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By Krantz S.G.

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Then the convex hull of the union of that collection of simplices will serve as K1 . 18 CHAPTER 1. BASIC IDEAS For our construction of K2 , we may assume that the origin lies in K. Using the Minkowski functional as our notion of distance, and denoting it by p, we may think of K = {x ∈ RN : p(x) ≤ 1} . Now let χ be a nonnegative Cc∞ function, supported in the unit ball, with integral equal to 1. Now define p (x) = p(x − t)χ(t) dt + |x|2 . We see that p is strictly convex. Also, for p (x) ≤ p(x) + C1 x ≤ 1 + C1 , > 0 small, all x ∈ K .

85] for further discussion of these ideas. Now let x, y lie in the convex domain of f. Assume without loss of generality that f(x) = 0 and we shall prove that f is continuous at x. For 0 ≤ t ≤ 1, we know that f((1 − t)x + ty) ≤ (1 − t)f(x) + tf(y) . We think of t as positive and small, so that (1 − t)x + ty is close to x. Then we see that |f((1 − t)x + ty)| ≤ t|f(y)| . That shows that lim f(z) = 0 = f(x) , z→x so that f is continuous at x. Of course it is also useful to consider convex functions on a domain.

The same reasoning shows that a strictly convex function on an open interval has no maximum. , a weakly convex) function? 21 We say that F : P, Q ∈ RN , Rn → R is midpoint convex if, for any F ((1/2)P + (1/2)Q) ≤ (1/2)F (P ) + (1/2)F (Q) . 1) Clearly any convex function is midpoint convex. The converse is true for any function that is known to be continuous. We omit the details, but refer the reader to [SIM, p. 3]. 22 Consider the function f(x) = ex . 1) amounts to 1 1 eP/2+Q/2 ≤ eP + eQ , 2 2 and this is true because 2ab ≤ a2 + b2 .

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