A Beckenbach - Dresher type inequality in uniformly complete f-algebras

An easy modification of the continuous functional calculus on unitary f -algebras as defined in [3] makes it possible to translate the Fenchel–Moreau duality to f -algebra setting and to produce some envelope representations results, see [8]. This machinery, often called quasilinearization (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(R N ) the f -algebra of continuous functions on R N with polynomial growth; i. e., y E B(R N ) if and only if y 6 C (R + ) and there are n E N and M E R + satisfying | y ( t ) | 6 M ( 1 + w ( t )) ⁿ ( t E R N ), where t := ( ti,..., I n ), w ( t ):= | ti | + ... + \ 1 n | and 1 is the function identically equal to 1 on R N . Denote by B o (R ^ ) the set of all functions in B(R ++ ) vanishing at zero. Let A (R N ) stands for the set of all y E B(R N ) such that lim _a^ o a ^{- 1} y ( a t ) exists uniformly on bounded subsets of R N . Evidently, A (R ++ ) C B ₀ (R N ). Finally, let H (R ++ ) denotes the set of all continuous positively homogeneous functions on R N .

Lemma 1. The sets B(R N ) , B o (R N ) , and A (R N ) are uniformly complete f -algebras with respect to pointwise operations and ordering. Any y E A (R N ) admits a unique decomposition y = yi + wy ₂ with yi E H (R N ) and y ₂ E B o (R N ) , i. e.

A (R + ) = H (R + ) ® wB o ( R + ) .

Moreover, ^i ( t ) = y ⁰ (0) t := lim _a^ o a ^{- 1} y ( a t ) for all t E R N .

C See [3, Lemma 4.8, Section 5]. B

2 ° . Consider an f -algebra E. Denote by H ( E ) the the set of all nonzero R-valued lattice homomorphisms on E and by H _m ( E ) the subset of H ( E ) consisting of multiplicative functionals. We say that ω ∈ H ( E ) is singular if ω ( xy ) = 0 for all x, y ∈ E . Let H _s ( E ) denotes the set of singular members of H ( E ). Given a finite tuple x = ( x 1 , . . . , x N ) ∈ E ^N , denote by hh x ii := hh x 1 , . . . , x N ii the f -subalgebra of E generated by { x 1 , . . . , x N } .

Definition. Let E be a uniformly complete f -algebra and x 1 , . . . , x N ∈ E + . Take a continuous function y : R N ^ R. Say that the element p( x1,..., x n ) exists or is well-defined in E provided that there is y ∈ E satisfying

ω ( y ) = ϕ ( ω ( x1 ) , . . . , ω ( x N )) ( ω ∈ H m ( hh x1, . . . ,x N ,y ii ) , ω ( y ) = ϕ1 ( ω ( x1 ) , . . . , ω ( x N )) ( ω ∈ H s ( hh x1, . . . , x N , y ii ) ,

cp. [3, Remark 5.3 (ii)]. This is written down as y = p(x ₁ ,..., x n ).

Lemma 2. Assume that E is a uniformly complete f -algebra and x1 , . . . , x N ∈ E+, and x := ( x ₁ ,..., x n ) . Then x( ^ ) :=

( x ₁ ,..., x n ) exists for every y G A (R N ) , and the mapping x : y ^ x( ^ ) = ^ ( x1 ,...,x n ) is the unique multiplicative lattice homomorphism from A (R N ) to E such that dt j ( x ₁ ,..., x n ) = x j for all j := 1 ,..., N . Moreover, x(A(R N )) = hh x1 , . . . , x N ii .

C Take ϕ ∈ A (R ₊ ^N ). In view of Lemma 1 ϕ = ϕ1 + wϕ2 with ϕ1 ∈ H (R ^N ₊ ), ϕ2 ∈ B 0 (R ₊ ^N ), and w ( t ) = | t1 | + . . . + | t N | . For x ∈ E denote by x ˙ ∈ Orth( E ) the multiplication operator y → xy ( x ∈ E ). According to [5, Theorem 3.3] and [8, Theorem 2.10] we can define correctly p 1 ( x ₁ ,..., x n ) in E and p ₂ ( x 1,..., x n ) in Orth( E ), respectively. Now, it remains to put

x₁,..., xn) :=

₁(x₁,..., xn) +

₂(x 1,..., xn)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 (R^N), see [3]. B

Lemma 3. Assume that ϕ ∈A (R₊^N) is convex. Then for all x := (x1, . . . , xN) ∈E^N, y := (y₁,..., yN) G EN, and n, p G Orth(E)₊with n + p = Ie we have p(nx + py) 6 n

x) + pp(y), where nx := (nx₁,..., nxN). The reverse inequality holds whenever p is concave.

