The problem is $$ (P) \hspace{1cm} \begin{array}{ll}\min & e^{-x} \\ \text{s.t.} & \frac{x^2}{y} \leq 0 \end{array} $$ over the domain $\mathcal{D}= \{(x,y)| y>0\}$, which I tried to rewrite as $$ (\tilde{P}) \hspace{1cm} \begin{array}{ll}\min & e^{-x} \\ \text{s.t.} & \frac{x^2}{y} \leq 0 \\ & -y<0 \end{array} $$ So the Lagrangian is $\mathcal{L}(x,y,\lambda,\mu) = e^{-x} + \lambda \frac{x^2}{y} - \mu y$
To get the dual problem I tried to find $(\lambda,\mu)$ such as $$ q(\lambda, \mu) = \inf_{x,y} \mathcal{L}(x,y,\lambda,\mu) \neq -\infty $$ But I cant seem to find any because
- $\lambda \geq 0 , \mu \geq 0$, by taking $x=0,y \to \infty$, $q(\lambda, \mu) = -\infty$
- $\lambda \geq 0 , \mu < 0$, by taking $y<0, |y|<|x|,x \to \infty$, $q(\lambda, \mu) = -\infty$
- $\lambda < 0$, by taking $y=1,x \to \infty$, $q(\lambda, \mu) = -\infty$
My intuition is that on the second case I shouldn't be able to take $y<0$ 'cause is off the domain $\mathcal{D}$ at $(P)$ but since I put that as a restriction it shouln't be a constraint over $y$ when looking at $q$ for $(\tilde{P})$ (the definition of $q$ is the infimum of $\mathcal{L}$ over the points on the domain of the restrictions).
This problem is an exercise on Boyd's Convex Optimization (5.21) where the solution is obtained from the problem $(P)$ and I asume they consider the domain $D$ as a restriction over the infimum (it's not specified).
So are $(P)$ and $(\tilde{P})$ not equivalente problems? or am I missing something?
Thanks in advance for your help.
Hint:At first place you can ignore $y$ it dose not play role in problem P. Secondly: You need only consider cases $\mu , \lambda \geq 0$ and what if $ \mu = 0 $ ?
$D$ is an open set and you do not need incorporate it as a sign inequality constraint. Even more the standard definition of Lagrange Duality does not involve any strict inequality.