# KKT conditions strict inequality constraints

Some people asked questions about KKT conditions with strict inequality constraints, such as

Kuhn Tucker conditions with strict inequality constraints?

Questions about constraints and KKT conditions

https://www.researchgate.net/post/Is_KKT_condition_applicable_for_optimality_when_the_constraints_is_strictly_less_than_type

Let us see an example.

$\min\quad \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a}$

$\mathrm{s.t.}\quad a>0,\ b>0,\ c>0,\ d>0,\ a^4+b^4+c^4+d^4 = 4.$

The minimum $4$ is attained at $a=b=c=d=1$. The constraints $a,b,c,d>0$ are strict inequalities. How to deal with them in KKT conditions?

Just as some people said (e.g., the 3rd link above), we simply ignore the strict inequality constraints and use KKT conditions. If the minimum is attainable (that is, min not inf), the solution will satisfy the strict inequalities. For this example, it is the Lagrange multiplier method $L = \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} + \lambda (a^4+b^4+c^4+d^4 - 4)$.

I found the lecture note:

https://ocw.mit.edu/courses/sloan-school-of-management/15-084j-nonlinear-programming-spring-2004/lecture-notes/lec6_constr_opt.pdf

In the lecture note (2nd page), the following problem is considered:

$\min\quad f(x)$

$\mathrm{s.t.}\quad h(x) = 0,\ g(x) \le 0, \ x \in X.$

Here $X$ is an open set. Then KKT conditions are used without including $x\in X$.

I want to make sure if this is the case. Thanks.

• That's what I mean. I read the book "Nonlinear programming" by Dimitri P. Bertsekas, but the problem including an open set $x\in X$ as constraint is not considered. So I need more textbooks etc. to support this. By the way, in some case, we may replace the strict inequality by the "less than and equal to" and use KKT conditions as usual then choose the desired. Aug 15, 2018 at 14:53
A counterexample. Convexity alone is not enough to guarantee strong duality. Consider for example the convex problem $$\min _{x, y>0} e^{-x}: x^{2} / y \leq 0,$$ with variables $$x$$ and $$y$$, and domain $$\mathcal{D}=\{(x, y) \mid y>0\}$$. We have $$p^{*}=1$$. The Lagrangian is $$L(x, y, \lambda)=e^{-x}+\lambda x^{2} / y$$, and the dual function is $$g(\lambda)=\inf _{x, y>0}\left(e^{-x}+\lambda x^{2} / y\right)= \begin{cases}0 & \lambda \geq 0 \\ -\infty & \lambda<0\end{cases}$$ so we can write the dual problem as $$d^{*}=\max _{\lambda} 0: \lambda \geq 0$$ with optimal value $$d^{\star}=0$$. The optimal duality gap is $$p^{\star}-d^{\star}=1$$. In this problem, Slater's condition is not satisfied, since $$x=0$$ for any feasible pair $$(x, y)$$.
• Thanks for your answer. (+1) In this example, if we consider $L = e^{-x} + \lambda x^2/y$, the equations $\frac{\partial L}{\partial x} = \frac{\partial L}{\partial y} = 0$ have no solution. Apr 4 at 8:00