dual feasibility of Kuhn-Tucker condition?

734 Views Asked by At

minimize $f(x)$

subject to \begin{align} f_i(x) & \le 0, \quad i \in \left\{ 1,\ldots,m \right\} \\ h_i(x) & = 0, \quad i \in \left\{ 1,\ldots,p \right\} \end{align}

Then the Lagrange function is given by $$L(x,\lambda ,v) = f(x) + \sum \lambda _i f_i(x) + \sum v_i h_i(x)$$ and $$g\left( {\lambda ,v} \right) = \inf L\left( {x,\lambda ,v} \right)$$ where $g$ is the lagrange dual function,

and for any $v$ and any $\lambda\ge0$ then we have $g(\lambda ,v) \le {p^*}$


This was what I found from the wikipedia for the nonlinear optimization technique. Everything was understandable except the last line saying that $\lambda\ge0$.

so my question is that why we have to set lambda to be non-negative for inequality constraint while the sign of lagrangian multiplier does not matter for equality constraints