����������� ������ 45 (1993), 29–33

UDC 517.982.2

���������� ������ ��� research paper

AN INEQUALITY RELATED TO THE UNIFORM CONVEXITY IN BANACH SPACES

Miroslav Pavlović

Abstract. We prove an inequality that implies that a 2-concave and p-convex Banach lattice is “more” uniformly convex than L^{p}.

1. Introduction

In this note we prove the following

Theorem. Let X be a 2-concave Banach lattice with 2-concavity constant equal to one, and let 1 ⩽ p ⩽ 2. Then

∥(|x + y|^{p} + |x − y|^{p})^{1/p}∥ ⩾ {(∥x∥ + ∥y∥)^{p} + |∥x∥ − ∥y∥|^{p}}^{1/p} ,  (1)

for all x, y ∈ X. In particular, inequality (1) holds in an arbitrary L^{q} space with 1 ⩽ q ⩽ 2.

For the definition of the expression (|u|^{p}+|v|^{p})^{1/p} and other notions concerning abstract Banach lattices we refer to [3], Ch. 1 (especially Theorem 1.d.1). In the case where X = L^{p} (1 < p < 2) inequality (1) becomes

∥x + y∥^{p} + ∥x − y∥^{p} ⩾ (∥x∥ + ∥y∥)^{p} + |∥x∥ − ∥y∥|^{p},  (2)

which was used by Hanner [2] to calculate the precise value of the modulus of convexity of L^{p}. Moreover, it follows from [4] that the validity of (2) in some normed spaces X implies that X is “more” uniformly convex than L^{p} (where L^{p} is at least two-dimensional). An immediate consequence of Theorem is that (1) holds in a large class (denoted by Δ(p, 2); see Section 2) containing, for example, L^{q} for p ⩽ q ⩽ 2 as well as certain Orlicz and mixed normed Lebesgue spaces. Note that, in [4], the validity of (2) in L^{q} (p ⩽ q ⩽ 2) was deduced from the case q = p by using the fact that L^{q} can be embedded into L^{p}(0, 1) isometrically (see [3], pp. 181–182). The proof in the present note is quite elementary and lies on the fact that (for 1 ⩽ p ⩽ 2) the function

F_{p}(ξ, η) := {(ξ^{1/2} + η^{1/2})^{p} + |ξ^{1/2} − η^{1/2}|^{p}}^{2/p}  (ξ ⩾ 0, η ⩾ 0)  (3)

is convex. Before proving the result we mention a generalization of F_{p} that could be of some independent interest. Let r_{j} (j = 0, 1, 2,…) denote the Rademacher functions,

r_{j}(t) = sign(sin(2^{j}πt)) (t real).

Define the functions Φ_{p} on the positive cone l_{+}^{1} of the sequence space l^{1} by

Φ_{p}(ξ) = {∫_{0}^{1}| ∑_{j=0}^{∞} r_{j}(t)ξ_{j}^{1/2}|^{p} dt}^{2/p} (ξ = (ξ_{j})_{0}^{∞} ⩾ 0).

That the definition is correct follows from the well known fact that if (a_{j})_{0}^{∞} ∈ l^{2}, then the series ∑ a_{j}r_{j}(t) converges almost everywhere, and from Khintchine’s inequality [3], Theorem 2.b.3, which says that

A_{p}∥ξ∥_{l^{1}} ⩽ Φ_{p}(ξ) ⩽ B_{p}∥ξ∥_{l^{1}} (A_{p}, B_{p} = const > 0).

Starting from the observation that Φ(ξ_{1}, ξ_{2}, 0, 0,…) = const F_{p}(ξ_{1}, ξ_{2}) we conjecture that Φ_{p} is a convex function on l_{+}^{1} (for 1 ⩽ p ⩽ 2). (We shall also prove that if p > 2, then F_{p} is concave, and we conjecture that Φ_{p} is concave if p > 2).

This would lead to the inequality

∥Φ_{p}(x_{1}, x_{2},…)∥ ⩾ Φ_{p}(∥x_{1}∥, ∥x_{2}∥,…),

where x_{1}, x_{2}, … are elements of a Banach lattice whose 2-concavity constant is eual to one. Further remarks are at the end of the paper.

2. Definitions and examples

We denote by Δ(p, q), where 1 ⩽ p ⩽ q ⩽ +∞, the class of (real) Banach lattices X such that

∥(|u|^{p} + |v|^{p})^{1/p}∥ ⩽ (∥u∥^{p} + ∥v∥^{p})^{1/p}  (4)

and

∥(|u|^{q} + |v|^{q})^{1/q}∥ ⩾ (∥u∥^{q} + ∥v∥^{q})^{1/q}  (5)

for all u, v ∈ X. In other words, X is in Δ(p, q) if it is p-convex and q-concave and its p-convexity and q-concavity constants are equal to one. It is clear that Δ(1, ∞) is just the class of all Banach lattices. And by [3], Proposition 1.d.5, Δ(p, q) is contained in Δ(r, s) for r ⩽ p ⩽ q ⩽ s. In particular, L^{q} ∈ Δ(r, s) if r ⩽ q ⩽ s, a fact which can easily be verified by direct calculations.

