One general method in operator theory

An order bounded operator with target a Dedekind complete vector lattice is determined up to an orthomorphism from the kernels of its strata. Some applications to 2-disjoint operators are briefly discussed.

A linear functional on a vector space is determined up to a scalar from its zero hyperplane. In contrast, a linear operator is recovered from its kernel up to a simple multiplier on a rather special occasion. Fortunately, Boolean valued analysis [1] prompts us that some operator analog of the functional case is valid for each operator with target a Kantorovich space, a Dedekind complete vector lattice. The present expository talk addresses some opportunities that are opened up along the lines of this rather promising approach.

Let X be a Riesz space, and let Y be a Kantorovich (or Dedekind complete Riesz) space Y with base a complete Boolean algebra B . Without loss of generality, we may assume that Y is a nonzero space embedded as an order dense ideal in the universally complete Kantorovich space R^ which is the descent of the reals R inside the separated Boolean valued universe V ^(B) over B (cp. [1, Theorem 5.2.4]).

We further let X ^л stand for the standard name of X in V ^(B) . Clearly, X ^л is a Riesz space over R ^л inside V ^(B) . Denote by l := T f the ascent of T to V ^(B) . Clearly, l acts from X ^л to the ascent Y f of Y in the sense of the Boolean valued universe V ^(B) . Therefore,

l(xл) = Tx inside V(B) for all x G X, which means in terms of truth values that

[ l : X ^л ^ R ]] = 1,   ( V x G X) [ l(x ^л ) = Tx]] = 1.

Since l is defined up to a scalar from ker(l) , we infer the following analog of the Sard theorem.

Theorem 1. Let S and T be linear operators from X to Y . Then ker(bS) D ker(bT) for all b G B if and only if there is an orthomorphism a on Y such that S = aT .

We see that a linear operator T is in sense determined up to an orthomorphism from the family of the kernels of the strata bT of T . This remark opens a possibility of studying some properties of T in terms of the kernels of the strata of T .

Clearly, T is a Riesz homomorphism if and only if so is its ascent l = T f . Since the ascent of the sum is the sum of the ascents of the summands, we reduce the proof of Theorem 2 to the case of the functionals.

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S. S. Kutateladze

From now on we will consider an order bounded operator T : X ^ Y. Straightforward calculations of truth values show that T+f = l+ and T-f = l- inside V(B). Moreover, [ker(l) is a Riesz subspace of Xл] = 1 whenever so are ker(bT) for all b G B. Indeed, given x, y G X, put b := [Tx = 0л] л [Ty = 0л].

This means that x,y G ker(bT) . Hence, we see by condition that bT ( x V y) = 0 . In other words,

[Tx = 0 ^л] Л [Ty = 0 ^л] 6 [ T(x V y) = 0 ^л].

Whence

[ker(l) is a Riesz subspace of X^л]

= [( V x, y G X ^л )(l(x) = 0 ^л Л l(y) = 0 ^л ^ l(x V y) = 0 ^л )]

= A [l(x ^л ) = 0 ^л Л 1(у ^л ) = 0 ^л ^ l((x V у) ^л ) = 01 = 1. x,y ∈ X

Recall that a subspace H of a Riesz space X is a G -space or Grothendieck subspace (cp. [2, 3]) provided that H enjoys the following property:

( V x, y G H ) (x V y V 0 + x Л y Л 0 G H ).

By analogous calculations of truth values we infer that

[ker(l) is a Grothendieck subspace of X^л]

= [ ( V x, y G X ^л )(l(x) = 0 ^л Л l(y) = 0 ^л ^ l(x V y V 0 + x Л y Л 0) = 0 ^л )]

= ^A [l(x ^л )=o ^л Л 1(у ^л )=0 ^л ^ l((x V y V 0 + x Л y Л 0) ^л ) = 0 ^л J.

x,y ∈ X

Assuming that the kernel of each stratum bT is a Grothendieck subspace, take x, y ∈ X and put b := [Tx = 0л] Л [Ty = 0л].

This means that x,y G ker(bT) . By hypothesis bT (x V y V 0 + x Л y Л 0) = 0 . In other words,

[Tx = 0 ^л] Л [ Ty = 0 ^л] 6 [ T(x V y V 0 + x Л y Л 0) = 0 ^л].

It follows now that

[ ker(l) is a Grothendieck subspace of X 1 = 1.

By way of example, we may now assert that the following theorems appear as the descents of their scalar analogs.

Theorem 2. An order bounded operator T from X to Y may be presented as the difference of some Riesz homomorphisms and only if the kernel of each stratum bT of T is a Riesz subspace of X for all b ∈ B .

Theorem 3. The modulus of an order bounded operator T : X ^ Y is the sum of some pair of Riesz homomorphisms if and only if the kernel of each stratum bT of T with b ∈ B is a Grothendieck subspace of the ambient Riesz space X .

One General Method in Operator Theory

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To prove the relevant scalar claims, we use one of the formulas of subdifferential calculus: Theorem 4 (of decomposition) . Assume that H 1 , . . . , H N are cones in a Riesz space X.

Assume further that f and g are positive functionals on X . The inequality f (h1 V • • • V hN) > g(h1 V • • •V hN)

holds for all hk E Hk (k := 1,..., N) if and only if to each decomposition of g into a sum of N positive terms g = gi + • • • + gN there is a decomposition of f into a sum of N positive terms f = fi + • • • + fN such that fk(hk) > gk(hk) (hk E Hk; k :=1,...,N).

Remark 1. The complete proofs of Theorems 2 and 3 are given in [4, 5]. Theorem 4 appeared in this form in [6].

Remark 2. Note that the sums of Riesz homomorphisms were first described by S. J. Bernau, C. B. Huijsmans, and B. de Pagter in terms of n -disjoint operators in [7]. A survey of some conceptually close results on n -disjoint operators is given in [8, § 5.6].