Topics in Inequalities - Theorems and Techniques Hojoo ... - Index of

Chapter 4
Convexity
Any good idea can be stated in fifty words or less. S. M. Ulam
4.1 Jensen’s Inequality
In the previous chapter, we deduced the weighted AM-GM inequality from the AM-GM inequality. We use
the same idea to study the following functional inequalities.
Proposition 4.1.1. Let f : [a, b] −→ R be a continuous function. Then, the followings are equivalent.
(1) For all n ∈ N, the following inequality holds.
ω 1 f(x 1 ) + · · · + ω n f(x n ) ≥ f(ω 1 x 1 + · · · + ω n x n )
for all x 1 , · · · , x n ∈ [a, b] and ω 1 , · · · , ω n > 0 with ω 1 + · · · + ω n = 1.
(2) For all n ∈ N, the following inequality holds.
r 1 f(x 1 ) + · · · + r n f(x n ) ≥ f(r 1 x 1 + · · · + r n x n )
for all x 1 , · · · , x n ∈ [a, b] and r 1 , · · · , r n ∈ Q + with r 1 + · · · + r n = 1.
(3) For all N ∈ N, the following inequality holds.
( )
f(y 1 ) + · · · + f(y N ) y1 + · · · + y N
≥ f
N
N
for all y 1 , · · · , y N ∈ [a, b].
(4) For all k ∈ {0, 1, 2, · · · }, the following inequality holds.
f(y 1 ) + · · · + f(y 2 k)
2 k ≥ f
( )
y1 + · · · + y 2 k
2 k
for all y 1 , · · · , y 2 k ∈ [a, b].
(5) We have 1 2 f(x) + 1 2 f(y) ≥ f ( )
x+y
2 for all x, y ∈ [a, b].
(6) We have λf(x) + (1 − λ)f(y) ≥ f (λx + (1 − λ)y) for all x, y ∈ [a, b] and λ ∈ (0, 1).
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒ (5) is obvious.
(2) ⇒ (1) : Let x 1 , · · · , x n ∈ [a, b] and ω 1 , · · · , ω n > 0 with ω 1 + · · · + ω n = 1. One may see that there
exist positive rational sequences {r k (1)} k∈N , · · · , {r k (n)} k∈N satisfying
lim r k(j) = w j (1 ≤ j ≤ n) and r k (1) + · · · + r k (n) = 1 for all k ∈ N.
k→∞
By the hypothesis in (2), we obtain r k (1)f(x 1 ) + · · · + r k (n)f(x n ) ≥ f(r k (1) x 1 + · · · + r k (n) x n ). Since f is
continuous, taking k → ∞ to both sides yields the inequality
ω 1 f(x 1 ) + · · · + ω n f(x n ) ≥ f(ω 1 x 1 + · · · + ω n x n ).
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