My book in linear programming states two things that I do not understand. We are working with the lagrange method with linear constraints.:
From multivariate calculus we have that at a critical point we must have:
, these equations are called 17.3 in the proof below.
Now comes the part I do not understand. In the proof that follow:
1.[GREEN LINE] What is it that makes the proof hold for linear constraints, why do they need to specify that, if the g functions were not linear wouldn't the proof still hold?
2.[RED SQUARE] Why do they get a unique global maximum if 17.4 holds?, I can't see how the unique part just follows?
What requires linear constraints is the proof. The assertion the only displacements that are relevant are those that lie in the feasible set is easily true if the constraints are linear, that is, the feasible set is a linear subspace and (17.5) just says $\xi$ is parallel to the set. If the set has some curvature (degree of constraints $>1$), then (17.5) says $\xi$ is tangent to the set at $x^*$, and it may very well happen that any displacement in that direction escapes the set. But in that setting teh result is also true, although the proof requires a more careful understanding of Hessian. This enters in fact the realm of Calculus on manifolds, which I guess beyond the scope here.
For the red box, suppose we have $f(y^*)>f(x^*)$. Since the constraints are linear the line $L$ joining $y^*$ to $x^*$ is contained in the feasible set, and then we can work with the restriction of $f|L$. This is a one-variable function with one local maximum and a higher value, hence it must have a local minimum $z^*\in L$ between them. At that local minimum the second derivative of the restriction is $\ge0$, and after some computation one gets $\xi^THf(z^*)\xi\ge0$ for the direction $\xi$ of $L$.