2011, Том 13, Выпуск 1, С. 38–43
УДК512.555+517.982
A BECKENBACH–DRESHER TYPE INEQUALITY IN UNIFORMLY COMPLETEf-ALGEBRAS
A. G. Kusraev
To the memory of Gleb Akilov on the occasion of the 90th anniversary of his birth
A general form Beckenbach–Dresher inequality in uniformly completef-algebras is given.
Mathematics Subject Classification (2000): 06F25, 46A40.
Key words:f-algebra, vector lattice, lattice homomorphism, positive operator.
An easy modification of the continuous functional calculus on unitary f-algebras as de-fined in [3] makes it possible to translate the Fenchel–Moreau duality tof-algebra setting and to produce someenvelope representations results, see [8]. This machinery, often called quasi-linearization (see [2, 9]), yields the validity of some classical inequalities in every uniformly complete vector lattice [4, 5]. The aim of this note is to give general forms of Peetre–Persson and Beckenbach–Dresher inequalities in uniformly complete f-algebras.
The unexplained terms of use below can be found in [1] and [6].
1◦. We need a slightly improved version of continuous functional calculus on uniformly complete f-algebras constructed in [3, Theorem 5.2].
Denote by B(RN+) thef-algebra of continuous functions onRN+ with polynomial growth; i. e., ϕ ∈ B(RN
+) if and only if ϕ ∈ C(RN+) and there are n ∈ N and M ∈ R+ satisfying
|ϕ(t)|6M(1+w(t))n (t∈RN
+), where t:= (t1, . . . , tN),w(t) :=|t1|+. . .+|tN|and 1is the function identically equal to 1 onRN+. Denote by B0(RN+) the set of all functions in B(RN+) vanishing at zero. LetA(RN
+) stands for the set of allϕ∈B(RN+) such that limα↓0α−1ϕ(αt)
exists uniformly on bounded subsets of RN+. Evidently, A(R+N) ⊂ B0(RN+). Finally, let H(RN+) denotes the set of all continuous positively homogeneous functions onRN+.
Lemma 1. The sets B(RN+), B0(RN+), and A(RN+) are uniformly complete f-algebras with respect to pointwise operations and ordering. Anyϕ∈A(RN+)admits a unique decom-positionϕ=ϕ1+wϕ2 withϕ1 ∈H(RN+) and ϕ2 ∈B0(RN+), i. e.
A(RN
+) =H(RN+)⊕wB0(RN+).
Moreover, ϕ1(t) =ϕ′(0)t:= limα↓0α−1ϕ(αt) for all t∈RN.
⊳See [3, Lemma 4.8, Section 5]. ⊲
c
2◦.Consider an f-algebraE. Denote byH(E) the the set of all nonzeroR-valued lattice homomorphisms on E and byHm(E) the subset of H(E) consisting of multiplicative func-tionals. We say thatω ∈H(E) is singular if ω(xy) = 0 for all x, y∈E. Let Hs(E) denotes the set of singular members of H(E). Given a finite tuplex= (x1, . . . , xN)∈EN, denote by hhxii:=hhx1, . . . , xNii thef-subalgebra of E generated by{x1, . . . , xN}.
Definition. Let E be a uniformly complete f-algebra and x1, . . . , xN ∈ E+. Take a continuous functionϕ:RN+ →R. Say that the elementϕ(xb 1, . . . , xN)exists or iswell-defined inE provided that there is y∈E satisfying
ω(y) =ϕ(ω(x1), . . . , ω(xN)) (ω ∈Hm(hhx1, . . . , xN, yii), ω(y) =ϕ1(ω(x1), . . . , ω(xN)) (ω∈Hs(hhx1, . . . , xN, yii),
(1)
cp. [3, Remark 5.3 (ii)]. This is written down asy=ϕ(xb 1, . . . , xN).
Lemma 2. Assume that E is a uniformly complete f-algebra and x1, . . . , xN ∈E+, and
x:= (x1, . . . , xN). Thenbx(ϕ) :=ϕ(xb 1, . . . , xN)exists for everyϕ∈A(RN+), and the mapping b
x : ϕ 7→ bx(ϕ) = ϕ(xb 1, . . . , xN) is the unique multiplicative lattice homomorphism from
A(RN+) to E such that dtbj(x1, . . . , xN) =xj for all j:= 1, . . . , N. Moreover, bx(A(RN+)) = hhx1, . . . , xNii.
