By Legendre A.-M.

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1 implies that, in particular, g (0) ≥ 0 and thus 1 1 (y − x)T H (x)(y − x) = 2 2 d f xi ,xk (x)(yi − xi )(yk − xk ) = i,k=1 1 g (0) ≥ 0, 2 concluding the proof of (1). 4, it is sufﬁcient to show the following: (2) Let x ∈ C. Then f has an afﬁne support at x. Since f is of class C 2 , Taylor’s theorem, for functions of d variables, implies that d 1 f xi ,xk x + ϑ(y − x) (yi − xi )(yk − xk ) 2 i,k=1 1 = f (x) + u · (y − x) + (y − x)T H x + ϑ(y − x) (y − x) 2 ≥ f (x) + u · (y − x) for y ∈ C, f (y) = f (x) + u · (y − x) + where u = grad f (x), 0 < ϑ < 1 is chosen suitably, depending on y, and we have used the assumption that the Hessian form of f at x + ϑ(y − x) is positive semideﬁnite.

Let x ∈ C and choose y ∈ int C. Then there is a δ > 0 such that y + δ B d ⊆ int C ⊆ C. The convexity of C now implies that (1 − λ)x + λy + λδ B d = (1 − λ)x + λ(y + δ B d ) ⊆ C. Thus (1 − λ)x + λy ∈ int C for 0 < λ ≤ 1. Since (1 − λ)x + λy → x as λ → 0, it follows that x ∈ cl int C. 2. Let A ⊆ Ed be bounded. Then cl conv A = conv cl A. Proof. 1 shows that cl conv A is convex too. cl conv A is a closed set which contains A. Thus it also contains cl A. Since cl conv A is a convex set which contains cl A, it also contains conv cl A.

2 Alexandrov’s Theorem on Second-Order Differentiability In the last section it was shown that a convex function is differentiable almost everywhere, and we remarked that the same holds with respect to Baire category and metric. A deep result of Alexandrov says that, in the sense of measure theory, 28 Convex Functions even more is true: a convex function is twice differentiable, except on a set of measure zero. In contrast, results of Zamﬁrescu [1037, 1038] imply that, from the Baire category viewpoint, a typical convex function is twice differentiable only on a small set.

### Elements de geometrie by Legendre A.-M.

by Charles

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