Prove that every local minimum of a convex function on a convex set is a global minimum, and give a convex function with infinitely many global minima to show uniqueness needs a stronger assumption.

Prove that every local minimum of a convex function on a convex set is a global minimum, and give a convex function with infinitely many global minima to show uniqueness needs a stronger assumption.

Approach: Suppose a strictly better point exists elsewhere and evaluate the function on the segment joining it to the supposed local minimum, using the convexity inequality on that segment.

Every local minimum of a convex function on a convex set is a global minimum. Let x be a local minimum, so f(x) <= f(z) for all z within some radius r of x, and suppose some y in the set has f(y) < f(x). The set is convex, so the point z_t = (1-t) x + t y lies in it for every t in [0, 1], and the convex function inequality gives f(z_t) <= (1-t) f(x) + t f(y) < f(x) for every t in (0, 1]. Choosing t small enough that z_t is within r of x contradicts local optimality, so no such y exists and x is a global minimum. Uniqueness is a separate matter: f(x) = max(0, |x| - 1) is convex and every point of [-1, 1] is a global minimum. Strict convexity rules that out, since a segment of minimisers would force f to be affine on it, and this is exactly why a positive definite covariance matrix gives one optimal portfolio while a singular one gives a whole face of them.

Follow-up: Does the same argument work for a quasiconvex function, and what fails if it does not?

Key concepts: convex function, local minimum, global minimum, strict convexity.