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I am studying the meaning of the multiplier in Lagrange Multiplier Method and got the following question. I tried it myself, but I am not sure if it is correct. I would really appreciate it if someone could help me check. In order to state the question, we need some background information first:

Background Information

Theorem$\space\space\space\space$ Let $f$ and $h$ be $C^1$ functions of two variables. For any field value of the parameter $a$, let $(x^*(a), y^*(a))$ be the solution of the following maximization problem \begin{equation} Max\space\space\space\space f(x, y)\\ \space\space\space\space\space\space\space\space\space s.t.\space\space\space\space h(x, y) = a \end{equation} with corresponding multiplier $\mu^*(a)$. Suppose that $x^*$, $y^*$, and $\mu^*$ are $C^1$ functions of $a$ and that NDCQ holds at $(x^*(a), y^*(a), \mu^*(a))$. Then, \begin{equation} \mu^*(a) = \frac{d}{da}f(x^*(a),y^*(a)). \end{equation}

Question

Use the Implicit Function Theorem to write out a specific inequality which would guarantee the validity of the assumption in the above theorem that the solution $(x(a), y(a))$ of the maximization problem depends smoothly on $a$.

My Attempt

Consider the maximization problem as in the above theorem: \begin{equation} Max\space\space\space\space f(x, y)\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space s.t.\space\space\space\space h^*(x, y; a) \equiv h(x, y) - a = 0. \end{equation} Assume that the constraint qualification holds at $(x^*(a), y^*(a))$, the solution to the maximization problem: \begin{equation} \nabla h^*(x, y; a) = \begin{pmatrix} \frac{\partial h^*}{\partial x}(x^*(a), y^*(a); a)\\ \frac{\partial h^*}{\partial y}(x^*(a), y^*(a); a). \end{pmatrix} \end{equation} Form the Lagrangian: \begin{equation} L(x, y, \mu; a) = f(x, y, ;a) - \mu h^*(x, y; a). \end{equation} The constrained maximizer $(x^*(a), y^*(a))$ must satisfy the F.O.C.s: \begin{equation} \frac{\partial L}{\partial x}(x, y, \mu; a) = 0\\ \frac{\partial L}{\partial y}(x, y, \mu; a) = 0\\ \frac{\partial L}{\partial \mu}(x, y, \mu; a) = 0, \end{equation} a system of 3 equations in 3 unknowns.
By the Implicit Function Theorem, we need the matrix \begin{equation} \begin{pmatrix} \frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial x \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial y^2} & \frac{\partial^2 L}{\partial y \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial \mu} & \frac{\partial^2 L}{\partial y \partial \mu} & \frac{\partial^2 L}{\partial \mu^2} \end{pmatrix} = \begin{pmatrix} \frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial x \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial y^2} & \frac{\partial^2 L}{\partial y \partial \mu}\\ \frac{\partial^2 L}{\partial x \partial \mu} & \frac{\partial^2 L}{\partial y \partial \mu} & 0 \end{pmatrix}, \end{equation} which is the Hessian matrix $D^2L$ of the Lagrangian, to be nonsingular. Therefore, the inequality we are looking for is that \begin{equation} det(D^2L) \neq 0. \end{equation}

Could someone please help me check if this is correct? Thanks a lot for any help!

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  • $\begingroup$ What is NDCQ? ${}{}$ $\endgroup$
    – FShrike
    Aug 19, 2023 at 22:58
  • $\begingroup$ @AShrike NDCQ is Non-Degenerate Constraint Qualification. Please refer to this link "math.stackexchange.com/questions/4754105/…" for details. (The NDCQ is stated in a Theorem in the Background Information.) $\endgroup$
    – Beerus
    Aug 19, 2023 at 23:38
  • $\begingroup$ No, NDCQ is equivalent to LICQ, and It does not mean that the Hessian of the Lagrangian is nonsingular. The Implicit function theorem allows us to see that $(x(a),y(a))$ is differentiable. So, it is only necessary to use the Chain Rule to differentiate the objective function $f(x(a),y(a))$ $\endgroup$ Aug 26, 2023 at 10:17

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