 2.7-8 in Erwin Kreyszig's book.
If $\{e_1,\cdots,e_n\}$ is a basis for $X$, then each $x\in X$ is written as
$x=\sum_{k=1}^{n} \xi_k e_k$
and because $f\in X'$ is linear
\begin{equation}
f(x)=f\left(\sum_{k=1}^{n} \xi_k e_k\right) = \sum_{k=1}^{n} \xi_k \underbrace{f(e_k)}_{=\gamma_k} = \sum_{k=1}^{n} \xi_k \gamma_k
\end{equation}
Next,
\begin{equation}
\left | f(x) \right |=\left | \sum \xi_k \gamma_k \right |\leq \sum \left | \xi_k \gamma_k \right | \leq \left \| x \right \| \sum \left | \gamma_k \right |
\end{equation}
and because $f\in X'$ is bounded, 
$\left \| f \right \| = \sup_{\left \| x \right \|\neq 0} \frac{\left | f(x) \right |}{\left \| x \right \|}\leq \frac{\left \| x \right \|\sum \gamma _{k}}{\left \| x \right \|}=\sum \gamma _{k}$  
which gives the 1st direction
$\left \| f \right \|\leq \sum \left | \gamma _{k} \right |$
on the other hand if we consider $x=\left ( \xi_k \right )$ with
$\xi_{k}=
\left\{\begin{matrix}
\left | \gamma_{k} \right | /\gamma_{k} & \text{if}\ \gamma _{k}\neq 0\\ 
 0 & \text{if}\ \gamma _{k}= 0
\end{matrix}\right.$
we observe that $\left \| x \right \| = 1$ and we also observe that
$\left | f(x) \right |=\sum \left | \gamma _{k} \right |\leqslant \left \| f \right \| \left \| x \right \|=\left \| f \right \|$
which gives the second direction
$\sum \left | \gamma _{k} \right |\leq \left \| f \right \|$
comparing the 1st direction and the second direction, we conclude
$\left \| f \right \| = \sum \gamma _{k}$ which is the norm on $l^1$
A: The result is correct. The reasoning has a gap:
You have the inequalities $|f(x)|\le \sum_k|\gamma_k|\cdot \|x\|$ and $|f(x)|\le \|f\|\cdot\|x\|$. 
The problem is, that, in general, from $A\le B$  and $A\le C$ we cannot conclude neither $B\le C$ nor $C\le B$, not even $C=B$, as for example take $(A,B,C)=(0,2,3)$ or $(0,3,2)$.
Well, in this particular case, one direction follows from your calculations, namely $\|f\|\le\sum_k|\gamma_k|$ because $\|f\|=\sup_{\|x\|=1}(|f(x)|)$.
For the other direction ($\|f\|\ge\sum_k|\gamma_k|$) take certain elements $x$ that make $|f(x)|$ as big as possible.
