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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.

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    $\begingroup$ You've misunderstood the relationship between the constraints that have been handled in the Lagrangian and the domain $D$. Since the constraints that force $x$ to be in the upper right section of the unit disk have been incorporated into the Lagrangian they don't have to be incorporated into the domain. $\endgroup$ – Brian Borchers Oct 1 '18 at 3:03
  • $\begingroup$ I'm not entirely sure I follow. Do you mean that my interpretation of $D$ is incorrect? If so, what should it be? $\endgroup$ – BSplitter Oct 1 '18 at 3:06
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    $\begingroup$ Try $D=R^{2}$. Random text added for length... $\endgroup$ – Brian Borchers Oct 1 '18 at 3:08
  • $\begingroup$ Note Brian's comment: The $\inf$ should be over all $x$. $\endgroup$ – copper.hat Oct 1 '18 at 3:14
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    $\begingroup$ Find conditions under when the $\inf$ is finite. This will give the dual constraints. $\endgroup$ – copper.hat Oct 1 '18 at 3:16
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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 $d^*=\sup_{\lambda \ge 0} g(\lambda) = - { 1\over 2}$.

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  • $\begingroup$ I apologize for not responding sooner; I haven't had a chance to look at this problem until today. Correct me if I'm wrong, but in the $\lambda_1>\frac{1}{2}$ case, wouldn't $g(\lambda) = -\lambda_1$, since the first and second terms of $g(\lambda)$ would cancel out? I still agree with the conclusion. $\endgroup$ – BSplitter Oct 4 '18 at 2:25
  • $\begingroup$ @BSplitter: I think you might be missing a factor of ${1 \over 2}$ so it doesn't exactly cancel. (Of course, I am prone to making mistakes!) $\endgroup$ – copper.hat Oct 4 '18 at 2:27
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    $\begingroup$ I wrote it out slightly differently. I just called the matrix in question $M$, and its inverse $M^{-1}$. Then the minimizing $x$ would be $M^{-1} \begin{bmatrix}\lambda_2 \\ \lambda_3\end{bmatrix}$. Plugging that in allows some cancellation, I think. $\endgroup$ – BSplitter Oct 4 '18 at 2:29

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