}}\in\Lambda_0}((1-\delta_1)f({{\lambda }})-\theta_0((1-\delta_1)f({{\lambda }}))g_{{\lambda }}(\mu)\\
&=&\theta_0((1-\delta_1)f(\mu))=\theta(f(\mu)),\end{aligned}$$ which verifies (\[RU4\]). Moreover, we have, by (\[U12\]) and (\[U11\]), $$\begin{aligned}
\label{U13}
\lefteqn{\left\|\sum_{{{\lambda }}\in\Lambda_0}((1-\delta_1)f({{\lambda }})-\theta_0((1-\delta_1)f({{\lambda }}))g_{{\lambda }}\right\|  }\nonumber\\
&\le&\left\|\sum_{{{\lambda }}\in\Lambda_0}|(1-\delta_1)f({{\lambda }})-\theta_0((1-\delta_1)f({{\lambda }}))| g_{{\lambda }}\right\|\le  \min(M_3,1)\delta_1.\end{aligned}$$ Hence, by (\[U10\]), (\[U13\]), and (\[U14\]) $$\|f-\Theta(f)\|\le \delta_1\|f\|+\delta_1 \le (M_3+1) \delta_1=\delta\le \delta_0,$$ which proves (\[RU2\]). Furthermore, it shows that in case of condition (i), $\Theta(f)\in F$ for all $f\in F_0$. If condition (ii) is fulfilled, that is, $F=a_0B_X$ for some $a_0>0$, we argue as follows: $$\|\Theta(f)\|\le\|(1-\delta_1)f\|+M_3\delta_1\le (1-\delta_1)a_0+\delta_1 a_0=a_0,$$ thus, again, $\Theta(f)\in F$ for all $f\in F_0$.

\[pro:1\] Let $K={{\mathbb K}}$. Assume that $S,{\widetilde }{S},R,\Psi$ are as above (\[XG3\]), (\[XG4\]), that $F$ is a bounded subset of a normed space $X$, and $\Lambda$ is a linearly independent subset of $X^\#$. Let $F_0$ be a nonempty subset of $F$ and assume that either\
(i) $F_0^{\delta_0}\subseteq F$ for some $\delta_0>0$, or\
(ii) $F$ is a non-zero multiple of the unit ball of $X$.\
Furthermore suppose $\sup_{f\in F_0}|f({{\lambda }})|<\infty$ for each ${{\lambda }}\in \Lambda$ and $S$ is uniformly continuous on $F$. Then for all $n\in {{\mathbb N}}_0$, $$e_{2\kappa n}^{{\rm q }}(S,F_0)\le \|\Psi\|_{\rm Lip}\, e_n^{{\rm q }}({\widetilde }{S},{\widetilde }{F}).$$

This is a direct consequence of Lemmas \[lem:1\], \[lem:2\], and the uniform continuity of $S$.

In 