[Bimos]

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

[graycat]

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.

[ubj]

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.