Question About Fritz John Theorem and Slater Constraint Qualification

Question

Background Information

I am studying constraint qualifications. Here are two theorems leading to my question:

Theorem 1$\space\space\space\space$ [Fritz John Theorem] Suppose that $f, g_1, \dots, g_k$ are $C^1$ functions of $n$ variables. Suppose that $x^*$ is a local maximizer of $f$ on the constraint set defined by the $k$ inequalities \begin{equation} g_1(x_1, \dots, x_n) \leq b_1, \dots, g_k(x_1, \dots, x_n) \leq b_k. \end{equation} Form the Lagrangian \begin{equation} L(x_1, \dots, x_n, \lambda_0, \lambda_1, \dots, \lambda_k) \equiv \lambda_0f(\mathbf{x}) - \lambda_1[g_1{\mathbf{x}} - b_1] - \dots - \lambda_k[g_k{\mathbf{x}} - b_k], \end{equation} with a multiplier $\lambda_0$ for the objective function. Then, there exist multipliers $\lambda^* = (\lambda_0^*, \lambda_1^*, \dots, \lambda_k)$ such that:

(a) $\frac{\partial L}{\partial x_1}(\mathbf{x}^*, \lambda^*) = 0, \dots, \frac{\partial L}{\partial x_n}(\mathbf{x}^*, \lambda^*) = 0$,
(b) $\lambda_1^*[g_1(\mathbf{x}^*) - b_1] = 0, \dots, \lambda_k^*[g_k(\mathbf{x}^*) - b_k] = 0$,
(c) $\lambda_1^* \geq 0, \dots, \lambda_k^* \geq 0$,
(d) $g_1(\mathbf{x}^*) \leq b_1, \dots, g_k(\mathbf{x}^*) \leq b_k$,
(e) $\lambda_0^* = 0$ or $1$, and
(f) $(\lambda_0^*, \lambda_1^*, \dots, \lambda_k^*) \neq (0, 0, \dots, 0)$.

Theorem 2$\space\space\space\space$ Let $f, g_1, \dots, g_k$ be as in Theorem 1, and suppose that $\mathbf{x}^* \in \mathbb{R}^n$ is a local maximizer of $f$ on the constraint set defined by \begin{equation} g_1(\mathbf{x}) \leq b_1, \dots, g_k(\mathbf{x}) \leq b_k. \end{equation} For ease of notation, suppose that $g_1, \dots, g_h$ yield binding constraints at $\mathbf{x}^*$ and that $g_{h+1}, \dots, g_k$ are not binding at $\mathbf{x}^*$. Suppose that the binding constraint functions satisfy one of the following properties:

(a) (Slater Constraint Qualification) There is a ball $U$ about $\mathbf{x}^*$ in $\mathbb{R}^n$ such that $g_1, \dots, g_h$ are convex functions on $U$ and there exists $\mathbf{z} \in U$ so that each $g_i(\mathbf{z} < b_i)$.
(b) $g_1, \dots, g_h$ are concave functions.
(c) $g_1, \dots, g_h$ are linear functions.

Then we can take $\lambda_0^* = 1$ in the conclusion of Theorem 1.

Problem

I was asked to determine which of the three constraint qualifications in Theorem 2 hold for the constraint functions in the following maximization problem: \begin{equation} Max\space\space\space\space f(x, y, z) = xyz + z\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space s.t.\space\space\space\space g_1(x, y, z) \equiv x^2 + y^2 + z \leq 6\\ \space\space\space\space\space\space\space\space\space\space\space\space\space g_2(x, y, z) \equiv -x \leq 0\\ \space\space\space\space\space\space\space\space\space\space\space\space\space g_3(x, y, z) \equiv -y \leq 0\\ \space\space\space\space\space\space\space\space\space\space\space\space\space g_4(x, y, z) \equiv -z \leq 0 \end{equation}

My Question

The answer to this problem is (a). I was very confused about it. One can check that there are two local maximizer of the $f$ on the constraint set:
(1) $x = 0$, $y = 0$, and $z = 6$, and
(2) $x = 1$, $y = 1$, and $z = 4$.
In case (1), the binding constraint functions would be $g_1$, $g_2$, and $g_3$. In case (2), the binding constraint function would be $g_1$ only. I cannot figure out why in each case, the binding constraint functions are convex functions on a ball $U$ about the local maximizer of $f$. Could someone please help me with it? Thanks a lot in advance!

Answer 1

All of the constraints are convex everywhere. Together, $g_1, g_4$ show that the feasible set is compact, so local maxima exist.

According to the theorem, since $(1,1,2)$ is strictly feasible for all constraints, the Slater CQ holds and so we can take $\lambda_0 = 1$. This holds for any local $\max$. In particular, you do not care, in this instance, which constraints are binding or not.

However, determining the local maxima requires more work as we need to determine which constraints are active to apply the optimality conditions.

Note that the cost and constraints are unchanged if we swap $x,y$, so, for simplicity, we can assume that $0 \le y \le x$ at a local $\max$.

Suppose $(x,y,z)$ is a local maximiser. We must have $x^2+y^2+z = 6$, otherwise we could increase the cost (by increasing $z$) which would contradict $(x,y,z)$ being a local maximiser.

Suppose $z=0$ (and so $x>0$). Then, for small positive $t$, $(x-t,y, t(2-t))$ is feasible with increased cost, hence another contradiction, and so $z>0$.

If $x=y=0$ (and so $z=6$), then $(t,t,6-2t^2)$ is feasible for small $t$ and the cost is $6+4t^2-2t^4$ which is greater than $6$ for small $t$, hence a contradiction. Hence we cannot have both $x=y=0$.

Finally, if $y=0$, then $(x,t,z-t^2)$ is feasible for small $t$ and the cost increases (again for small $t$) hence we have $x>0$.

In particular, at any local $\max$, the only active constraint is $g_2$. Theorem 1 shows that $yz=2 \lambda x$, $xz=2 \lambda y$ and $xy+1 = \lambda$. The first two show that $x=y$, and this gives $z=2x^2+2$, which when combined with the $g_2$ constraint shows that $z=4$.

Hence the only local maximiser is $(1,1,4)$.

$g_2$ is not concave (and hence not linear) and so only (a) of the CQs apply.