Problem 5.21 in Boyd's Convex Optimization about strong duality failing Exercise 5.21 in Boyd & Vandenberghe asks us to study the problem of minimizing $e^{-x}$ subject to $x^2/y \leq 0$ with domain $\mathcal{D} = \{(x,y)\ |\ y > 0\}$. Now, $x^2/y \leq 0$ with $y > 0$ clearly forces $x = 0$, so the optimal value is $e^{-0} = 1$. The Lagrangian for this problem is
$$L(x, \lambda) = e^{-x} + \lambda \frac{x^2}{y}$$
and so we have the dual function
$$g(\lambda) = \inf_{x} \left( e^{-x} + \lambda \frac{x^2}{y} \right) $$
We have $g(0) = 0$, and for $\lambda > 0$ since the Lagrangian is convex in $x$ we can obtain the minimum by differentiation. I get the result
$$g(\lambda) = \frac{\lambda W\left( \frac{y}{2\lambda} \right) \left( W\left( \frac{y}{2\lambda} \right) + 2 \right)}{y}$$
which I have checked with Wolfram so I believe it's correct. Miraculously, Wolfram can compute this derivative and tells me
$$g'(\lambda) = \frac{W\left( \frac{y}{2\lambda} \right)^2}{y}$$
This is strictly positive for $\lambda > 0$, and $g(0) = 0$ is also clearly not the maximum, so (since theory tells us $g$ is concave) I expect to find the maximum as $\lim_{\lambda \to \infty} g(\lambda)$.
To compute that, I noted (by setting $u = y/(2\lambda)$ in what follows) that
$$\lim_{\lambda \to \infty} \lambda W\left( \frac{y}{2\lambda} \right) = \lim_{u \to 0} \frac{y}{2u} W(u) = \frac{y}{2}$$
since Wolfram tells me $\lim_{x \to 0} W(x)/x = 1$. Now since $ W \left( \frac{y}{2\lambda} \right) + 2$ goes to $2$ as $\lambda \to \infty$, it seems I get $$\lim_{\lambda \to \infty} g(\lambda) = 1!$$
This would vindicate strong duality, which wasn't supposed to hold. Furthermore Boyd asks me to compute the optimal solution to the dual problem, which doesn't seem to be attained for any finite value. Am I missing something here?
Note: There is a previous question about this exercise but it does not answer my question.
 A: The Lagrangian is
$$L(x, y, \lambda) = \mathrm{e}^{- x} + \lambda \frac{x^2}{y}.$$
The dual function is
$$g(\lambda) = \inf_{\mathcal{D}} \left(\mathrm{e}^{- x} + \lambda \frac{x^2}{y}\right)
 = \left\{\begin{array}{ll}
  0 & \lambda \ge 0\\
  -\infty & \lambda < 0.
 \end{array}\right.$$
where $\mathcal{D} = \{(x,y) : ~ y > 0\}$ is the domain.

Remarks: We may deal with $\inf_{\mathcal{D}} \left(\mathrm{e}^{- x} + \lambda \frac{x^2}{y}\right)$ in another way.
For $\lambda > 0$, we have
\begin{align*}
 \inf_{\mathcal{D}} \left(\mathrm{e}^{- x} + \lambda \frac{x^2}{y}\right)
 &= \inf_{y > 0} 
 \inf_{x} \left(\mathrm{e}^{- x} + \lambda \frac{x^2}{y}\right)\\
 &= \inf_{y > 0}  \frac{\lambda W\left( \frac{y}{2\lambda} \right) \left( W\left( \frac{y}{2\lambda} \right) + 2 \right)}{y}\\
 &= \lim_{y\to \infty}  \frac{\lambda W\left( \frac{y}{2\lambda} \right) \left( W\left( \frac{y}{2\lambda} \right) + 2 \right)}{y} \tag{1}\\
 &= \lim_{u\to \infty} \left[\frac{W(u)^2}{2u} + \frac{W(u)}{u}\right]\\
 &= 0 \tag{2}
\end{align*}
where we obtain (1) by using
$$\frac{\partial }{\partial y}\frac{\lambda W\left( \frac{y}{2\lambda} \right) \left( W\left( \frac{y}{2\lambda} \right) + 2 \right)}{y}
= - \frac{\lambda [W\left(\frac{y}{2\lambda}\right)]^2}{y^2} < 0,$$
and we obtain (2) by using
$\lim_{u \to \infty} \frac{W(u)}{u} = 0$
and $\lim_{u \to \infty} \frac{W(u)}{\sqrt{u}} = 0$
(using L'Hopital's rule and $W'(u) = \frac{1}{u}\, \frac{W(u)}{1 + W(u)}$).
