I'm studying convex optimization and had a question regarding dual constraints while solving an exercise. The problem is as follows:
$$\begin{array}{ll} \text{minimize} & x^2 + 1\\ \text{subject to} & (x-2)(x-4)\le0\end{array}$$
with variable $x\in\Bbb{R}$.
Find the Lagrangian and Lagrange dual function $g$. Verify the lower bound property (i.e. $p^* \ge \text{inf}_x(L(x, \lambda))$.
The problem is relatively easy but I was confused about a small detail. My approach is as follows:
$$ \begin{align} L(x, \lambda) & = f_0(x) + \lambda f_1(x) \\ & = (x^2 + 1) + \lambda(x^2-6x+8) \\ & = (1 + \lambda)x^2 - 6\lambda x + (1 + 8\lambda) \end{align} $$
and so, $$\nabla_x L(x, \lambda) = 2(1+\lambda)x - 6\lambda$$ which gives us an optimal value:
$$\tilde{x} = \frac{3\lambda}{1 + \lambda}$$
We can use this to get the Lagrange dual function:
$$ \begin{align} g(\lambda) & = \text{inf}_x(L(x, \lambda)) \\ & = L(3\lambda / (1 + \lambda), \lambda) \\ & = -\frac{9\lambda^2}{1 + \lambda}+ (1 + 8\lambda) \end{align} $$
So for the Lagrange dual function I write the final form as:
$$ g(\lambda) = \begin{cases} \begin{align} -\frac{9\lambda^2}{1 + \lambda} + (1 + 8\lambda)\quad & (\lambda \ge 0)\\ -\infty \quad \quad \quad \ \quad & \text{otherwise} \end{align} \end{cases} $$
But apparently according to the solution manual the condition in the dual function is not $\lambda \ge 0$, but $\lambda \gt -1$.
How were the conditions for this dual function came to? Did I miss something in the process?
Thank you.
