# Thread: prove that solution of min problem is unique

1. ## prove that solution of min problem is unique

Hi

I have such exercise:
Let U be an open convex set of R^n and f be a strictly quasi-convex function.
Show that if there exists a solution of minimization problem, then it is a unique solution.

The problem Q is: min f(x) for all x belonging to U

I know that in 1dim f is strictly quasi-convex <=> for all (x,y)belonging to U^2, x not=y, for all t belonging to (0,1)
f(tx+(1-t)y) < max (f(x), f(y))
but here we have n-dimensional problem, so I am horribly confused and really don't know how to do this :/
Do you know how to solve it?

2. Originally Posted by doma
Hi

I have such exercise:
Let U be an open convex set of R^n and f be a strictly quasi-convex function.
Show that if there exists a solution of minimization problem, then it is a unique solution.

The problem Q is: min f(x) for all x belonging to U

I know that in 1dim f is strictly quasi-convex <=> for all (x,y)belonging to U^2, x not=y, for all t belonging to (0,1)
f(tx+(1-t)y) < max (f(x), f(y))
but here we have n-dimensional problem, so I am horribly confused and really don't know how to do this :/
Do you know how to solve it?
Exactly what do you mean by "solution", the point that makes the fuction miminum or simply that minimum value? For a quasi-convex problem on a convex set, I would say that a "solution" to the minimization consists of a point, p, in the set such that f(p) is less than or equal f(q) for q any other point in the set. That is NOT unique. For example, let S be the square in $R^2$ with vertices at (0,0), (1,0), (1,1), and (0,1) and let f((x,y))= y. Then any point (x,0) with $0\le x\le 1$ is a "solution". Of course, the value of f at any of those points is the same: the set of real numbers is "linearly ordered"- if a set of real numbers has a minimum, that minimum is unique.