For g(y,V,w)=(y−w)TV(y−w), for y in Rn, V is positive definite (e.g. variance matrix) and w is in a closed convex cone C in Rn.
Obviously for every fixed y and V minw\in C g(y,V,w) exists
But my question is for every fixed y,V and y0, V0 does
minw\in C [ g(y,V,w) - g(y0,V0,w) ] exist?
As the only difference is the fixed added constant in the second problem's objective function - note that the feasible set is the same in the two problems - the set of optimal solutions must be identical.
Did you really mean to have a w_0? If so I'm not sure how it is to be regarded.
Assuming not, suppose you work in one dimensional space. Take V and V_0 to both be the matrix ((1)). Take y=0 and y_0 = 1. So the functions are x^2 and (x-1)^2. Their difference can be arbitrarily large or small depending on the convex cone region specified (a ray, in this case).
Thank you Michael again, and thank you Daniel for pointing out the typo. Yes everything should be w, no w0
My apologies that the question wasn't stated correctly, V hast to be positive definite. Unfortunately I think Daniel is right. I think the issue is that V-V0 must also be positive definite which is not the case in his example.
The way that I read the question, you are simply asking about the existence of the minimum value of a function. This is different from finding the minimum value. There is a theorem from point set topology which says that under certain circumstances you can be assured of the existence of extreme (min or max) values. To generalize, let Y=F(X) be a function on metric spaces, in your case, n dimensional spaces. If you have a subset S of X that is closed and bounded, then you are guaranteed of the existence of extreme values. "Closed" is a term from topology which, intuitively means, "contains boundary points". For example, [0,1] is closed, while (0,1) is not closed. "Bounded" means, intuitively, that the subset S can be enclosed in a large enough (but finite) n dimensional sphere.
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