Affine mappings and convex functions. Examples of convex functions In this section, X; Y denote real vector spaces, unless otherwise specified. Affine mappings. Definition 0.1. Let X; Y be vector spaces, A ⊂ X an affine set. A mapping F : A ! Y is affine if F ((1 − t)x + ty) = (1 − t)F (x) + tF (y) whenever t 2 R and x; y 2 A. Proposition 0.2. Let F : X ! Y . (a) F is linear if and only if F is affine and F (0) = 0. (b) F is affine if and only if there exist a linear mapping T : X ! Y and a vector y0 2 Y such that F (x) = T x + y0 (x 2 X). Moreover, T is unique in this case. Proof. Exercise. Corollary 0.3. Let A ⊂ X be an affine set. If F : A ! Y is affine, then Pn Pn Pn F ( 1 λixi) = 1 λiF (xi) whenever xi 2 A; λi 2 R; 1 λj = 1. Proof. Fix a0 2 A and consider the linear set L = A − a0. Since A = L + a0, we can define an affine mapping G: L ! Y by G(y) = F (y + a0). By Proposition 0.2, we can write G(y) = T y + y0, where T : L ! Y is linear and y0 2 Y is fixed. Then our F is of the form F (x) = G(x − a0) = T (x − a0) + y0 : Now, an easy calculation completes the proof. Now, we can consider the following notion of affine mappings defined on convex (not necessarily affine) sets. Definition 0.4. Let X; Y be vector spaces, C ⊂ X a convex set. We say that a mapping F : C ! Y is c-affine if F ((1 − t)x + ty) = (1 − t)F (x) + tF (y) whenever t 2 [0; 1] and x; y 2 X. Lemma 0.5. Let A ⊂ X be an affine set and F : A ! Y a mapping. Then F is affine if and only if F is c-affine. Proof. Each affine mapping is clearly c-affine. To prove the vice-versa, suppose that F is c-affine and consider x; y 2 A, t 2 R, z = (1 − t)x + ty. We want to show that F (z) = (1 − t)F (x) + tF (y). For t 2 [0; 1] this is true. Suppose t > 1. Then y is a convex combination of x and z: 1 t−1 y = t z + t x : 1 t−1 Consequently, F (y) = t F (z) + t F (x) which easily implies F (z) = (1 − t)F (x) + tF (y). The case t < 0 can be proved in a similar way by expressing x by means of y; z. Lemma 0.6. Let C be a convex set in a vector space X. Then aff(C) = fαu − βv : α; β ≥ 0; α − β = 1; u; v; 2 Cg: 1 2 Proof. Let us prove the inclusion \⊂" (the other one is trivial). Take x 2 aff(C) Pn and write it as an affine combination x = i=1 λiui of elements ui 2 C (1 ≤ i ≤ n). We can suppose that λi 6= 0 for each i. Denote I = fi : λi > 0g ;J = fi : λi < 0g : Clearly, I 6= ;. If J = ;, we have x 2 C, and hence x = αx − βx with α = 1; β = 0. P P Now, let J 6= ;. Then x = αu − βv where α = j2I λj, β = j2J (−λj), u = P λi P −λi i2I α ui 2 C and v = i2J β ui 2 C. Theorem 0.7. Let X; Y be vector spaces, C ⊂ X a convex set, A = aff(C). If F : C ! Y is c-affine, then there exists an affine mapping G: A ! Y such that GjC = F . Moreover, such G is unique. Proof. By Lemma 0.6, each x 2 A can be written in the form (1) x = αu − βv where α; β ≥ 0; α − β = 1; u; v 2 C: Then we put G(x) = αF (u) − βF (v). Claim. G(x) does not depend on the representation (1). To see this, consider two such representations αu−βv = α0u0−β0v0. Then αu+β0v0 = α0u0 + βv and α0 + β = α + β0 =: ∆. We must have ∆ > 0, since otherwise we would 0 0 α β0 0 α0 0 β get α = α = β = β = 0 which is impossible. Since ∆ u + ∆ v = ∆ u + ∆ v are α β0 0 α0 0 β convex combinations, we get ∆ F (u) + ∆ (v ) = ∆ (u ) + ∆ (v) which easily implies αF (u) − βF (v) = α0F (u0) − β0F (v0). Our Claim is proved. Thus we have defined a mapping G: A ! Y . Moreover, G is an extension of F . Since any affine