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The feasible region {x | f(x) = 0} is nonconvex no matter whether f is convex.
by Bimos 3y ago
The feasible region {x | f(x) = 0} is nonconvex no matter whether f is convex.
- graycat 3y agoConsider 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 3y agoI'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.