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Владикавказский математический журнал 2011, Том 13, Выпуск 1, С. 38-43
УДК 512.555+517.982
A BECKENBACH-DRESHER TYPE INEQUALITY IN UNIFORMLY COMPLETE /-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 complete /-algebras is given.
Mathematics Subject Classification (2000): 06F25, 46A40.
Key words: /-algebra, vector lattice, lattice homomorphism, positive operator.
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 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) the f-algebra of continuous functions on with polynomial growth; i. e., p £ B(R+) if and only if p £ C(RN) and there are n £ N and M £ R+ satisfying |p(t)| < M (1 + w(t))n (t £ RN), where t := (t1,...,tN), w(t):= |ti| + ... + | tN | and 1 is the function identically equal to 1 on RN. Denote by B0(RN) the set of all functions in B(RN) vanishing at zero. Let A (RN) stands for the set of all p £ B(RN) such that lima^0 a-1p(at ) exists uniformly on bounded subsets of RN. Evidently, A (RN) C B0(RN). Finally, let H(RN) denotes the set of all continuous positively homogeneous functions on RN.
Lemma 1. The sets B(R N), B0(RN), and A (RN) are uniformly complete f -algebras with respect to pointwise operations and ordering. Any p £ A (RN) admits a unique decomposition p = p1 + wp2 with p1 £ H(RN) and p2 £ B0(RN), i■ e.
A (RN) = H (RN) e wB0(R N).
Moreover, p1(t) = p'(0)t := limaj0 a-1p(at) for all t £ RN.
< See [3, Lemma 4.8, Section 5]. >
© 2011 Kusraev A. G.
2°. Consider an f-algebra E. Denote by H(E) the the set of all nonzero R-valued lattice homomorphisms on E and by Hm(E) the subset of H(E) consisting of multiplicative functional. We say that u G H(E) is singular if u(xy) = 0 for all x,y G E. Let Hs(E) denotes the set of singular members of H(E). Given a finite tuple x = (xi,..., xN) G EN, denote by ((x)) := ((x1,... ,xN)) the f-subalgebra of E generated by {x1,... , xN}.
Definition. Let E be a uniformly complete f-algebra and x1,...,xn G E+. Take a continuous function p : R+ — R. Say that the element </?(x1,..., xN) exists or is well-defined in E provided that there is y G E satisfying
u(y) = p(u(X1),..., u(XN)) (u G Hm(((X1,...,XN ,y))), (^
u(y) = ^1(u(X1), . . . ,u(XN)) (u G HS(((X1, . . . ,xn,y))),
cp. [3, Remark 5.3 (ii)]. This is written down as y = (p(x1,... , xN).
Lemma 2. Assume that E is a uniformly complete f -algebra and x1,..., xN G E+, and x:= (x1,..., xN). Then x(p) := </?(x1,..., xN) exists for every p G A(RN), and the mapping x : p — x(p) = <x(x1 , ...,xN) is the unique multiplicative lattice homomorphism from N) to E such that dj(x1,..., xN) = x^ for all j := 1,..., N. Moreover, X(A(RN)) =
((X1, . . . ,XN)).
< Take p G A (RN). In view of Lemma 1 p = p1 + wp2 with p1 G H (R+), p2 G and w(t) = |t1| + ... + |tN|. For x G E denote by X G Orth(E) the multiplication operator y — xy (x G E). According to [5, Theorem 3.3] and [8, Theorem 2.10] we can define correctly p1 (x1;... ,xN) in E and p2(X 1;... ,XN) in Orth(E), respectively. Now, it remains to put p(x1;..., xN) := p1(x1,..., xN) + p2(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 p G A(Rn), see [3]. >
Lemma 3. Assume that p G A (RN) is convex. Then for all x := (x1,...,xn ) G EN, y := (y1;..., yN) G En, and n, p G Orth(E)+ with n + p = we have p(nx + py) ^ np(x) + pp(y), where nx := (nx1;... ,nxN). The reverse inequality holds whenever p is concave.