extension of F has to satisfy the same definition, F has at most one affine extension to A. In view of Lemma 0.5, it suffices to show that G is c-affine on A. Let x; y 2 A, t 2 (0; 1), p = (1 − t)x + ty. By Lemma 0.6, we can write x = αu − βv ; y = γw − δz ; where α; β; γ; δ ≥ 0, α − β = γ = −δ = 1, u; v; w; z 2 C. Then p = (1 − t)αu + tγw − (1 − t)βv + tδz: Put a = (1 − t)α + tγ and b = (1 − t)β + tδ and observe that a − b = 1. If a 6= 0 and b 6= 0, we can write h (1−t)α tγ i h (1−t)β tδ i p = a a u + a w − b b v + b z : The expressions in square brackets are points of C. Thus (1 − t)α tγ (1 − t)β tδ G(p) = aF u + w − bF v + z a a b b (1 − t)α tγ (1 − t)β tδ = a F (u) + F (w) − b F (v) + F (z) a a b b = (1 − t)[αF (u) − βF (v)] + λ[γF (w) − δF (z)] = (1 − t)G(x) + tG(y) : 3 By the above theorem, every c-affine mapping on a convex set is a restriction of an affine mapping. From this reason, c-affine mappings on a convex set will be called affine. Exercise 0.8. Let X; Y be vector spaces, C ⊂ X a convex set, and F : C ! Y an affine mapping. Let A ⊂ C and B ⊂ Y be sets. (a) If A is convex, then F (A) is convex. If B is convex, then F −1(B) is convex. (b) Is it true that conv(F (A)) = F (conv(A)) ? (b) Is it true that conv(F −1(B)) = F −1(conv(B)) ? Convex functions. By R we mean the extended real line [−∞; +1]. Definition 0.9. Let C ⊂ X be a convex set, f : C ! R a function. (a) f is called proper (on C) if its effective domain dom(f) = fx 2 C : f(x) 2 Rg is nonempty. (b) f is called convex (on C) if its epigraph epi(f) = f(x; t) 2 C × R : t ≥ f(x)g is a convex set in X × R. (c) f is called concave (on C) if −f is convex. It is clear that f is concave on C if and only if its hypograph hypo(f) = f(x; t) 2 C × R : t ≤ f(x)g is a convex set. Moreover, f : C ! R is both convex and concave if and only if f is affine. Notice that if f : C ! [−∞; +1] is a convex function, then the function ( f(x)(x 2 C) f^(x) = +1 (x 2 X n C) is convex on X. Observation 0.10. If fi : C ! R (i 2 I) are convex functions, then also their pointwise supremum f(x) = sup fi(x) i2I T is a convex function. Indeed, epi(f) = i2I epi(fi) . Theorem 0.11. Let C be a convex set in a vector space X, f : C ! R a function. Then f is convex if and only if f((1 − λ)x + λy) ≤ (1 − λ)f(x) + λf(y) whenever x; y 2 C, λ 2 (0; 1) and the right-hand side is defined. 4 Proof. \)" Fix x; y 2 C and λ 2 (0; 1) such that (1 − λ)f(x) + λf(y) is defined (i.e., f(x); f(y) are not infinite with opposite signs). If f(x) = +1 of f(y) = +1, the inequality is obviously satisfied. Let now f(x) < +1 and f(y) < +1. For any t; s 2 R such that t ≥ f(x) and s ≥ f(y), the points (x; t) and (y; s) belong to the convex set epi(f). Thus also ((1 − λ)x + λy; (1 − λ)t + λs) 2 epi(f), which implies f((1 − λ)x + λy) ≤ (1 − λ)t + λs : The formula follows since t ≥ f(x) and s ≥ f(y) were arbitrary. \(" Let (x; t) and (y; s) be two points of epi(f), and λ 2 (0; 1). Then f(x) ≤ t < +1 and f(y) ≤ s < +1. By our formula, f((1−λ)x+λy) ≤ (1−λ)f(x)+λf(y) ≤ (1−λ)t+λs. Hence (1−λ)(x; t)+λ(y; s) = ((1−λ)x+λy; (1−λ)t+λs) 2 epi(f). Observation 0.12. If f : C ! R is a convex function on a convex set C, then the sets dom(f), fx 2 C : f(x) ≤ αg and fx 2 C : f(x) < αg (α 2 R) are convex. Convex functions that assume the value −∞ are not particularly interesting. For −1 example, if f : R ! R is convex with J := f (−∞) 6= ;, then J is an interval, dom(f) ⊂ @I, and f is identically +1 on R n J.