It was proved by Figiel [1] (see also [3], pp. 80–81) that if X ∈ Δ(p, q) with p > 1 and q < +∞, then X is uniformly convex in the sense that

δ_{X}(ε) = inf \{ 1 − ∥\frac{x + y}{2} ∥ : ∥x − y∥ = ε, ∥x∥ = ∥y∥ = 1 \} > 0

for ε > 0. The function δ_{X} is called the modulus of convexity of X. Let δ_{p} denote the modulus of convexity of L^{p}, dim(L^{p}) ⩾ 2. (It follows from [2] that δ_{p} is independent of a particular choice of L^{p}.) As noted in Introduction, the following fact follows immediately from (1) and (4).

Corollary 1. If X ∈ Δ(p, 2) (in particular, X = L^{q} for 2 ⩾ q ⩾ p), then inequality (2) holds.

As noted in Introduction, this implies the following

Corollary 2. If X ∈ Δ(p, 2), then δ_{X}(ε) ⩾ δ_{p}(ε) (ε > 0).

Mixed normed spaces. For technical reasons we define only sequence spaces. Let 1 ⩽ r, s ⩽ 2. The space X = l^{r,s} consists of those scalar sequences x = {x_{j,k}}_{j,k=0}^{∞} such that

∥x∥ = { ∑_{j=0}^{∞}[ ∑_{k=0}^{∞} |x_{j,k}|^{s}]^{r/s}}^{1/r} < ∞.

It is not hard to show that l^{r,s} ∈ Δ(p, q), where p = min(r, s) and q = max(r, s). Hence, by Corollary 2, δ_{X}(ε) ⩾ δ_{p}(ε). Since l^{r,s} contains an isometric copy of l^{p}, we conclude that δ_{X} = δ_{p}.

Orlicz spaces. Let M be a convex, strictly increasing function on the interval [0, ∞) with M(0) = 0. The space l^{M} consists of the scalar sequnces x = {x_{j}}_{0}^{∞} for which

∥x∥ = ∥x∥_{M} = inf\{ λ > 0 : ∑_{j=0}^{∞} M (\frac{|x_{j}|}{λ}) ⩽ 1 \} < ∞.

One can prove that l^{M} ∈ Δ(p, q) provided that the function M(t^{1/p}) is convex and the function M(t^{1/q}) is concave. Therefore, inequality (1) holds in l^{M} if the function M(t^{1/q}) is concave. Estimates for the moduli of convexity of Orlicz spaces can be found in [1].

3. Proofs

Our proof is based on the following lemma.

Lemma. Let F_{p} be defined by (3). Then, if 1 ⩽ p ⩽ 2, the function F_{p} is convex, and if p > 2, it is concave. In all the cases F_{p}(ξ, η) increases with ξ and η.

Before proving the lemma we use it to prove the theorem. Let x, y ∈ X, where X ∈ Δ(1, 2), and 1 ⩽ p ⩽ 2. Then

(|x + y|^{p} + |x − y|^{p})^{1/p} = ((|x| + |y|)^{p} + ||x| − |y||^{p})^{1/p}

(this is deduced from the case where x, y are real scalars, by using Theorem 1.d.1 of [3]) and we may assume that x ⩾ 0, y ⩾ 0. Assuming this we have

(|x + y|^{p} + |x − y|^{p})^{1/p} = F_{p}(x^{2}, y^{2})^{1/2}

(see [3], Theorem 1.d.1). Since F_{p} is convex, homogeneous and “increasing”, there is a set A ⊂ { (α, β) : α ⩾ 0, β ⩾ 0 } such that

F_{p}(ξ, η) = sup{ αξ + βη : (α, β) ∈ A },

whence F_{p}(x^{2}, y^{2})^{1/2} ⩾ (αx^{2} + βy^{2})^{1/2}, (α, β) ∈ A, and hence, by (5) with q = 2,

∥F_{p}(x^{2}, y^{2})^{1/2}∥ ⩾ (α∥x∥^{2} + β∥y∥^{2})^{1/2}

for all (α, β) ∈ A. Taking the supremum over (α, β) ∈ A we obtain

∥F_{p}(x^{2}, y^{2})^{1/2}∥ ⩾ F_{p}(∥x∥^{2}, ∥y∥^{2})^{1/2},

which concludes the proof.