< Let L be the order ideal generated by ((x1 ,...,yN)). Clearly, L is an f-subalgebra of E. If n0 := n|L and p0 := p|L then n0, p0 Orth(L). For any u G H(L) there exists a unique u G Hm(Orth(L)) such that u(nx) = u(n)u(x) for all x G L and n G Orth(L), [3, Proposition 2.2 (i)]. If u is nonsingular then au is multiplicative for some a > 0 [3, Corollary 2.5 (i)], and thus we may assume without loss of generality that u G Hm(L). By using (1), the convexity of p, and the relation u(n) + u(p) = 1 we deduce
u(x(nx + py)) = p(u(n0)u(x) + u(p0)u(y)) ^ u(n0)p(u(x)) + u(p0)p(u(y)) = u(n0)u(px(x)) + u(p0)u(px(y)) = u(np(x) + pp(y)),
where u(x) := (u(x1),..., u(xN)). If u is singular then by above definition we have u(p(x)) = u(p1(x)), u(p(y)) = u(p1(y)), and u(p(nx + py)) = u(pX1(nx + py)). At the same time p1 is sublinear, since it coincides with the directional derivative of the convex function p at zero, see Lemma 3. Thus, by replacing p by p1 in the above arguments we again obtain u(p(nx + py)) ^ u(np(x) + pp(y)). It remains to observe that every u0 G H(((x1,... ,xN))) admits an extension to u G H(L) and thus H(L) separates the points of ((x 1,... ,xN)). >
Lemma 4. If p G A (RN) is isotonic, then p is also isotonic, i. e. x ^ y implies X(x) ^ p(y) for all x, y G EnN. (The order in EN is defined componentwise.)
< Follows immediately from the above definition (1). >
3°. Everywhere below (G, +) is a commutative semigroup, while E is a uniformly complete f-algebra and f1;..., fN : 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) n-1 exists in Orth(E) for every n £ F(u),
(ii) F(u) + F(v) C F(u + v) - Orth(E) + for all u, v £ G, and
(iii) the infimum (the supremum) of {np(n-1f(u)) : n £ F(u)} exists in E for each u £ G, where f(u):= (f1(u),... ,fN(u)) £ Ef and n-1f(u):= (n-1 f1(u),..., n-1 fN(u)).
Lemma 5. Given a function p : 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 W(n-1 f(u))}, (h(u) := sup (np(n-1f(u))} ). (2)
neF(u) v neF(u) 7
< By 3 (i) and Lemma 2 p (n 1f (u)) exists in E and by 3 (iii) g and h are well defined. > 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) ^ 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 p £ A(R+) is increasing and convex;
(2) h is superadditive whenever f1;..., fN are superadditive and p £ A (R+) is increasing and concave.
< 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 n £ F(u) and p £ F(v). By 3 (ii) we can choose a £ F(u + v) with a ^ n + p. In view of 3 (i) n, p, and a are invertible. Taking subadditivity of f : G ^ En and some elementary properties of orthomorphisms into account we have
a-1 f (u + v) ^ a-1 (f(u) + f (v)) ^ na-1 (n-1 f (u)) + pa-1 (p-1f(v)).
Putting t := a — n — p and making use of Lemmas 3, 4 and 5 we deduce
g(u + v) ^ ap(a-1f(u + v)) ^ ap(na-1 (n-1 f(u))) + pa-1 (p-1f(v) + Ta-10)
^ n<^(n-1 f(u)) + pp(p-1 f(v)) + a-1 tp(0) = n<^(n-1 f(u)) + p<^(p-1 f(v)).
By taking infimum over n £ F(u) and p £ F(v) we come to the required inequality. >
Remark 1. Suppose that the hypotheses of 3 (i-iii) are fulfilled for some fixed u, v £ G. Then the inequality g(u + v) ^ 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) = {f0(x)} (x £ G) with f0 : G ^ Orth(E)+ we have the following particular case of the above Theorem, see [8].
