$sup_{f\in X^* , ||f||\le1}\sum_{n=1}^N |f(x_n)| = max _{\epsilon_n \in \{\pm1\}}||\sum_{n=1}^N \epsilon_nx_n ||$ Let $\{x_n\}_{n=1}^N $ be elements in a normed space $X$.
I want to prove that 
$$sup_{f\in X^* , ||f||\le1}\sum_{n=1}^N |f(x_n)| = max _{\epsilon_n \in \{\pm1\}}||\sum_{n=1}^N \epsilon_nx_n|| $$
im not sure how. 
I thought it may be related to the following theorem: 
$$ Y\subset X , \overline{Y} \ne X  $$ then 
$$ d(x,Y) = max_{f\in X^* , f(Y)=0 , ||f||=1} |f(x)|$$
But im not sure how to use this theorem in my problem(if it's related at all). 
Thanks for helping!
 A: If the underlying field is complex, the statement is false. Take $X=\Bbb C$ and $x=1, y=i$. Then, $$
\sup_{\|f\|_{X^*}\le 1}|f(x)|+|f(y)|=2,\quad \ \ \ \max_{\epsilon_n\in \{-1,1\}^2}\|\epsilon_1x+\epsilon_2y\|=\sqrt2.
$$
However, in real case, the equality holds.
$(\ge)$ For each choice of $\epsilon_n\in \{\pm 1\}^n$, we have 
$$
\left\|\sum_n \epsilon_nx_n\right\| = \sup_{\|f\|_{X^*}\le 1}\left|f\left(\sum_n \epsilon_nx_n\right)\right|\le\sup_{\|f\|_{X^*}\le 1} \sum_n |f(x_n)|.
$$ by Hahn-Banach theorem, i.e. $\|x\| = \sup_{\|f\|_{X^*}\le 1}|f(x)|$. This shows $ \max\limits_{\epsilon_n\in\{\pm 1\}^n}\left\|\sum_n \epsilon_nx_n\right\|\le \sup\limits_{\|f\|_{X^*}\le 1}\sum_n |f(x_n)|.$ Note that $\ge$ is also true on the complex field. 
$(\le)$ We have for each $\|f\|_{X^*}\le 1$, that
$$\begin{align*}
\sum_n |f(x_n)| &= \sum_n f(x_n)\cdot \text{sgn}(f(x_n))\\&=f\left(\sum_n \text{sgn}(f(x_n))\cdot x_n\right)\\&\le \left\|\sum_n \text{sgn}(f(x_n))\cdot x_n\right\|\\&\le \max_{\epsilon_n\in \{\pm 1\}^n}\left\|\sum_n \epsilon_nx_n\right\|.
\end{align*}$$ Thus we obtain $\sup\limits_{\|f\|_{X^*}\le 1}\sum_n |f(x_n)|\le \max\limits_{\epsilon_n\in\{\pm 1\}^n}\left\|\sum_n \epsilon_nx_n\right\|$.
