Kuhn Tucker condition is sufficient for a global optimum?

Question

$L$ is the variable and $s,r$ are parameters. The question asks to solve $max_{L\geq0}rf(L)-wL$ where $f(L)$ is twice continuously differentiable, strictly increasing and strictly concave. Then how can I show that the Kuhn Tucker condition is sufficient for a global optimum?

Answer 1

There are neccesary and sufficient conditions. Let $g(L)=rf(L)-wL$ and $h(L)=L$. These are concave $C^1$ functions. Also $\exists \hat{L}$ such that $h(\hat{L})>0$ (Slater condition). These conditions are part of those needed for the KT (sufficiency) theorem to hold. But note that if the parameters $w$ and $r$ are such that $g(L)>0$ for $L>0$ then it must be that at the optimum $L>0$ and the restriction is not active. In that case the problem is an unrestricted one. Since $g(L)$ is strictly concave and the domain is a convex set, then every local maximum is a global maximum.$\Box$

Answer 2

KKT conditions are just necessary conditions for optimality, maximum or minimum. If your objective function is convex (for a minimum)/concave (for a maximum) and also it is your domain, KKT becomes a sufficient and necessary condition. So you only have to proof that:

• The domain $L\ge 0$ is convex/concave (it is)
• The function $r f(L)-wL$ is convex/concave and $C^1$

To proof the second one is easy. First $r f(L)-wL$ is obviusly $C^2$ because the derivative operator is linear and $wL$ is $C^\infty$. Second, to test if your objective function is convave we test $$ r f(\lambda L_0+(\lambda -1)L_1)-w(\lambda L_0+(\lambda -1)L_1) \le \lambda \Bigg[rf(L_0)- wL_0\Bigg]+(\lambda -1)\Bigg[rf(L_1)- wL_1\Bigg] $$ $$ f(\lambda L_0+(\lambda -1)L_1) \le \lambda f(L_0)+(\lambda -1)f(L_1) $$ which is true because $f(L)$ is concave.