denoted by $B_X$.

\[lem:2\] Let $K={{\mathbb K}}$, let $F$ be a bounded subset of a normed space $X$, and let $\emptyset \ne F_0\subseteq F$. Assume that either\
(i) there is a $\delta_0>0$ such that $F_0^{\delta_0}\subseteq F$ or\
(ii) $F$ is a non-zero multiple of the unit ball of $X$.\
Furthermore, let $\Lambda_0\subset X^{\#}$ be a finite, linearly independent set with $$sup_{f\in F_0}|f({{\lambda }})|<\infty \quad ({{\lambda }}\in \Lambda_0).$$ Then for each $\delta>0$ there are mappings $$\theta:{{\mathbb K}}\to {{\mathbb K}},\quad \Theta:F_0\to F$$ such that $\theta({{\mathbb K}})$ is a finite set, $$\label{RU4}
(\Theta(f))({{\lambda }})=\theta(f({{\lambda }}))\quad(f\in F_0,\, {{\lambda }}\in \Lambda_0),$$ and $$\label{RU2}
\sup_{f\in F_0}\|f-\Theta(f)\|\le \delta.$$

We can assume $\delta\le \delta_0<1$. The linear independence of $\Lambda_0$ implies that for each ${{\lambda }}\in\Lambda_0$ there is a $g_{{\lambda }}\in X$ with $g_{{\lambda }}({{\lambda }})=1$ and $g_{{\lambda }}(\mu)=0$ for $\mu\in \Lambda_0\setminus\{{{\lambda }}\}$. Define $$\label{U12}
M_1=\max_{{{\lambda }}\in\Lambda_0}\|g_{{\lambda }}\|,\quad M_2=\sup_{{{\lambda }}\in\Lambda_0, f\in F_0}|f({{\lambda }})|,\quad M_3=\sup_{f\in F_0} \|f\|,$$ $$\label{U14}
\delta_1=\delta/(M_3+1),$$ and choose any $\theta_0: {{\mathbb K}}\to {{\mathbb K}}$ such that $\theta_0({{\mathbb K}})$ is finite and $$\label{U11}
|a-\theta_0(a)|\le M_1^{-1}|\Lambda_0|^{-1}\delta_1\min(M_3,1)\quad(|a|\le M_2).$$ Now we define $\theta:{{\mathbb K}}\to{{\mathbb K}}$ by setting for $a\in{{\mathbb K}}$, $$\theta(a)=\theta_0((1-\delta_1)a),$$ and $\Theta:F_0\to X$ by $$\label{U10}
\Theta(f)=(1-\delta_1)f-\sum_{{{\lambda }}\in\Lambda_0}((1-\delta_1)f({{\lambda }})-\theta_0((1-\delta_1)f({{\lambda }}))g_{{\lambda }}.$$ Then for $f\in F_0$, $\mu\in \Lambda_0$, $$\begin{aligned}
\lefteqn{ (\Theta(f))(\mu)}\\
&=&(1-\delta_1)f(\mu)-\sum_{{{\lambda 