Let $U$ and $V$ be Euclidean spaces. Let $X\subseteq U$ and $Y\subseteq V$ be open, bounded subsets.
Let $G:=\operatorname{BLip}(X\times Y,U)$ be the (Banach) space of bounded Lipschitz functions from $X\times Y$ to $U$ with norm given by the $\max$ of the $\sup$ norm and the Lipschitz modulus. (See e.g. Chapter 8 of "Lipschitz Functions" by Stefan Cobzas, Radu Miculescu and Adriana Nicolae.)
Let $H\subseteq G$ be open and suppose:
- For all $h\in H$, $h(X\times Y)\subseteq X$.
- There exists $C>0$ such that for all $g_1,g_2,h_1,h_2\in H$, all $x_1,x_2\in X$ and all $y_1,y_2,z\in Y$:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le C \max{\{\|g_1-g_2\|,\|h_1-h_2\|\}} \max{\{\|x_1-x_2\|,\|y_1-y_2\|\}}. \end{align}$$
Can $H$ be non-empty? What is the largest possible $H$ with this property (in some loose sense)? Could $H=\{g\in G|g(X\times Y)\subseteq X\}$?
Note 1: In any case for all $g_1,g_2,h_1,h_2\in G$, all $x_1,x_2\in X$ and all $y_1,y_2,z\in Y$:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le \|g_1(h_1(x_1,y_1),z)-g_1(h_1(x_2,y_2),z)\| + \|g_2(h_2(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le [\|g_1\| \|h_1\| + \|g_2\| \|h_2\|]\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}} \end{align}$$
and:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le { \|g_1(h_1(x_1,y_1),z)-g_1(h_2(x_1,y_1),z)+g_1(h_2(x_1,y_1),z)-g_2(h_2(x_1,y_1),z)\| \\ + \|g_2(h_2(x_2,y_2),z)-g_2(h_1(x_2,y_2),z)+g_2(h_1(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)\| } \\ & \le { \|g_1(h_1(x_1,y_1),z)-g_1(h_2(x_1,y_1),z)\|+\|g_1(h_2(x_1,y_1),z)-g_2(h_2(x_1,y_1),z)\| \\ + \|g_2(h_2(x_2,y_2),z)-g_2(h_1(x_2,y_2),z)\|+\|g_2(h_1(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)\| } \\ & \le [\|g_1\|+\|g_2\|] \|h_1 - h_2\| + 2\|g_1-g_2\| \\ & \le [2+\|g_1\|+\|g_2\|] \max{\{\|g_1-g_2\|,\|h_1-h_2\|\}}. \end{align}$$
Note 2: Suppose that $g_1,g_2,h_1,h_2\in G\cap B(X\times Y,U)$, where $B(X\times Y,U)$ is the space of bounded linear operators from $X\times Y$ to $U$.
Then for all $x_1,x_2\in X$ and all $y_1,y_2,z\in Y$:
$$\begin{align} & g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z) \\ & = g_1(h_1(x_1-x_2,y_1-y_2),0)-g_2(h_2(x_1-x_2,y_1-y_2),0) \\ & = g_1((h_1-h_2)(x_1-x_2,y_1-y_2),0)+(g_1-g_2)(h_2(x_1-x_2,y_1-y_2),0). \end{align}$$
So:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le [\|g_1\|\|h_1-h_2\|+\|g_1-g_2\|\|h_2\|]\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}}. \end{align}$$
Thus by symmetry:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le {\min{\{\|g_1\|+\|h_2\|,\|g_2\|+\|h_1\|\}} \cdot \\ \max{\{\|g_1-g_2\|,\|h_1-h_2\|\}}\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}}.} \end{align}$$
Hence, were it not for the requirement that $H$ be open, $H$ could contain $G\cap B(X\times Y,U)$.
Note 3: A sufficient assumption.
Suppose that $g_1,g_2,h_1,h_2\in G$.
