# Problem calculating the Dual of a convex problem

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.

• The solution is $x=0$. Why are you trying to take the dual? Commented Jun 20, 2019 at 0:43
• 'cause is a pedagogical exercise to see a case when there's not strong duality. And also I was trying to undersand the procedure of the excercise itself which ask for 4 things (a) determine is a convex problem and find the optimal value. (b) compute the dual and find the optimal value of the dual problem. (c) Check that Slater's condition doesn't hold. (d) Study a penalized version of the problem. And I got stuck on part (b). Commented Jun 20, 2019 at 0:56
• Strict constraints ($y>0$ here) are a bit unusual. Commented Jun 20, 2019 at 1:23
• Do not add $y>0$. That is the domain of the function $f(x,y)=x^2/y$, so let it remain implicit to that constraint, and minimize over $y>0$ in the Lagrangian. Commented Jun 20, 2019 at 2:12
• That book has 2 authors. Commented Jun 20, 2019 at 21:38

## 1 Answer

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.

• but on this case, if i were to take $y<0$ and then $x\to \infty$, $q(\lambda, \mu=0) = -\infty$. or not? Commented Jun 20, 2019 at 0:11
• $e^{-x} + \lambda \frac{x^2}{y}$ is always positive term for $\lambda \geq 0$ Commented Jun 20, 2019 at 0:31
• then it would be $q(\lambda) = \inf_D e^{-x} + \lambda \frac{x^2}{y}$ as $\lambda \geq 0$? Commented Jun 20, 2019 at 0:58
• You explicitly calculate that by setting derivative = 0 Commented Jun 20, 2019 at 1:03