As far as I know, the dual problem is defined as
$g(\lambda,\nu)=\inf_{x\in\mathbb{R}^n}L(x,\lambda,\nu)$ subject to $\lambda\geq0$
where $\lambda$ is corresponding to the inequality constraints and $\nu$ is corresponding to the equality constraints.
Why in some cases when converting primal to dual problem, we will have other constraints more than $\lambda\geq0$ ?
Thank you so much
The dual problem is $\max_{\lambda} \{ g(\lambda) : \lambda \geq 0\}$. If $g(\lambda) = -\infty$, we say that $\lambda$ is infeasible for the dual. Therefore, the constraints come from when $g(\lambda) = -\infty$.
As an example, consider $\min_x \{ c^Tx : Ax\geq b, x \geq 0\}$. The Lagrangian is $L(x,y)=c^Tx + y^T(b-Ax)$ and $g(y) = \min_{x \geq 0} L(x,y) = \min_{x \geq 0} b^Ty + x^T(c-A^Ty)$. If some component of $c-A^Ty$ is negative, then $g(y)=-\infty$ (by letting the corresponding $x_i \to \infty$). Therefore, $c-A^Ty \geq 0$ is a constraint in the dual problem. If $c-A^Ty \geq 0$, $g(y) = b^Ty$ ($x=0$ is optimal), so the dual is $\max_y \{ b^Ty : A^Ty \leq c, y \geq 0\}$.