Define $f:X\times Y\times Y\rightarrow U$ by:
$$f(x,y,z)=g_1(h_1(x,y),z)-g_1(h_2(x,y),z)$$
for all $x\in X$, $y,z\in Y$. So for all $x\in X$, $y,z\in Y$:
$$\|f(x,y,z)\|\le \|g_1\|\|(h_1-h_2)(x,y)\|\le \|g_1\|\|h_1-h_2\|.$$
Assumption: Suppose that there exists $K>0$ such that for all $x_1,x_2\in X$ and all $y_1,y_2,z\in Y$:
$$\|f(x_1,y_1,z)-f(x_2,y_2,z)\|\le K \|g_1\|\|h_1-h_2\|\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}}.$$
Then as for all $x_1,x_2\in X$ and all $y_1,y_2,z\in Y$:
$$\begin{align} & g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z) \\ & = (g_1(h_1(x_1,y_1),z)-g_1(h_1(x_2,y_2),z))-(g_2(h_2(x_1,y_1),z)-g_2(h_2(x_2,y_2),z)) \\ & = {(g_1(h_1(x_1,y_1),z)-g_1(h_1(x_2,y_2),z))-(g_1(h_2(x_1,y_1),z)-g_1(h_2(x_2,y_2),z)) \\ +(g_1(h_2(x_1,y_1),z)-g_1(h_2(x_2,y_2),z))-(g_2(h_2(x_1,y_1),z)-g_2(h_2(x_2,y_2),z))} \\ & = {(g_1(h_1(x_1,y_1),z)-g_1(h_2(x_1,y_1),z))-(g_1(h_1(x_2,y_2),z)-g_1(h_2(x_2,y_2),z)) \\ +((g_1-g_2)(h_2(x_1,y_1),z)-(g_1-g_2)(h_2(x_2,y_2),z))} \\ & = {(f(x_1,y_1,z)-f(x_2,y_2,z)) + ((g_1-g_2)(h_2(x_1,y_1),z)-(g_1-g_2)(h_2(x_2,y_2),z))}, \end{align}$$
we have that:
$$\begin{align} & \|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\| \\ & \le \|f(x_1,y_1,z)-f(x_2,y_2,z)\| + \|(g_1-g_2)(h_2(x_1,y_1),z)-(g_1-g_2)(h_2(x_2,y_2),z)\| \\ & \le [K\|g_1\|\|h_1-h_2\| + \|g_1-g_2\|\|h_2\|]\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}} \\ & \le [K\|g_1\| + \|h_2\|]\max{\{\|g_1-g_2\|,\|h_1-h_2\|\}}\max{\{\|x_1-x_2\|,\|y_1-y_2\|\}}, \end{align}$$
as required.
However, our assumption is false for general $g_1,h_1,h_2\in G$.
To see this, take $h_1(x,y)=x-c_1$, $h_2(x,y)=x-c_2$, $g_1(x,y)=\max{\{0,x\}}$ for all $x\in X$, $y\in Y$. Then $\|h_1-h_2\|=|c_1-c_2|$, but $\|f(x_1,y_1,z)-f(x_2,y_2,z)\|=\|x_1-x_2\|$ for $x_1,x_2\in(\min{\{c_1,c_2\}},\max{\{c_1,c_2\}})$.
Note 4: The answer to the final question appears to be "no" in general.
Much as before, take $h_1(x,y)=x-c_1$, $h_2(x,y)=x-c_2$, $g_1(x,y)=g_2(x,y)=\max{\{0,x\}}$ for all $x\in X$, $y\in Y$. Then $\max{\{\|g_1-g_2\|,\|h_1-h_2\|\}}=|c_1-c_2|$, but $\|g_1(h_1(x_1,y_1),z)+g_2(h_2(x_2,y_2),z)-g_1(h_1(x_2,y_2),z)-g_2(h_2(x_1,y_1),z)\|=\|x_1-x_2\|$ for $x_1,x_2\in(\min{\{c_1,c_2\}},\max{\{c_1,c_2\}})$.
Note 5: This question relates to my previous question here: "Doubly Lipschitz" functions
The older question can be viewed as a simplification of this one, but does not directly give an answer. (I was hoping for inspiration for the harder question. I now think I'm better off just asking the real question!)