i'm having problems trying to understand this. We have according to books that the necessary regularity conditions need to be verified for a given point x* to be a feasible stationary point. This is:
∇f(x∗) = 0
and
alpha1*∇g1(x∗)+ alpha2*∇g2(x∗)+...+alphan*∇gn(x∗) = 0
So my question is, if i have a problem with only one restriction, lets say:
Min f(x,y) = x^2 + y^2
s.t. x-y=0
and i have a candidate point for minimum x = (1,-1)
Do i need to check de regularity condition even knowing that if a plug this point into the restriction i doesnt satisfy? and also, i'm having issues with this specific problem, because when i want to check de regularity condition, given the fact that i only have one restricton i need to check if the gradient of g1(x) is not zero. Checking, i have that is not zero, so in theory the point is feasible, but when i plug it i don't see it feasible. So, could yo help me with that?
If you could give me an intuitive explanation of regularity conditions, it would be really helpsful.
I'm sorry if this is a silly question, and also my english is not perfect. Sorry for mistakes.
I'm making some guesses here based on what you wrote because much of what you wrote wasn't quite right. Basically, for an optimization problem $$ \min\limits_x \{ f(x) : g(x) = 0\} $$ the KKT conditions give necessary conditions for optimality $$\begin{array}{rl} \nabla f(x) + g'(x)^*y &= 0\\ g(x)&=0 \end{array}$$ If we use the inner product $\langle x,y\rangle=x^Ty$ and have $g$ of the form $$ g(x)=\begin{bmatrix}g_1(x)\\\vdots\\g_m(x)\end{bmatrix} $$ then, the optimality conditions become $$\begin{array}{rl} \nabla f(x) + \sum\limits_{i=1}^m y_i\nabla g_i(x) &= 0\\ g(x)&=0 \end{array}$$ Properly, these conditions have some regularity conditions called constraint qualifications. The most common constraint qualification is the linear independence constraint qualification (LICQ) that requires the $\nabla g_i(x)$ to be linearly independent at the optimal point $x$. Do they have to be linearly independent for a point to be optimal? No. There are other constraint qualifications with the weakest generally being that something called the linearized cone and the tangent cone at the point $x$ coincide. That's a pain to check or write down, so most people use the LICQ because it's easy to verify and implies this condition. It also makes the linear systems involved for algorithms much easier because they'll generally be nonsingular. Finally, does a point need to satisfy a constraint qualification to be optimal at all? No. Constraint qualifications are an artifact of the derivation of the KKT conditions. They're what allow optimality conditions based on limits to be rewritten as a linear system, which allows us to make algorithms.
For your particular question, you only have one constraint, so you'll automatically satisfy the LICQ. Therefore, your regularity condition is satisfied automatically. However, the point $x=(-1,1)$ is not optimal nor feasible. In your case, the optimal solution is $x=(0,0)$.