Let be a Banach space over . A function is called strongly convex with parameter iff the following equation holds for all and :
Proposition (existence and uniqueness of minimizers of strongly convex functions):
Let be a Banach space over and let be a strongly convex function with parameter , which additionally is bounded below (say by ) and continuous. Then admits a unique minimizer (ie. an element which realizes the infimum of , where ranges over ).
Proof: Since , the value exists. Choose a sequence in such that
is a Cauchy sequence because if is such that and is such that , then
in particular, if we show that a minimizer exists, then it will be unique, for if we set and call any other minimizer , the above estimate holds for arbitrary.
Since is Banach, is convergent, say to . If we show that
for all , then . By the continuity of , choose such that implies . By convergence of pick sufficiently large so that for all . Then choose such that . Then the triangle inequality implies
↑If is separable, so that arbitrary products of nonempty open sets are nonempty, the continuity of implies that the axiom of choice is not required for this construction.