How do I show that this mapping is Lipschitz continuous Let $X$ be a Banach space. Since metric spaces are paracompact we know that $X$ is too. Then given the following cover of $X$:
\begin{equation}
\mathcal{N}=\{N_x\ |\ x\in X\},
\end{equation} where $N_x$ is an open neighbourhood about $x$, there exists a locally finite open refinement:
\begin{equation}
\{M_{\iota}\}_{\iota\in I}.
\end{equation}
Define the functions
\begin{align}
\rho_{\iota}(u)&\equiv\text{dist}(u, X\setminus M_{\iota}),\\
\varphi_{\iota}(u)&\equiv\frac{\rho_{\iota}(u)}{\sum_{\iota'\in I}\rho_{\iota'}(u)}.
\end{align}
I would like to show that $\varphi_{\iota}$ is Lipschitz continuous on some neighbourhood $W$ of $x$, for every $x\in X$.
Since $\{M_{\iota}\}_{\iota\in I}$ is locally finite then given $x\in X$, there exists an open neighbourhood $W$ about $x$ such that $W\cap M_{\iota'}\neq\emptyset$ for finitely many $\iota'\in I$. Let $u, v\in W$ then
\begin{align}
|\varphi_{\iota}(u)-\varphi_{\iota}(v)|&=\left|\frac{\rho_{\iota}(u)}{\sum_{\iota'\in I}\rho_{\iota'}(u)}-\frac{\rho_{\iota}(v)}{\sum_{\iota'\in I}\rho_{\iota'}(v)}\right|
\end{align}
How do I proceed form here? I know that $\rho_{\iota}$ is Lipschitz for every $\iota$ but I'm not sure how to use this in the argument.
 A: Write $$q(u)=\sum_t \rho_t (u)$$
First note that $q$ is Lipschitz on $W$. Also, there exist constants $\delta_1,\delta_2,\delta_3>0$ (depending on $W$) such that $$\tag{1}\delta_1\leq q(u)\leq \delta_2,\ \forall\ u\in W$$
$$\tag{2}\rho_t(u)\leq\delta_3,\ \forall\ u\in W$$
Note
\begin{eqnarray}
 |\varphi_t(u)-\varphi_t(v)| &=& \left|\frac{\rho_t(u)}{q(u)}-\frac{\rho_t(v)}{q(v)}\right|      \nonumber \\
   &=& \left|\frac{1}{q(u)q(v)}\left\{q(v)\left[\rho_t(u)-\rho_t(v)\right]-\rho_t(v)\left[q(u)-q(v)\right]\right\}\right| \nonumber \\
   &\le& \frac{q(v)}{q(u)q(v)}\left|\rho_t(u)-\rho_t(v)\right|+\frac{\rho_t(v)}{q(u)q(v)}\left|q(u)-q(v)\right|
\end{eqnarray}
If $u,v\in W$ we conclude from $(1),(2)$ and the last inequality that $\varphi_t$ is Lipschitz in $W$.
A: Note that $\sum_{\iota'\in I}\rho_{\iota'}(u)$ is a locally finite sum of
Lipschitz functions, so it is locally Lipschitz. Also, the $M_\iota$ are an
open cover, so the sum is positive everywhere.
Further note that $(x, y) \mapsto x/y$ is continuously differentiable (where
$y \neq 0$), therefore it is locally Lipschitz.
To complete the proof, use the fact that a composition of locally Lipschitz functions is locally Lipschitz.
