Let f be a real function on the interval I . Prove that the following state- ments are equivalent:

6.2.1. Let f be a real function on the interval I . Prove that the following state- ments are equivalent:

(i) f is convex;

266 6 Convex Functions

(ii) for any x , y, z ∈ I with x <y<z ,

z −y (iii) for any a ∈I , the mapping

is nondecreasing; (iv) for any x , y, z ∈ I with x <y<z ,

f (x) f (y) f (z)

(v) the set epi f = {(x,y) ∈ R 2 ;x ∈ I and f (x) ≤ y} is convex. Solution. (i) =⇒ (ii) Fix x, y, z ∈ I with x < y < z. There exists λ ∈ (0,1) such

that y = λ x + (1 − λ )z. Then

λ x −z

and 1 − −y λ = .

x −z

x −z

Thus, since f is convex,

which implies

The second inequality in (ii) is proved with a similar argument, and we leave the details to the reader.

(ii) =⇒ (iii) Fix t 1 ,t 2 ∈ I \ {a} with t 1 <t 2 . If t 1 <t 2 < a, we apply (ii) with (x, y, z) = (t 1 ,t 2 , a). If t 1 <a<t 2 , we apply (ii) with (x, y, z) = (t 1 , a,t 2 ). If a <t 1 <t 2 , we apply (ii) with (x, y, z) = (a,t 1 ,t 2 ).

(iii) =⇒ (iv) Assume that x, y, z ∈ I and x < y < z.

6.2 Basic Properties of Convex Functions and Applications 267 Then

f (y) − f (x) = (y − x)(z − x)

y −x (iv) =⇒ (i) Fix x, y ∈ I and take λ ∈ [0,1]. Thus, by (iv),

z −x

λ x + (1 − λ )y

y ≥ 0.

f (x) f ( λ x + (1 − λ )y) f (y)

But

f (x) f ( λ x + (1 − λ )y) − λ f (x) − (1 − λ ) f (y) f (y)

This implies f ( λ x + (1 − λ )y) − λ f (x) − (1 − λ ) f (y) ≤ 0; hence f is convex. (i)=⇒ (v) Fix (x 1 ,y 1 ), (x 2 ,y 2 )∈ R 2 such that x 1 ,x 2 ∈ I and f (x 1 )≤ y 1 ,f (x 2 )≤y 2 .

We prove that if λ ∈ [0,1] then ( λ x 1 + (1 − λ )x 2 ,f( λ x 1 + (1 − λ )x 2 )) ∈ epi f . We first observe that since I is an interval and x 1 ,x 2 ∈ I, then λ x 1 + (1 − λ )x 2 ∈ I. Next, since f is convex, we have

f ( λ x 1 + (1 − λ )x 2 )) ≤ λ f (x 1 ) + (1 − λ ) f (x 2 ), which concludes the proof.

(v) =⇒ (i) Fix x, y ∈ I and λ ∈ [0,1]. Since epi f is a convex set and (x, f (x)) ∈ epi f , (y, f (y)) ∈ epi f , we deduce that ( λ x + (1 − λ )y, λ f (x) + (1 − λ ) f (y)) ∈ epi f .

Therefore f ( λ x + (1 − λ )y) ≤ λ f (x) + (1 − λ ) f (y).

The property stated in Exercise 6.1.1 (ii) is also known as the three chords lemma (see Figure 6.2). In particular, this result establishes that if f is a real-valued function on the interval I, then for any a , x, y ∈ I with a < x < y,

Geometrically, this property asserts that the slope of a chord increases if the left end (a, f (a)) of the chord is fixed and the right end is moved to the right. ⊓ ⊔

6.2.2. Let x 1 ,x 2 ,...,x n

be positive real numbers. Show that (x x1+x2+···+xn

1 x 2 ··· x n )

≤x 1 x 2 ···x n .

268 6 Convex Functions

Fig. 6.2 A function is convex if and only if its sequential secants have increasing slopes.

Solution. The function f (x) = x ln x, x > 0, is convex. Indeed, f ′ (x) = 1 + ln x,

f ′′ (x) = x −1 > 0. So, by Jensen’s inequality we obtain

1 + ··· + x n

x1+···+xn

≤ ln 1 ··· x n , that is,

ln x 1 x n 1 n

1 + ··· + x n

x 1 +···+x n

x ≤x 1 ···x x n .

Applying now the mean value inequality, we obtain

(x 1 ··· x n ) n

x1+···+xn

1 + ··· + x n

x 1 +···+x n

≤x 1 ··· x n .⊓ ⊔