Consider the following optimization problem: \begin{align*} &\min_{x_1,x_2 \in \mathbb{R}}x_1x_2\\ &\text{Subject to } x_1^2 + x_2^2\le 1, x_1\ge 0, x_2 \ge 0\\ \end{align*} I have been tasked with find the optimal value of the dual problem here. The definition for the optimal value of the dual problem that I have been given is $$d^* = \max_{\vec{\lambda} \in \mathbb{R}^m, \vec{\lambda} \ge 0} g(\vec{\lambda})$$ where $$g(\vec{\lambda}) = \inf_{\vec{x}\in D}\bigg(x_1x_2 + \lambda_1 (x_1^2+x_2^2-1)-\lambda_2x_1-\lambda_3x_2\bigg)$$ and $D$ is the domain under consideration. Here, $D$ is $\mathbb{R}^2$.
NOTE: Normally $g$ is a function of $\lambda$ and $\mu$, but since there are no equality constraints, this is just a function of $\lambda$.
My issue here is I have no idea how to find $g(\vec{\lambda})$, much less how to find the maximum over all $\vec{\lambda} \in \mathbb{R}^3$. Are there any steps that I should consider here? Any help would be greatly appreciated.
Look for additional conditions on $\lambda$ such that $g$ is finite.
It is a little easier to do if we write $g$ as a quadratic.
Let $S= {1 \over 2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ (note the eigenvalues are $\pm {1 \over 2}$), then $g(\lambda) = \inf_x x^T (\lambda_1 I + S) x -(\lambda_2, \lambda_3)^T x -\lambda_1$.
Note that we need $\lambda_1 \ge {1 \over 2}$, as otherwise the $\inf$ is $-\infty$.
If $\lambda_1 = {1 \over 2}$, we can choose $x_2 = -x_1$, and then we can see that if $\lambda_2 \neq \lambda_3$, then the $\inf$ is $-\infty$. Hence if $\lambda_1 = {1 \over 2}$ we must have $\lambda_2 = \lambda_3$ in order to have a finite $g$.
If $\lambda_1 > {1 \over 2}$ then we see that the minimising $x$ satisfies $x = {1 \over 2} (\lambda_1 I + S)^{-1} \begin{bmatrix} \lambda_2 \\ \lambda_3 \end{bmatrix}$, and so $g(\lambda) = -{1 \over 4} \begin{bmatrix} \lambda_2 & \lambda_3 \end{bmatrix}(\lambda_1 I + S)^{-1} \begin{bmatrix} \lambda_2 \\ \lambda_3 \end{bmatrix} -\lambda_1 < -{1 \over 2}$.
If $\lambda_1 = {1 \over 2}$ and $\lambda_2 = \lambda_3$, then we can write $x^T (\lambda_1 I + S) x -(\lambda_2, \lambda_3)^T x = {1 \over 2} (e^T x)^2 - \lambda_2 (e^T x)$, where $e$ is the vector of ones. In particular, the $\min$ occurs when $e^T x = \lambda_2$ and hence $g(\lambda) = -{1 \over 2}\lambda_2^2 -{1 \over 2} \le - {1\over 2}$.
Hence $d^*=\sup_{\lambda \ge 0} g(\lambda) = - { 1\over 2}$.