$ W^{1,\infty}(D)\ni f:D\to\mathbb R, (x,y)\mapsto f(x,y)$, is a strictly increasing on both dimensions (i.e. if $x_1>x_2$ then $f(x_1,y)>f(x_2,y)$), lipschitz continuous function defined on a convex set $D\subseteq\mathbb R^2$.
Univariate function $g_c:x\mapsto y $ s.t. $(x,y)\in D$ and $f(x,y)=c$. Given $g_c$ is locally convex on its domain and $c$ is a constant. What can we imply about the convexity of the function $g_{c+\epsilon}$ within $D$?
$g_{c+\epsilon}$ is also implicitly defined by $f(x,g_{c+\epsilon}(x))=c+\epsilon$
Can we say that, for any $f$, $\exists a>0$ s.t. $g_{c+\epsilon}$ is also convex for $(x,y)\in D'$, $\forall|\epsilon|<a$, and for some $D'\subseteq D$?
Intuitively I don't think this will always work for functions on $W^{1,\infty}$; but the perservation of quasi-convexity could work. So my other question is, what are the classifications of functions $f$ such that we can always find a set of functions $g_{c+\epsilon}$ is convex given $g_c$ is convex?