Corollary 1. Suppose that f1;..., fN are subadditive, f0 : G ^ Orth(E)+ is superadditive, and f0(u) is invertible in Orth(E) for every u £ G. Then, given an increasing continuous convex function p £ A(R+), the Peetre-Persson inequality holds:
The reverse inequality holds in (3) whenever f0,f1;...,fN are superadditive, and p 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 < a ^ 1 or a < 0, then for all x1,...,xn £ E we have
' N \ 1/a \ ( N \ 1/a
ENaJ Ef(wrj . (4)
The reverse inequality holds if f : E+ — F is superlinear and a ^ 1.
< The function 0a(t) = (if + ... + if)1/a (t £ Rf) is superlinear if 0 < a < 1 and sublinear if a ^ 1. In case a < 0 we define 0a(t) = 0 whenever t1 ■ ... ■ tN = 0 and then
is superlinear on int(Rf). In all cases £ H (R+) and (4) follows from the generalized
Jensen inequality in vector lattices, see [4, Theorem 5.2] and [7, Theorem 4.2]. >
Let A and B be uniformly complete unitary f-algebras, while E C A is a vector sublattice.
For every x £ A+ and 0 < p £ R the p-power xp is well defined in A, see [3, Theorem 4.12].
If x £ A+ is invertible and p < 0, then we can also define xp := (x-1 )-p. It can be easily seen
that u(xp) = u(x)p for any u £ Hm(A0) with an f-subalgebra A0 C containing x. Assume
i
that R : E —> B is a positive operator. Given x £ A with xp £ E, we define Rp(x) := R(xp)p. This definition is sound provided that x is invertible in A and R(xp) is invertible in B.
Lemma 7. If p ^ 1 and x1;..., xN £ A+ are such that xp",..., xf £ E and (x1 + ... + xN)p £ E, then the inequality holds:
Rp(x1 + ... + xn) ^ Rp(x) + ... + Rp(XN). (5)
The reversed inequality is true whenever p ^ 1, p = 0. (In case p < 0 the positive elements x » and R(xP) are assumed to be invertible in A.)
< Denote Ui := x1-, a:= 1/p, and observe that (uf + ... + u%) a = (pa(ui,..., un) where 0a(u1;..., uN) is understood in the sense of homogeneous functional calculus. In particular,
{x\ + ... + xn)p = (uf + ... + ufj) a £ 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(x i + ...+xN) = + ... + = (R((pa(ui,...,uN)))a
with uf £ E replaced by x» and making use of Rp(xj) = R(u»)a (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,
again by Lemma 6, we have i?((u" + ... + u%)<*) ^ (R(u\)a + ... + R(uN)a)a and rising
i. . i
both sides of this inequality to the ath power we get the reversed inequality (5). >
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 x1,...,xn £ A+ such that xf,xf, (£N=1 xj)a, (£N=1 x»)^ £ E (i := 1,...,N). If p ^ 1,
ß ^ 1 ^ a, ß = 0, and T(xß) are invertible in Orth(F) whenever ß < 0, then
^ Z-, ^^(p-l)//?' W
xi
(T((Eil,x,)ß^(p-1)/ß fei (T(xft)(
< Put G = E, f (x) := f(x) := S(xa)1/a, fo(x) := T(xß)1/ß, N = 1, and p(t) = in Corollary 1. By Lemma 7 f is subadditive, f0 is superadditive, and f0(x») is invertible in Orth(F). Moreover, p e 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 p(t) = 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 p e ) in Peetre-Persson's inequality (3) and thus the reversed
inequality is true in (6) whenever 0 < p ^ 1, a, ß ^ 1, and a, ß = 0.
References
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AnATOLY G. Kusraev Southern Mathematical Institute Vladikavkaz Science Center of the RAS, Director RUSSIA, 362027, Vladikavkaz, Markus street, 22 E-mail: kusraev@smath.ru
НЕРАВЕНСТВО ТИПА БЕККЕНБАХА - ДРЕШЕРА В РАВНОМЕРНО ПОЛНЫХ f-АЛГЕБРАХ
А. Г. Кусраев
Установлено общее неравенство типа Беккенбаха — Дрешера в равномерно полных /-алгебрах.
Ключевые слова: /-алгебра, векторная решетка, решеточный гомоморфизм, положительный оператор.
