I am posing this question in conjunction with the one posted here. Let $F:\mathbb{R}^{n}\rightarrow \mathbb{R}$ be a differentiable function, and let $L$ denote the following subset of $\mathbb{R}^{n}$: \begin{eqnarray} L=\{x\in \mathbb{R}^{n}:a^{T}x=b\} \end{eqnarray} for some nonzero vector $a\in \mathbb{R}^{n}$ and a scalar $b\in \mathbb{R}$.
It is known that $F$ attains a unique minimum in $\mathbb{R}^{n}$, i.e., there exists a unique point $x^{0}\in \mathbb{R}^{n}$ where $F$ attains a minimum. Further, it is also known that $F$ attains a unique minimum in $L$, i.e., there exists a unique point $x^{*}\in L$ where $F$ attains a minimum in $L$. In addition, it is known that $x^{0}\notin L$.
From Lagrange's theorem, we know that there exists a scalar $\mu^{*}\in \mathbb{R}$ such that the Lagrangian \begin{eqnarray} l(x,\mu)=F(x)+\mu(a^{T}x-b) \end{eqnarray} satisfies the following conditions:
- $\nabla l(x^{*},\mu^{*})=\textbf{0}$
- For all $y\in \mathbb{R}^{n}$ such that $a^{T}y=0$, we have $y^{T}\nabla^{2}l(x^{*},\mu^{*})y\geq 0$.
Since $x^{0}$ is a point of minimum of $F$ in $\mathbb{R}^{n}$, clearly, the following are true:
- $\nabla F(x^{0})=\textbf{0}$
- For all $y\in \mathbb{R}^{n}$, we have $y^{T}\nabla^{2}F(x^{0})y\geq 0$.
In the above mentioned conditions, $\nabla$ stands for the partial derivative operator wrt $x$.
I am interested to know if it is possible to argue that $\mu^{*}\neq 0$. Suppose that $\mu^{*}=0$. Then, the conditions 3 and 4 mentioned above translate to saying that $x^{0}$ is a point of minimum of $l(x,\mu^{*})$.
This necessitates the question of whether or not it is true that the Lagrangian cannot have two or more points of minima. I'd be happy to have this clarified.
Thanks in advance.