Exercise
Solution:
Could someone please explain why the reached contradiction solves the exercise?
I can understand the solution but I don't know how does that implies that $\overline x$ is a KKT point.
Here is how I interprete the solution.
We suppose $u_0=0,$ hence if $\sum_{i\in I} u_i\nabla g_i(\overline x)=0$ has solution then it must be nonzero. As we found a contradiction, then do we must have that the system $\sum_{i\in I} u_i\nabla g_i(\overline x)=0$ has no solution?
or what could we say about it?
Thank you in advance.
$\bar x$ is FJ point. It means, in particular, that not all among $u_0,u_i$, $i\in I$, are zeros.
Assume $u_0=0$ then not all $u_i$, $i\in I$, are zeros. It leads to contradiction with Cottle's constraint qualification.
Since $u_0\ne 0$ we can divide the first FJ equation by $u_0$ to get $$ \nabla f(\bar x)+\sum_{i\in I}\Big(\frac{u_i}{u_0}\Big)\nabla g_i(\bar x)=0. $$ This is exactly the first KKT condition with the new coefficients $\tilde u_i=\frac{u_i}{u_0}$.
P.S. The wording "hence if $\sum_{i\in I} u_i\nabla g_i(\overline x)=0$ has solution then it must be nonzero" is not correct. The system always has a trivial solution $u_i=0$, and the conclusion "then it is nonzero" is not true for the trivial solution. The correct conclusion is that there exists at least one more solution except for the trivial one.