Proof of Lemma. Let 1 < p ⩽ 2. (The case p = 1 is similar.) Since F_{p}(λξ, λη) = λF_{p}(ξ, η) for λ ⩾ 0, the convexity of F_{p} will follow from the convexity of the function f(t) = F_{p}(1, t), t > 0. To prove that f is convex observe first that f(t) = tf(1/t), whence f′′(t) = t^{−3}f′′(1/t) for t ≠ 1. And since f′(1) exists, it remains to prove that f′′(t) ⩾ 0 for 0 < t < 1. To prove this write f as

f(t) = g(t^{1/2})^{2/p}, g(t) = (1 + t)^{p} + (1 − t)^{p} (0 < t < 1).

We have

2pf′′(t) = t^{−2/3}g(t^{1/2})^{(2/p)−2}[(\frac{2}{p} − 1) g′(t^{1/2})^{2}t^{1/2}
+ g(t^{1/2})g′′(t^{1/2})t^{1/2} − g(t^{1/2})g′(t^{1/2})].

Hence, f′′(t) > 0 if and only if A(t) > 0, where

A(t) = \frac{1}{p} [(\frac{2}{p} − 1) g′(t)^{2}t + g(t)g′′(t)t − g(t)g′(t)]
= 4(p − 1)t(1 − t^{2})^{p−2} − [(1 + t)^{2p−2} − (1 − t)^{2p−2}].

If 3/2 ⩽ p ⩽ 2, the function φ(t) = (1+t)^{2p−2} −(1−t)^{2p−2} is concave and therefore

φ(t) ⩽ φ(0) + φ′(0)t = 4(p − 1)t ⩽ 4(p − 1)t(1 − t^{2})^{p−2},

which implies A(t) > 0. If 1 < p ⩽ 3/2, then

A′(t) = 4(p − 1)(1 − t^{2})^{p−3}[1 + (3 − 2p)t^{2}] − 2(p − 1)[(1 + t)^{2p−3} + (1 − t)^{2p−3}].

Since 0 ⩽ 3 − 2p ⩽ 1, the function t ↦ t^{3−2p} is concave, hence

\frac{(1 + t)^{2p−3} + (1 − t)^{2p−3}}{2} = \frac{1}{2} [(\frac{1}{1 + t})^{3−2p} + (\frac{1}{1 − t})^{3−2p}] ⩽ (1 − t^{2})^{2p−3}.

Hence

A′(t) ⩾ 4(p − 1)(1 − t^{2})^{2p−3}[1 + (3 − 2p)t^{2} − 1] ⩾ 0.

This implies A(t) ⩾ A(0) = 0, which concludes the proof in the case 1 < p ⩽ 2. If p > 2, proving that F_{p} is concave reduces to proving that A(t) ⩽ 0 (0 < t < 1). In this case the function φ is convex which implies that

φ(t) ⩾ φ(0) + φ′(0)t = 4(p − 1)t ⩾ 4(p − 1)t(1 − t^{2})^{p−2},

and this completes the proof.

Remark. The discussion of the case 1 < p ⩽ 2 can be made simplier. Namely, it is easy to see that the function g(t^{1/2}) is convex (0 < t < 1), which implies that f(t) = g(t^{1/2})^{2/p} is convex (since 2/p > 1). This trick can also be used if 2 < p < 3, because then the function g(t^{1/2}) is concave. However, if p > 3, g(t^{1/2}) is convex.

4. Dual results

Using the case p > 2 of Lemma one proves that if x, y ∈ X, where X ∈ Δ(2, ∞) (which means that X satisfies (4) with p = 2), then there holds the reverse of (1). A consequence is that the reverse of (2) is valid in every lattice of class Δ(2, p) (p > 2) and, in particular, in L^{q} for 2 ⩽ q ⩽ p. (The latter was proved in [4] by using the Riesz-Thorin interpolation theorem.) Combining this with Hanner’s results we see that if X ∈ Δ(2, p), then X is “more” uniformly convex than L^{p} (dimension ⩾ 2) in the sense that ρ_{X}(τ) ⩽ ρ_{p}(τ), where

ρ_{X}(τ) = sup \{ \frac{∥x + τy∥ + ∥x − τy∥}{2} − 1 : ∥x∥ = 1, ∥y∥ = 1 \} ,

and ρ_{p} = ρ_{L^{p}}. The function ρ_{X} is called the modulus of smoothness of X (see [3], Ch. 1, for further information).

REFERENCES

[1] T. Figiel, On the moduli of convexity and smoothness, Studia math. 56 (1976), 121–155.

[2] O. Hanner, On the uniform convexity of L^{p} and l^{p}, Arkiv f. Math 3:3 (1956), 239–244.

[3] J. Lindenstaruss and L. Tzafriri, Classical Banach Spaces II. Function Spaces, Ergebnisse 97, Berlin-Heidelberg-New York, Springer-Verlag 1979.

[4] M. Pavlović, Some inequalities in L^{p} spaces II, Mat. vesnik 10(23)(38) (1986), 321–326.

(received 05.03.1993)

Matematički fakultet, Studentski trg 16, 11000 Beograd, YUGOSLAVIA