If X has beta property, do $(x_i)_i$ points are LUR? Let $X$ be a Banach space, $S_X$ its unit sphere and $B_X$ its unit ball.
The space $X$ is said to has $\beta$ property if there exists a system $\{(x_i,f_i):i \in I\} \subset S_X \times S_{X^*}$ and $0 \leq \rho <1$ such that 
(i) $f_i(x_i)=1$, for all $i \in I$
(ii) $|f_i(x_j)| \leq \rho$, for all $i \neq j$
(iii) $||x|| = \sup\{|f_i(x)|:i \in I\}$, for all $x \in X$
A point $x \in S_X$ is called locally uniformly convex (LUR) if for every $\epsilon>0$ there exists $\delta(\epsilon)>0$ such that
$y \in S_X, \cfrac{||x+y||}{2}>1-\delta(\epsilon) \Rightarrow ||x-y||<\epsilon $
I'm trying to prove that if $X$ has $\beta$ property and $\dim X >1$, such points $(x_i)_{i}$ cannot be LUR points.
My first attempt at a solution was try to define a sequence $(x_n)$ in $S_X$ such that $||x_n+x_i||$ approaches to $2$ and $||x_n-x_i||$ does not approach to $0$. It is easy to see that if $x \in S_X \cap \ker{f_i}$, we have that $||x-x_i|| \geq 1$, then maybe it is a good idea to define such sequence in $S_X \cap \ker{f_i}$. 
Is it a good approach? Any other hints?
 A: Assume $X$ has the $\beta$ property and $\dim X>1.$ Fix $i\in I.$ The point $x_i$ fails the LUR property with $\epsilon=1-\rho.$
For any $k\in\ker f_i,$
$$\|k\|\leq 1-\rho\implies\|x_i+k\|=1\tag{*}$$
because $f_i(x_i+k)=f_i(x_i)=1$ and $|f_j(x_i+k)|\leq|f_j(x)|+|f_j(k)|\leq \rho+\|k\|\leq1.$ (The inequality $|f_j(k)|\leq \|k\|$ comes from property (iii).)
Since $\dim X>1,$ there exists $k\in \ker f_i$ with $\|k\|=1-\rho.$ We get a counterexample to the LUR property with $\epsilon=1-\rho$ and $y=x+k,$ because $\|(x+y)/2\|=1$ and $\|y\|=1$ by (*), but $\|x-y\|=\|k\|=\epsilon.$
A: Assume for contradiction that $X$ has the $\beta$ property and that the points $x_i$ are LUR,
define
$$ \varepsilon = \frac12 \, \sup_{x \in S_X} \left( \inf_{i \in I} \| \pm x - x_i\| \right), $$
and choose $x_{\varepsilon}$ such that
$$ \inf_{i \in I} \|\pm x_{\varepsilon} - x_i\| \geq \varepsilon, \quad $$
If $\varepsilon = 0$, then there exists a subsequence of $\{x_i\}$, say $x_{k_i}$,
such that $x_{k_i} \to \pm x_{\varepsilon}$ 
(either to $+ x_{\varepsilon}$, or to $- x_{\varepsilon}$) in $X$.
But then, using the $\beta$ property, (i),
\begin{align}
    |f_{k_i}(x_{k_{i-1}})| &= |f_{k_i}(x_{k_i}) + f_{k_i}(x_{k_{i-1}}-x_{k_i})| \\
                           &\geq |f_{k_i}(x_{k_i})| - \|f_{k_i}\| \, \|x_{k_i} - x_{k_{i-1}}\| \\
                           &= 1 - \|x_{k_i} - x_{k_{i-1}}\| \to 1 \quad \text{as} \quad i \to \infty,
\end{align}
contradicting the $\beta$ property, (ii); therefore, $\varepsilon > 0$.
Using the $\beta$ property, (iii),
for any $\alpha > 0$ there exists $i_{\alpha} \in I$ such that
\begin{align}
    |f_{i_{\alpha}}(x_{\varepsilon})| \geq 1 - 2\alpha = \|x_{\varepsilon}\| - 2\alpha.
\end{align}
Assume $f_{i_{\alpha}}(x_{\varepsilon}) \geq 0$ (renaming $x_{\varepsilon} \rightarrow -x_{\varepsilon}$ if necessary).
Then we have
\begin{align}
    \frac{1}{2}\| x_{\varepsilon} + x_{i_{\alpha}} \| &\geq \frac{1}{2}  f_{i_{\alpha}} (x_{\varepsilon} + x_{i_{\alpha}}) \\
                                           &\geq 1 - \alpha.
\end{align}
To obtain a contradiction using this reasoning,
there must exist $\delta(\varepsilon)$ independent of $i$ such that the convexity condition is satisfied.
If $X$ is finite dimensional,
this is possible,
since there can be only finitely many $x_i$.
If $X$ is infinite-dimensional, I suppose this argument isn't enough for non-uniformly convex norms...