⊳Takeϕ∈A(RN+). In view of Lemma 1ϕ=ϕ1+wϕ2 withϕ1 ∈H(RN+),ϕ2 ∈B0(RN+), and w(t) =|t1|+. . .+|tN|. For x ∈E denote by ˙x∈Orth(E) the multiplication operator y 7→ xy (x ∈ E). According to [5, Theorem 3.3] and [8, Theorem 2.10] we can define correctlyϕb1(x1, . . . , xN) inEandϕb2( ˙x1, . . . ,xN˙ ) in Orth(E), respectively. Now, it remains to
putϕ(xb 1, . . . , xN) :=ϕb1(x1, . . . , xN) +ϕb2( ˙x1, . . . ,x˙N)w(x1, . . . , xN) and check the soundness of this definition. Closer examination of the proof can be carried out as in the case of ϕ∈A(RN), see [3].⊲
Lemma 3. Assume that ϕ ∈ A(RN+) is convex. Then for all x:= (x1, . . . , xN) ∈ EN, y:= (y1, . . . , yN) ∈ EN, and π, ρ ∈ Orth(E)+ with π +ρ = IE we have ϕ(πxb +ρy) 6 πϕ(x) +b ρϕ(y),b where πx:= (πx1, . . . , πxN). The reverse inequality holds whenever ϕ is concave.
⊳ Let L be the order ideal generated by hhx1, . . . , yNii. Clearly, L is an f-subalgebra ofE. Ifπ0:=π|Land ρ0:=ρ|L thenπ0, ρ0Orth(L). For anyω∈H(L) there exists a unique e
ω∈Hm(Orth(L)) such thatω(πx) =ω(π)ω(x) for alle x∈Landπ ∈Orth(L), [3, Proposition 2.2 (i)]. Ifωis nonsingular thenαωis multiplicative for someα >0 [3, Corollary 2.5 (i)], and thus we may assume without loss of generality that ω∈Hm(L). By using (1), the convexity ofϕ, and the relation ω(π) +e ω(ρ) = 1 we deducee
ω(ϕ(πxb +ρy)) =ϕ ω(πe 0)ω(x) +ω(ρe 0)ω(y)6ω(πe 0)ϕ(ω(x)) +ω(ρe 0)ϕ(ω(y))
=ω(πe 0)ω(ϕ(x)) +b ω(ρe 0)ω(ϕ(y)) =b ω(πϕ(x) +b ρϕ(y)),b
whereω(x) := (ω(x1), . . . , ω(xN)). Ifωis singular then by above definition we haveω(ϕ(x)) =b ω(ϕb1(x)), ω(ϕ(y)) =b ω(ϕb1(y)), and ω(ϕ(πxb +ρy)) = ω(ϕb1(πx+ρy)). At the same time
ϕ1 is sublinear, since it coincides with the directional derivative of the convex function ϕat
zero, see Lemma 3. Thus, by replacing ϕ by ϕ1 in the above arguments we again obtain
ω(ϕ(πxb +ρy))6ω(πϕ(x) +b ρϕ(y)). It remains to observe that everyb ω0∈H(hhx1, . . . , xNii)
admits an extension toω ∈H(L) and thus H(L) separates the points of hhx1, . . . , xNii.⊲
Lemma 4. If ϕ ∈ A(RN+) is isotonic, then ϕb is also isotonic, i. e. x 6 y implies
b
ϕ(x)6ϕ(y)b for all x,y∈EN
+. (The order in EN is defined componentwise.)
3◦.Everywhere below (G,+) is a commutative semigroup, whileEis a uniformly complete f-algebra andf1, . . . , fN :G→E+. Let P(M) stands for the power set ofM. Assume that
some set-valued map F :G→P(Orth(E)+) meets the following three conditions: (i) π−1 exists in Orth(E) for every π ∈F(u), inequality holds for allu, v ∈G.
Theorem. Suppose that the operators g, h:G→E are defined as in(2). Then:
(1) g is subadditive whenever f1, . . . , fN are subadditive and ϕ ∈ A(RN+) is increasing
and convex;
(2)his superadditive wheneverf1, . . . , fN are superadditive andϕ∈A(RN+)is increasing
and concave.
⊳We restrict ourselves to the subadditivity ofg. The superadditivity ofh can be proved in a similar way. Take u, v ∈ G and let π ∈ F(u) and ρ ∈ F(v). By 3 (ii) we can choose σ ∈F(u+v) withσ>π+ρ. In view of 3 (i)π,ρ, andσare invertible. Taking subadditivity of f :G→EN and some elementary properties of orthomorphisms into account we have
σ−1f(u+v)6σ−1 f(u) +f(v)6πσ−1 π−1f(u)+ρσ−1 ρ−1f(v). Remark 1. Suppose that the hypotheses of 3 (i–iii) are fulfilled for some fixed u, v ∈G. Then the inequality g(u+v)6g(u) +g(v) (h(u+v)>h(u) +h(v)) holds.
