Convex sets and inequalities
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We consider a natural correspondence between a family of inequalities and a closed convex set. As an application, we give new types of power mean inequalities and the H¨oldertype inequalities. 1. Concept and fundamental result Given a natural correspondence between a family of inequalities and a closed convex set in a topological linear space, one might expect that an inequality corresponding to a special point (e.g., an extreme point) would be of special interest in view of the convex analysis theory. In this paper, we realize this concept. Let X be an arbitrary set and {ϕ0 ,ϕ1 ,ϕ} a triple of nonnegative real-valued functions on X. Set m = inf
ϕ0 (x)=0
ϕ(x) , ϕ0 (x)
ϕ(x) . ϕ ϕ1 (x)=0 1 (x)
M = sup
(1.1)
Suppose that 0 < m, M < ∞. Then we have mϕ0 (x) ≤ ϕ(x) ≤ Mϕ1 (x) ∀x ∈ X.
(1.2)
For each x ∈ X, put
Dϕ (x) = (α,β) ∈ R2 : ϕ(x) ≤ αϕ1 (x) + βϕ0 (x) .
(1.3)
We consider the intersection Dϕ = ∩x∈X Dϕ (x) of all such sets. Note that Dϕ is a nonempty closed convex domain in R2 and that each point (α,β) ∈ Dϕ corresponds to the inequality ϕ ≤ αϕ1 + βϕ0 on X. We want to investigate the closed convex domain Dϕ . To do this, we define the constant αϕ by αϕ =
Mϕ(x) − mMϕ0 (x) . Mϕ1 (x)=mϕ0 (x) Mϕ1 (x) − mϕ0 (x) sup
Copyright © 2005 Hindawi Publishing Corporation Journal of Inequalities and Applications 2005:2 (2005) 107–117 DOI: 10.1155/JIA.2005.107
(1.4)
108
Convex sets and inequalities
Clearly, 0 ≤ αϕ ≤ M. Also, we have the following three fundamental facts: (A) if (α,β) ∈ Dϕ and α/M + β/m = 1, then α ≥ αϕ , (B) {(α,β) ∈ R2 : α/M + β/m ≥ 1, α ≥ αϕ } ⊂ Dϕ , (C) Dϕ ⊂ {(α,β) ∈ R2 : α/M + β(mλ) ≥ 1} for some 1 ≤ λ ≤ ∞. In particular, if αϕ < M, then Dϕ ⊂ {(α,β) ∈ R2 : α/M + β/m ≥ 1}. These facts will be used in the later sections to realize our concept. Proof of (A). Suppose (α,β) ∈ Dϕ and α/M + β/m = 1. Then
ϕ(x) ≤ αϕ1 (x) + m 1 −
α ϕ0 (x), M
(1.5)
and hence Mϕ(x) − mMϕ0 (x) ≤α Mϕ1 (x) − mϕ0 (x)
(1.6)
for all x ∈ X with Mϕ1 (x) = mϕ0 (x). This implies that αϕ ≤ α.
Proof of (B). If t ≥ αϕ /M, then ϕ(x) − mϕ0 (x) ≤ t (Mϕ1 (x) − mϕ0 (x)) and so ϕ(x) ≤ tMϕ1 (x) + m(1 − t)ϕ0 (x) for all x ∈ X. Hence, we have
Dϕ ⊃ (α,β) ∈ R2 : α ≥ tM, β ≥ m(1 − t), t ≥
αϕ for some t ∈ R M
α β + ≥ 1, α ≥ αϕ . = (α,β) ∈ R : M m 2
(1.7)
Proof of (C). By the definition of M, we find a sequence {xn } in X such that
ϕ1 xn = 0, (n = 1,2,...),
ϕ xn M = lim . n→∞ ϕ1 xn
(1.8)
Of course, we can assume that ϕ(xn ) = 0 for all n = 1,2,.... Since {ϕ0 (xn )/ϕ(xn )} is a bounded sequence with bound 1/m, we can take a subsequence {ϕ0 (xn )/ϕ(xn )} converging to some real number t with 0 ≤ t ≤ 1/m. Set λ = 1/(tm) so that 1 ≤ λ ≤ ∞. We have Dϕ ⊂
n
ϕ1 xn ϕ0 xn (α,β) ∈ R2 : α + β ≥ 1 ϕ xn ϕ xn
(1.9)
β α ⊂ (α,β) ∈ R2 : ≥1 . + M mλ In particular, if αϕ < M, then λ must be 1 by an easy geometrical consideration on the αβ-plane R2 .
Sin-Ei Takahasi et al. 109 2. Application: Djokovic’s inequality Let H be a Hlawka space, that is, a Banach space in whi
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