By Raynaud M. (Ed), Shioda T. (Ed)

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1. The proof of the implication (ii)⇒(iii) is due to Rademacher and Schoenberg [819]. Proof. (i) It is sufﬁcient to prove Radon’s theorem for A = {x1 , . . , xd+2 } ⊆ Ed . There are µ1 , . . , µd+2 ∈ R, not all 0, such that µ1 + · · · + µd+2 = 0, µ1 x1 + · · · + µd+2 xd+2 = 0. We clearly may assume that µ1 , . . , µk ≥ 0 and −λk+1 = µk+1 , . . , −λd+2 = µd+2 ≤ 0. Then µ1 + · · · + µk = λk+1 + · · · + λd+2 > 0, µ1 x1 + · · · + µk xk = λk+1 xk+1 + · · · + λd+2 xd+2 and thus µ1 x1 + · · · + µk xk λk+1 xk+1 + · · · + λd+2 xd+2 = .

Bd } be the standard basis of Ed . Then f (x + tbi ) = f (x) + u i t + o(|t|) as t → 0 for i = 1, . . , d. Combined with Jensen’s inequality, this shows that (7) f (y) = f x + (y − x) = f ≤ 1 1 x + d(y1 − x1 )b1 + · · · + x + d(yd − xd )bd d d 1 1 f x + d(y1 − x1 )b1 + · · · + f x + d(yd − xd )bd d d = f (x) + u 1 (y1 − x1 ) + · · · + u d (yd − xd ) + o(|y1 − x y |) + · · · + o(|yd − xd |) = f (x) + u · (y − x) + o( y − x ) as y → x, y ∈ C. A similar argument yields the following: (8) f (2x − y) ≤ f (x) + u · (x − y) + o( y − x ) as y → x, 2x − y ∈ C.

Its derivative is v ·h, where v = G(x +th). 7. Hence t f (x + th) − f (x) = t G(x + sh) · h ds 0 1 u · h + h T H h s + o(s) ds = u · h t + h T H h t 2 + o(t 2 ) as t → 0 2 = 0 32 Convex Functions by (15), uniformly for all h. Hence 1 (16) f (y) = f (x) + u · (y − x) + (y − x)T H (y − x) + o( y − x 2 y → x, y ∈ C, 2 ) as uniformly for y of the form x + th and thus for almost all y ∈ C. Since f (y) and f (x) + u · (y − x) + 12 (y − x)T H (y − x) depend continuously on y, (16) holds for all y ∈ C.