A corollary of Hahn-Banach I have this exercice.

It's easy to prove that $1\Leftarrow 2$; 
Now I want to prove the other implication.
Let $f_0$ be a function such that:
$$f_0: span (x_1,...,x_n):=\left\{{\left.\sum _{i=1}^{n}\lambda _{i}x_{i}\right|\lambda _{i}\in \mathbf {R} }\right\}\rightarrow \mathbb{R}$$
such that : $f_0(\sum _{i=1}^{n}\lambda _{i}x_{i})=\sum_{i=1}^{n}\lambda_i.\left \| x_i \right \|$
Now, using Hahn-Banach theorem we can find the result. But how to prove that: $\left \| f_0 \right \|=1$???
PS: $f\in (F')_1$ means that $\left \| f \right \|_{F'}=1$
 A: First notice that if $\|x+y\| = \|x\| + \|y\|$ holds, then for all $\alpha, \beta \ge 0$ we also have $$\|\alpha x+\beta y\| = \alpha\|x\| + \beta\|y\|$$
Namely:
\begin{align}
\alpha\|x\| + \beta\|y\| &= \alpha(\|x\| + \|y\|) - (\alpha - \beta)\|y\| \\
&= \|\alpha(x+y)\| - \|(\alpha - \beta)y\|\\
&\le \|\alpha(x+y) - (\alpha - \beta)y\|\\
&= \|\alpha x + \beta y\|\\
&\le \alpha\|x\| + \beta\|y\|
\end{align}
This inductively generalizes to $$\left\|\sum_{i=1}^n x_i\right\| = \sum_{i=1}^n \|x_i\| \implies \left\|\sum_{i=1}^n \alpha_ix_i\right\| = \sum_{i=1}^n \alpha_i\|x_i\| = \sum_{i=1}^n \|\alpha_i x_i\|, \text{ for all } \alpha_i \ge 0$$

Now, your functional $f_0 : \operatorname{span}\{x_1, \ldots, x_n\} \to \mathbb{R}$ is acting as
$$f_0\left(\sum_{i=1}^n \alpha_i x_i\right) = \sum_{i=1}^n \alpha_i\|x_i\|, \text{ for all }\alpha_i \in \mathbb{R}$$
so we have:
$$\left|f_0\left(\sum_{i=1}^n \alpha_i x_i\right)\right| = \left|\sum_{i=1}^n \alpha_i\|x_i\|\right| \le \sum_{i=1}^n |\alpha_i| \|x_i\| = \sum_{i=1}^n \|\alpha_ix_i\| =\left\|\sum_{i=1}^n \alpha_ix_i\right\|$$
Therefore $\|f_0\| \le 1$.
For the reverse inequality, notice that $$\|f_0\|\ge \frac{|f_0(x_i)|}{\|x_i\|} = \frac{\|x_i\|}{\|x_i\|} = 1$$
We conclude $\|f_0\| = 1$.