C Let L be the order ideal generated by hhx1 , . . . , yN ii. Clearly, L is an f -subalgebra of E. If π0 := π|L and ρ0 := ρ|L then π0, ρ0 Orth(L). For any ω ∈H(L) there exists a unique l G H_m(Orth(L)) such that ^(nx) = e(n)w(x) for all x G L and n G Orth(L), [3, Proposition 2.2 (i)]. If ω is nonsingular then αω is multiplicative for some α > 0 [3, Corollary 2.5 (i)], and thus we may assume without loss of generality that ω ∈H_m(L). By using (1), the convexity of p, and the relation w(n) + w(p) = 1 we deduce

w(p(nx + py)) = ^(ш(по)ш(х) + w(po)w(y)) 6 w(no)p(w(x)) + w(po)y(w(y))

= .ein, ^'uGx n + ш^Ц^у'У) = ^;пр;х! + p^y)), where w(x) := (l(x1 ),..., l(xn)). If l is singular then by above definition we have w(p(x)) = w(

Lemma 4. If ϕ ∈A (R₊^N ) is isotonic, then p is also isotonic, i. e. x 6 y implies <p(x) 6 <b(y) for all x, y G EN. (The order in E^Nis defined componentwise.)

C Follows immediately from the above definition (1). B

3°. Everywhere below (G, +) is a commutative semigroup, while E is a uniformly complete f-algebra and f 1,..., Jn : G ^ E+. Let P (M) stands for the power set of M. Assume that some set-valued map F : G ^ P(Orth(E)₊) meets the following three conditions:

• (i)    п^-1exists in Orth(E) for every п E F(u),

• (ii)    F(u) + F(v) C F(u + v) — Orth(E)₊for all u,v E G, and

• (iii)    the infimum (the supremum) of {п(?(п^-1f(u)) : п E F(u)} exists in E for each u E G, where f(u): = (f1(u),... , fn(u)) E EN and п^-1f(u): = (п^-1fi(u),... ,п^-1Jn(u)).

Lemma 5. Given a function ( : A(RN) and a set-valued map F : G ^ P(Orth(E)₊) satisfying 3 (i–iii), we have the operator g : G → E (h : G → E) well defined as

• g^(u) := inf {ⁿb(^{n-1 f(u)})},   fh^(u):= sup {ⁿn^-1f^(u))}).          ⁽²⁾

ⁿ'^{(u)                            X}           nOF (u)                       '

C By 3 (i) and Lemma 2 ( (n^-1f (u)) exists in E and by 3 (iii) g and h are well defined. B

4°. Now we are able to state and prove our main result. A function g : G ^ F is said to be subadditive if g(u + v) 6 g(u) + g(v) for all u, v ∈ G and superadditive if the reversed inequality holds for all u, 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 ( E A(R+) is increasing and convex;

• (2)    h is superadditive whenever f₁,..., Jn are superadditive and ( E A(R+) is increasing and concave.

C We restrict ourselves to the subadditivity of g . The superadditivity of h 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) with σ > π + ρ. In view of 3 (i) π, ρ, and σ are invertible. Taking subadditivity of f : G → EN and some elementary properties of orthomorphisms into account we have a-1 f(u + v) 6 a-1 (f(u) + f(v)) 6 па-1 (п-1 f(u)) + pa-1 (p-1f(v)).

Putting τ := σ - π - ρ and making use of Lemmas 3, 4 and 5 we deduce

g(u + v) 6 a<p(a^-1f(u + v)) 6 ар(па^-1(п^-1f(u))) + pa^-1(p^-1f(v) + та^-10)

6 пр(п^-1f(u)) + pp(p^-1f(v)) + a^-1Tp(0) = п(?(п^-1f(u)) + pp(p^-1f(v)).

By taking infimum over π ∈ F (u) and ρ ∈ F (v) we come to the required inequality. B

Remark 1. Suppose that the hypotheses of 3 (i–iii) are fulfilled for some fixed u, v ∈ G. Then the inequality g(u + v) 6 g(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 map F : G ^ P(E₊) satisfies the condition 3 (i-iii) with Orth(E) replaced by E.

• 5°.    For a single-valued map F(x) = {fo(x)} (x E G) with fo : G ^ Orth(E)₊we have the following particular case of the above Theorem, see [8].

