Is a KKT point a local optimum for strictly convex objective functions and non-convex constraints?

Question

Consider the problem \begin{equation} \min_x f(x)~~~{\rm s.t.}~~~ g_i(x)\leq 0,~~i=1,\dots,I, \end{equation}

where $x$ is the optimization parameter vector, $f(x)$ is the objective function and $g_i(x)$ is the $i$th constraint function. $I$ denotes the number of constraints.

Consider now a point $x^\star$ that satisfies the Karush-Kuhn-Tucker conditions. In general, in non-convex optimization, a KKT point can be everything, i.e., a local optimum, a global optimum, a saddle point or even a maximum.

Can we claim that $x^\star$ is at least a local optimum if we assume that $f(x)$ is a strictly convex function and that $g_i(x)$ are non-convex functions? All functions are at least twice continuously differentiable.

I have read this claim in this paper: http://kang.nt.e-technik.tu-darmstadt.de/nt/fileadmin/nas/Publications/2012/Cheng_TSP_2012_Posted.pdf [Appendix C] but I'm not 100% sure if this is true. I also could not find any literature about it.

In the paper, do the authors give any proof or references? Can you tell me the name of the paper? — Alex Silva, Dec 04 '14 at 09:57
Do you mean [24]? Yes I did that but I could not really verify the result. — The Pheromone Kid, Dec 04 '14 at 11:34
I was looking for a counter example but gerw did that nicely. :) — Alex Silva, Dec 04 '14 at 12:40

score 4 · Accepted Answer · answered Dec 04 '14 at 11:50

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Consider the following problem: \begin{align*} \min \quad & x_1^2 + \frac12 x_2^2, \\ \text{such that} \quad & -x_1^2 - x_2^2 \le -1. \end{align*} The objective is strictly convex, but the constraint is strictly concave.

It is easy to check that $x = (0,1)$ is the global minimizer, and it should also be the only local minimizer. The point $x = (1,0)$ is, however, a KKT point with multiplier $\mu = 1$. But it is not a local minimizer.

answered Dec 04 '14 at 11:50

gerw

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Thanks a lot! Some times is easier to find a counter example than proving sth that is not right ;-). – The Pheromone Kid Dec 04 '14 at 12:14

score 0 · Answer 2 · answered Dec 04 '14 at 09:17

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If your point $x^*$ is at least a local minimum, then the KKT conditions are satisfied for some KKT multipliers if the local minimum, $x^*$, satisfies some regulatory conditions called constraint qualifications. E.g. if the constraint gradients be linearly independence (the LICQ condition).

There are many constraint qualifications, and some may apply to your constraints.

answered Dec 04 '14 at 09:17

Tommy L

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You are right in every word but I am assuming that $x^\star$ is a KKT point, not a local optimum. My question is: Is a KKT pointer under the formulated conditions also a local optimum? – The Pheromone Kid Dec 04 '14 at 09:21
Oh, ok. In the general case this is not so, because the KKT conditions are necessary, but not sufficient. For the KKT conditions to be sufficient, the constraints must also be smooth and convex. It may still be so for your functions, though. – Tommy L Dec 04 '14 at 09:27

score 0 · Answer 3 · answered Apr 07 '16 at 18:43

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Besides satisfying the KKT condition, SOSC is also required. Reference: Theorem 1 in http://arxiv.org/pdf/1106.0898.pdf

answered Apr 07 '16 at 18:43

whatever

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Please make an effort to give Readers a fuller explanation of what "SOSC" means in this context. Quoting or paraphrasing (in complete sentences) the most relevant part of a referenced (linked) explanation (with proper credit) is apprpriate. – hardmath Apr 07 '16 at 19:06

Is a KKT point a local optimum for strictly convex objective functions and non-convex constraints?

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