Is the convex function of a DC function convex ? Or is it DC?

Question

Suppose function $g(x)$ is convex in $x$ and $h(x)$ is DC in $x$ (i.e., is the Difference of two Convex functions), is the composite fuction $H(x):=g(h(x))$ convex in $x$? Here, DC is short for difference of convex functions, that is, $h(x)$ can be expressed as, \begin{equation} h(x) = h_1(x) - h_2(x), \end{equation} where, $h_1(x)$ and $h_2(x)$ are convex in $x$.

Answer 1

In general, the composition $H$ is not convex. As a counterexample, let $D=[2,3]\subset\mathbb{R}$. Let $h_1,h_2:D\to\mathbb{R}$ be given by

$$ h_1(x)=\frac{1}{x}\quad\text{and}\quad h_2(x)=x^2. $$

Then $h\equiv h_1-h_2$ is d.c. Let $g:D\to\mathbb{R}$ be given by

$$ g(x)=\frac{1}{x}. $$

The function $g$ is convex on $D$, but the function

$$ H(x)=g(h(x))=\frac{1}{\frac{1}{x}-x^2} $$

is concave on $D$.

In general, the composition is d.c., under some mild assumptions. See Theorem H in this paper.

Edit

As Kavi Rama Murthy notes in the comments, to disprove that $H$ is not convex in general, you could take any two convex functions $h_1,h_2$ such that $h_1-h_2$ is not convex (e.g. $h(x)=0-x^2$) then compose with $g(x)=x$ (the identity). This is algebraically much simpler than my counterexample above.