Corollary 1. Suppose that f₁,..., Jn are subadditive, fo : G ^ Orth(E)₊is superadditive, and fo(u) is invertible in Orth(E) for every u E G. Then, given an increasing continuous convex function ( E A(RN), the Peetre-Persson inequality holds:

f(u + v)                  f(u)                  f(v)

^fo^(u + ^v)(    .     .   6 ^fo^(u)(          + ^fo^(v)( -TTF .             ⁽³⁾

V fo(u + v)/           Vfo(u)/           Vfo(v)/

The reverse inequality holds in (3) whenever fo,f₁,..., fn are superadditive, and ( is an increasing concave function.

Remark 3. The above theorem in the particular case of E = R was 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. Let E and F be uniformly complete vector lattices, f : E+ ^ F an increasing sublinear operator. If either and 0 < а 6 1 or а < 0, then for all xi,..., xn G E we have f (XNa)    6 X f (lxii)a

The reverse inequality holds if f : E₊^ F is superlinear and а > 1.

• <1    The function ф_a(t) = (t“ + ... + tN )^i/a(t G RN) is superlinear if 0 < а < 1 and sublinear if а > 1. In case а < 0 we define ф_а(t) = 0 whenever ti • ... • tN = 0 and then ф_а is superlinear on int(RN). In all cases ф_а G H (RN) and (4) follows from the generalized Jensen inequality in vector lattices, see [4, Theorem 5.2] and [7, Theorem 4.2]. B

Let A and B be uniformly complete unitary f -algebras, while E ⊂A is a vector sublattice. For every x G A+ and 0 < p G R the p-power xp is well defined in A, see [3, Theorem 4.12]. If x G A+ is invertible and p < 0, then we can also define xp := (x^-i)^-p. It can be easily seen that w(x^p) = w(x)^pfor any ш G H_m(Ag) with an f-subalgebra Aq C containing x. Assume

that R : E → B is a positive operator. Given x ∈A with x^p∈E, we define R_p(x) := R(x^p) ^p. This definition is sound provided that x is invertible in A and R(x^p) is invertible in B.

Lemma 7. If p > 1 and xi,..., xn G A+ are such that xp,..., xN G E and (xi + ... + xN )^p∈E, then the inequality holds:

Rp(xi + ... + xn) 6 Rp(x) + ... + Rp(xN).

The reversed inequality is true whenever p 6 1, p — 0. (In case p < 0 the positive elements xi and R(x_i^p) are assumed to be invertible in A.)

• <    Denote ui :— xp, а :— 1/p, and observe that (u“ + ... + uN)^а= Ф_а(u_i, • • •, un) where ф_а(ui,..., un) is understood in the sense of homogeneous functional calculus. In particular, (xi + ... + xn)^p= (u? + ... + uN) ^аG E for every p = 0. We need consider three cases. If p > 0 then by applying Lemma 6 to the right-hand side of the equality

Rp(xi + . . . + xn) — R( (u^a+ . . . + uN) “ ) — (R(Фа(U1, . . . , uN)))

with u_i^α∈E replaced by xi and making use of R_p(xi) = R(ui)^α(i := 1, . . . , N), we arrive 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, 1¹

again by Lemma 6, we have R((ua + ... + uN) “) 6 (R(u_i)^a+ ... + R(un)^a) ^aand rising both sides of this inequality to the аth power we get the reversed inequality (5). B

• 7°.    Now, we can deduce a generalization of one more Beckenbach-Dresher type inequality due to Peetre and Persson [11].

Corollary 2. Let S : E → F and T : E → Orth(F) be positive operators. Take xi,..., xn G A+ such that xa,xe, (PN=1 xi)a, (PN=1 xi) G E (i :— 1,..., N). If p > 1,

β 6 1 6 α, β 6= 0, and T (x_i^β) are invertible in Orth(F) whenever β < 0, then

(S ((EN=1 Xi)^a))p/^a      * s '

(t((e^Xi)^e))^(p-1)/e6 i=1 (T(хв))^(p-1)/e.

C Put G = E, f (x) := f(x) := S(x^a)^1/a, f₀(x) := T(х^в)^1/в, N = 1, and ^(t) = tp in Corollary 1. By Lemma 7 f is subadditive, fo is superadditive, and fo(xi) is invertible in Orth(F). Moreover, ^ G A(R+) is convex and increasing whenever p > 1. Now, the desired inequality is deduced by induction. B

Remark 4. If 0 < p < 1 then the concave function ^(t) = tp is not in A(R+) and we cannot guarantee the reversed inequality in (6). Nevertheless, in the case that F is a unitary f-algebra one can take ^ G B(RN) in Peetre-Persson’s inequality (3) and thus the reversed inequality is true in (6) whenever 0 < p 6 1, α, β 6 1, and α, β 6= 0.