Is the example for convex programming problem (convex program) where Karush-Kuhn-Tucker conditions are met for every $\lambda = (\lambda_0, … ,\lambda_m)$, but the $\lambda_0 = 0$?
X – linear space. $f_i: X \rightarrow \mathbb{R}\cup \{+\infty\}$ – convex functions, $i \in \{0,m\}$
Convex programming problem:
$$ \cases{f_0(x) \rightarrow inf \\
f_i(x)\leq 0, i \in \{1,m\}}$$
For $\lambda = (\lambda_0, … ,\lambda_m)$ let $L(x, \lambda) = \sum f_i(x)\lambda_i$
$\dot{x} – solution$
Karush-Kuhn-Tucker conditions:
- $ \displaystyle\min_{x \in X} L(x, \lambda) = L(\dot{x}, \lambda)$
- $ \lambda_i \geq 0, \quad i \in \{0,m\}$
- $ \lambda_i f_i(\dot{x}) = 0, \quad i \in \{1,m\}$
Best Answer
Take $f_0(x)=x$ and $f_1(x) = x^2$. Then $\dot x =0$ is the only feasible point. The Lagrange-function $$ L(x,\lambda) = \lambda_0x + \lambda_1 x^2 $$ has no minima if $\lambda_1=0$, $\lambda_0\ne0$. It has a minimum at $x=-\frac{\lambda_0}{2\lambda_1}$ if $\lambda_1>0$. Hence $$ L(\dot x,\lambda ) = \inf_x L(x,\lambda) $$ is true if and only if $\lambda_0=0$ and $\lambda_1>0$.