# The perturbation of a convex function can also be convex?

$$W^{1,\infty}(D)\ni f:D\to\mathbb R, (x,y)\mapsto f(x,y)$$, 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 $$(x,y)\in D'$$, $$\forall|\epsilon|, and for some $$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?

• (1) What do you mean by strictly monotonic? (2) What is $L$-continuous? (3) I do not understand how the function $g_c$ is defined. $g_c$ seems to be a function of one variable, but a statement that it is locally convex in $D$ suggests it is a function of two variables. Please edit the question. – Piotr Hajlasz Oct 30 '18 at 13:18
• @PiotrHajlasz Thanks for your commments and I made the edits – High GPA Oct 31 '18 at 9:02
• math.stackexchange.com/questions/2975837/… – Michael Greinecker Oct 31 '18 at 23:55