Consider claim:

> The feasible region {x | f(x) = 0} is nonconvex no matter whether f is convex.

For some positive integer n and for the set of real numbers R, consider closed (in the usual topology for R^n), convex set A a subset of R^n. Define function f: R^n --> R so that f(x) = 0 for all x in A, f(x) > 0 for all x not in A, and for all x in R^n f infinitely differentiable at x. It is a theorem that such an f exists.

Then { x | f(x) = 0 } = A and is both closed and convex in contradiction to the claim.

I'm assuming you're referring to nonlinear f(x) because this statement is trivially false for linear f(x).

But consider the function f(x) = max(0, g(x) - c) where the following holds:

* g(x) is nonlinear, positive definite in x, and convex.

* c > 0

Then f(x) is nonlinear and convex (it's the pointwise maximum of convex functions), and the set {x : f(x) = 0} is a convex set.