MEASURE AND INTEGRATION: LECTURE 14 Convex functions. Let ϕ : (a, b) → R, where −∞ ≤ a < b ≤ ∞. Then ϕ is convex if ϕ((1 − t)x + ty) ≤ (1 − t)ϕ(x) + tϕ(y) for all x, y ∈ (a, b) and t ∈ [0, 1]. Looking at the graph of ϕ, this means that (t, ϕ(t)) lies below the line segment connecting (x, ϕ(x)) and (y, ϕ(y)) for x < t < y. Convexity is equivalent to the following. For a < s < t < u < b, ϕ(u) − ϕ(t) ϕ(t) − ϕ(s) ≤ . t−s u−t If ϕ is diﬀerentiable, then ϕ is convex on (a, b) if and only if, for a < s < t < b, ϕ� (s) ≤ ϕ� (t). If ϕ is C 2 (continuously twice diﬀerentiable), then ϕ� increasing ⇒ ϕ�� ≥ 0. Theorem 0.1. If ϕ is convex on (a, b), then ϕ is continuous on (a, b). Jensen’s inequality. Let (Ω, M, µ) be a measure space such that µ(Ω) = 1 (i.e., µ is a probability measure). Let f : Ω → R and f ∈ L1 (µ). If a < f (x) < b for all x ∈ Ω and ϕ is convex on (a, b), then �� ϕ f dµ ≤ (ϕ ◦ f )dµ. Ω Proof. Let t = Ω � f dµ. Since a < f < b, � a = a · µ(Ω) < f dµ < b · µ(Ω) = b, Ω Ω so a < t < b. Conversely, ϕ(u) − ϕ(t) ϕ(t) − ϕ(s) ≤ . t−s u−t Fix t, and let ϕ(t) − ϕ(s) B = sup . t−s a t. Thus ϕ(s) ≥ ϕ(t) + B(s − t) for any a < s < b. Let s = f (x) for any x ∈ Ω. Then ϕ(f (x)) − ϕ(t) − B(f (x) − t) ≥ 0 for all x ∈ Ω. Now ϕ convex ⇒ ϕ continuous, so ϕ ◦ f is measurable. Thus, inte grating with respect to µ, � � � (ϕ ◦ f )dµ − ϕ(t) dµ − B f dµ ≥ 0, X X X and the inequality follows. � Examples. (1) Let ϕ(x) = ex be a convex function. Then �� exp f dµ ≤ ef dµ. Ω Ω (2) Let Ω = {p1 , . . . , pn } be a ﬁnite set of points and deﬁne µ({pi }) = 1/n. Then µ(Ω) = 1. Let f : Ω → R with f (pi ) = xi . Then � f dµ = Ω = n � f (pi )µ({pi }) i=1 1 (x1 + · · · + xn ). n Thus � exp � � 1 (x1 + · · · + xn ) ≤ ef dµ n Ω 1 x1 ≤ (e + · · · + exn ). n Let yi = exi . Then (y1 + · · · + yn )1/n ≤ 1 (y1 + · · · + yn ), n which is the inequality between arithmetic � and geometric means. We also could take µ({pi }) = αi > 0 and ni=1 αi = 1. Then y1α1 y2α2 · · · ynαn ≤ α1 y1 + α2 y2 + · · · + αn yn . MEASURE AND INTEGRATION: LECTURE 14 3 Hölder’s and Minkowski’s inequalities. We deﬁne numbers p and q to be conjugate exponents if 1/p + 1/q = 1. The conjugate exponent of 1 is ∞. Conjugate exponents are the same if and only if p = q = 2. Theorem 0.2. Let p and q be conjugate exponents with 1 < p < ∞. Let (X, M, µ) be a measure space and f, g : X → [0, ∞] measurable functions. Then �� 1/p �� 1/q � p q f g dµ ≤ f dµ g dµ (Hölder’s) X and �� X X �1/p �� 1/p �� 1/p p p (f + g) dµ ≤ f dµ + g dµ (Minkowski’s). p X X X Proof. �H¨older’s. Without loss � ofp generality� weq may assume that q � 0, then let 1 and X g = 1. Indeed, if f �= 0 and g = f = �� f fp X �1/p , g = �� g fp X � X fp = �1/p . � (Otherwise, if f p = 0, then f p = 0 a.e., and both sides of the inequal ity are equal to zero.) We claim that (0.1) 1 1 ab ≤ ap + bq for all a, b ∈ [0, ∞]. p q It is easy to check if a or b equals 0 or ∞. Assume 0 < a < ∞ and 0 < b < ∞, and write a = es/p and b = et/q for some s, t ∈ R. Let Ω = {x1 , x2 }, µ(x1 ) = 1/p, and µ(x2 ) = 1/q. We have � � exp f dµ ≤ ef dµ, Ω Ω where f (x1 ) = s and f (x2 ) = t. Thus, � � 1 1 s t + ≤ es + et , exp p q q p so (0.1) follows. Thus, � � � 1 1 1 1 p a dµ + bq dµ = + = 1. (f g)dµ ≤ p X q X p q X Minkowski’s. Observe that (f + g)p = f (f + g)p−1 + g(f + g)p−1 . 4 MEASURE AND INTEGRATION: LECTURE 14 Since p and q are conjugate exponents, q = p/(p − 1). Thus, �� 1/p �� (p−1)/p � p−1 p (p−1)p/(p−1) f (f + g) ≤ f (f + g) �� = f p �1/p �� p �(p−1)/p . (f + g) Similarly, � p−1 f (f + g) �� ≤ f p �1/p �� p �(p−1)/p (f + g) . Let Ω = {x1 , x2 }, µ(x1 ) = 1/2 = µ(x2 ), and ϕ = tp . Then �� p � f dµ ≤ f p dµ, Ω so � a+b 2 Ω �p ≤ ap b p + . 2 2 Thus, � � � 1 1 1 p p f + g p < ∞. (f + g) ≤ 2p 2 2 Since 1 − (p − 1)/p = 1/p, �� 1/p �� 1/p �� 1/p p p p (f + g) dµ ≤ f dµ + g dµ . X X X �

Measure and Integration: Lecture 14 Notes

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