Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 6 2 Solution Created 2026-10-03 Updated 2026-10-14
Here is a finite real convex function on the whole real vector space; no topology or continuity is required. We prove the convex domination form of the Hahn-Banach theorem by extending across one vector. Suppose a linear functional is already dominated on a vector subspace and choose . For and positive , the proposed extension must satisfy the boundsTo prove compatibility, take and . The convex combination of and with weights and lies in . Domination there and convexity giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates showThus both endpoints are finite, and some exists. For the upper bound gives domination of ; for the lower bound gives it; and for domination was assumed. This defines a dominated linear functional on extending .
Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By Zorn's lemma there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is . Hence there is a full linear extension with . A convex function here need not be positively homogeneous; this is why the bounds included all positive .
Now fix . The translated convex function vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on to obtain a linear functional with for every . SetThis is an affine function, satisfies for every , and has . Let . Every member is an affine minorant, while the member indexed by attains at . Thus the exact representation isThis is the finite convex function as supremum of affine minorants property. The linear functionals are algebraic; without a topology no assertion of continuous affine minorants is intended.