I am self-studying Convex Optimization by Boyd and Vandenberghe. I am currently at page 111 where the author states:
If $g : R^n → R^p$ is K-convex, $h : R^p → R$ is convex, and $\tilde{h}$ (the extended-value extension of h) is K-nondecreasing, then $ h \circ g$ is convex. This generalizes the fact that a nondecreasing convex function of a convex function is convex. The condition that $\tilde{h}$ be K-nondecreasing implies that dom h − K = dom h.
I don't really get the last part. Can someone explain why dom h and K must be disjoint?
If I think of a really straightforward example:
$g(x_1,...,x_{n-1},x_n) = [x_1,...,x_{n-1}]$, where ${\bf dom }\:{\it g} = R^{n}$ and
$h(x_1,...,x_{n-2},x_{n-1}) = x_1$, where ${\bf dom }\:{\it h} = R^{n-1}$.
Let the cone be $K = R^{n-1}_{+}$, the $n-1$ dimensional nonnegathive orthant. Then g is K-convex, h is convex and K-nondeacreasing, so $ h \circ g$ is convex, but ${\bf dom }\:{\it h} − {\it K} \neq {\bf dom }\:{h}$.