Remark 2. An f-algebra E can be identified with Orth(E) if and only if E has a unit element. Thus, above theorem remains true if E is a uniformly complete unitary f-algebra and the mapF :G→P(E+) satisfies the condition 3 (i–iii) with Orth(E) replaced byE.
5◦.For a single-valued map F(x) ={f
0(x)} (x∈ G) with f0 :G→ Orth(E)+ we have
the following particular case of the above Theorem, see [8].
Corollary 1. Suppose thatf1, . . . , fN are subadditive,f0 :G→Orth(E)+ is
superaddi-tive, andf0(u)is invertible inOrth(E)for everyu∈G. Then, given an increasing continuous
Remark 3.The above theorem in the particular case ofE =Rwas obtained by Persson [12, Theorems 1 and 2], while Corollary 2 covers the “single-valued case” by Peetre and Persson [11]. A short history of the Beckenbach–Dresher inequality is presented in [13]. Some instances of the inequality are also addressed in [9, 10].
6◦.We need two more auxiliary facts. First of them is a generalized Minkowski inequality. Lemma 6.LetEandF be uniformly complete vector lattices,f :E+ →F an increasing
sublinear operator. If either and0< α61or α <0, then for all x1, . . . , xN ∈E we have
Jensen inequality in vector lattices, see [4, Theorem 5.2] and [7, Theorem 4.2].⊲
LetAandBbe uniformly complete unitaryf-algebras, whileE ⊂Ais a vector sublattice. For every x∈A+ and 0< p∈R thep-power xp is well defined in A, see [3, Theorem 4.12].
Ifx∈A+ is invertible andp <0, then we can also definexp:= (x−1)−p. It can be easily seen
that ω(xp) = ω(x)p for any ω ∈ H
m(A0) with an f-subalgebra A0 ⊂ containing x. Assume
thatR :E→B is a positive operator. Givenx∈Awithxp ∈E, we defineRp(x) :=R(xp)1p.
This definition is sound provided thatx is invertible inA and R(xp) is invertible inB. Lemma 7. Ifp >1 and x1, . . . , xN ∈ A+ are such that xp1, . . . , xpN ∈E and (x1 +. . .+
xN)p∈E, then the inequality holds:
Rp(x1+. . .+xN)6Rp(x) +. . .+Rp(xN). (5)
The reversed inequality is true wheneverp61,p6= 0. (In casep <0the positive elementsxi and R(xpi) are assumed to be invertible inA.)
⊳Denote ui:=xpi,α:= 1/p, and observe that uα1 +. . .+uαN
1
α =φ
α(u1, . . . , uN) where φα(u1, . . . , uN) is understood in the sense of homogeneous functional calculus. In particular, (x1+. . .+xN)p = uα1 +. . .+uαN
1
α ∈E for every p6= 0. We need consider three cases. If
p>0 then by applying Lemma 6 to the right-hand side of the equality
Rp(x1+. . .+xN) =R immediately at the desired inequality (5). The same arguments involving the reversed version of (4) leads to the reversed inequality in (5) whenever 0< p <1. Finally, in the case p <0, again by Lemma 6, we have R (uα1 +. . .+uαN)α1
6 R(u1)α+. . .+R(uN)α
1
α and rising
both sides of this inequality to theαth power we get the reversed inequality (5). ⊲
β 616α,β 6= 0, and T(xβi) are invertible in Orth(F) whenever β <0, then
S PNi=1xi
αp/α
T PNi=1xi
β(p−1)/β 6 N
X
i=1
S xαip/α
T xβi(p−1)/β. (6)
⊳ Put G = E, f(x) := f(x) := S(xα)1/α, f
0(x) := T(xβ)1/β, N = 1, and ϕ(t) = tp in
Corollary 1. By Lemma 7 f is subadditive, f0 is superadditive, and f0(xi) is invertible in
Orth(F). Moreover, ϕ∈A(R+) is convex and increasing whenever p>1. Now, the desired inequality is deduced by induction.⊲
Remark 4. If 0 < p < 1 then the concave function ϕ(t) = tp is not in A(R+) and we cannot guarantee the reversed inequality in (6). Nevertheless, in the case thatF is a unitary f-algebra one can takeϕ∈B(RN+) in Peetre–Persson’s inequality (3) and thus the reversed inequality is true in (6) whenever 0< p 61,α, β61, andα, β 6= 0.
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Anatoly G. Kusraev
Southern Mathematical Institute
НЕРАВЕНСТВО ТИПА БЕККЕНБАХА — ДРЕШЕРА В РАВНОМЕРНО ПОЛНЫХf-АЛГЕБРАХ
А. Г. Кусраев
Установлено общее неравенство типа Беккенбаха — Дрешера в равномерно полныхf-алгебрах.
