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.
Best Answer
It is something weird with the solution manual. The Lagrange multipliers for inequalities ($\lambda$ in this example) are non-negative, so your solution is correct. I would even answer simply as $$ g(\lambda)=-\frac{9\lambda^2}{1 + \lambda} + (1 + 8\lambda), \quad\lambda\ge 0. $$ Negative values of $\lambda$ are infeasible for the dual problem, therefore, $g$ does not need to be defined there.
P.S. It may worth to note in the solution that the Lagrange function is convex (for all $\lambda\ge 0$) in $x$, hence, $\nabla_xL=0$ gives indeed the minimum.