THE UNIFORM BOUNDEDNESS PRINCIPLE FOR ORDER BOUNDED OPERATORS

Under appropriate hypotheses on the spaces, it is shown that a sequence of order bounded llnear operators which is poJntwJse order bounded is uniformly order bounded on order bounded subsets. This result is used to establish a Banach-Stelnhaus Theorem for order bounded operators.


INTRODUCTION
In this note we consider the problem of obtaining a version of the classical Uniform Boundedness PrlncJple of functional analysis for linear operators between vector lattices.
If X is a Banach space and Y is a normed linear space, the Uniform Boundedness Principle then asserts that any sequence (Tj) of continuous linear operators from X into Y which is poJntwise bounded on X is such that the sequence of operator norms (,Till) is bounded {[1] t4).It is easy to see that the condition that the (llTi.)are bounded is equivalent to the condition that the sequence (T i) Js uniformly bounded on bounded subsets of X {see the discussion in [1] 14).Thus, a possible version of the Uniform Boundedness Principle for order bounded linear operators (Ti) between vector lattices X and Y might be that if (Tj) is pointwise order bounded on X, then (Ti} is uniformly order bounded on order bounded subsets of X.
We will show below in Example that such d straightforward analogue of the Uniform Boundedness Principle does not hold even Jf both X and Y are Banach lattices.However, by Imposing additional conditions on C. SWARTZ the spaces and by employing the matrix methods of [1], we will obtain an order version of the Uniform Boundedness Principle in Theorem 3 below.Using the Uniform Boundedness Principle, we also establish a version of the Banach-Stetnhaus Theorem for order bounded operators which generalizes a result of Nakano.

RESULTS
First, conslder the following example which shows that a straightforward analogue of the Uniform Boundedness Principle does not ho]d for order bounded linear operators between Banach lattices.
(A linear operator T between vector lattices or Riesz spaces X and is order bounded If T carries order bounded subsets of X into order bounded subsets of , where a subset A of a vector lattice X order bounded if there is an order Interval we conform to the notation and terminology of [5]).)EXAMPLE [Tkf [fjJej e c 0 (note ([fj[) e c o by the Rlemann-Lebesgue Lemma).j=l However, (Tk} is not uniformly order bounded on order bounded subsets of X since if Vk(t) sin kt, then (Fk) is order bounded in X (IFkl I) but (Tl(i)} (el/2) is not order bounded in c O Note that both X and Y in this example are Banach lattices and both are Dedekind complete.
In order to obtain our version of the Uniform Boundedness Principle for ordered spaces, we first obtain a matrix theorem which is the analogue for ordered spaces of the matrix result given in [I] 2.1.Throughout the remainder of this note we let X and Y denote Riesz spaces (vector lattices), h sequence (x k) in X Is u-convergent to x, where u O, if there exists a scalar sequence tk40 such that IXk x tku; we write u lim x k x.The element u is called a convergence regulator for (Xk).The sequence {x k) is reltivel niformly convergent to x if (x k) is u-convergent to x for some convergence regulator u O; we write r llm x k x.
We now prove our Uniform Boundedness Principle for order bounded operators.
Recall that if Y is Dedekind complete, then a linear map T X Y is order bounded if and only if T is regular ([5] VIII.2.2).THEOREM 3. Let g be Dedekind o-complete and let Y be Dedeklnd complete and have an order unit u.
If T 1 X Y is a sequence of order bounded linear maps which is pointwise order bounded on X, then (T i) is uniformly order bounded on order bounded subsets of g.

PROOF.
If the conclusion fails, there is an interval I-w, w] in X such that (Tt([-w, w]) e ) is not order bounded.Thus, for each there exist x i e [-w, w] and m i such that TmtX l 14j-u, u].For notational convenience, assume that m I i.Then 14j-u, u]. (2.1)

Tlx1
Now consider the matrix M [(1/t)Tl(xj/J2)].For each J, (Tl(xj/j2)) 1 is order bounded so the jth column of M is relatively uniformly convergent to O.The sequence (xj/J 2) is relatively uniformly convergent to 0, and since each T t is sequentially continuous with respect to relative uniform convergence ([5] VIII .2), the I th row of M is also relatively uniformly convergent to 0. Thus, the rows and columns of M are u-convergent to O.By Lemma 2 there is an increasing sequence of positive integers (pi) such that I(1/pi)Tpi(Xpj/pj2) 2-l-Jr for # J. Again for notational convenience, we assume that Pi i.Since X is -complete and Ixjl w, the series j/j2 is absolutely order convergent to an element x X ([5] IV. 9); moreover, this series is actually w-convergent in X n since Ix xj/j21 Ixjl/J2 ( 1/J2)w.From the continuity of T where we have used the order completeness of Y to insure that the series on the right hand side of (2) are convergent.Both terms on the right hand side of (2) are order bounded.Hence, there exists k such that (1/t)lTl(xi/t2) k[-u, u] for all t.Putting 1 k contradicts (1) and the result Is established.
Note that the range space c O in Example is Dedeklnd complete, but does not have an order unit.
It is perhaps worthwhile noting that if (T1) is uniformly order bounded on order bounded subsets of X, then the sequence of modull (ITII) also has this property.For if 0 x u and ITt[0, u] w, then ITllx sup(lTizl 0 z $ x) w ([5] VIII.2) so ITil[0, u] From Theorem 3 we can also obtain an order analogue of the equicontlnuity conclusion of the classical Uniform Boundedness Principle.Recall that if (T 1) is a sequence of continuous linear operators from a normed space X into a normed space Y, then (T i) is equlcontlnuous if and only if (TI) is uniformly bounded on bounded subsets of X if and only if Tix llm Tix 0 uniformly in 1 whenever xj 0 ([1] 4.5).It is easy to establish J order analogues of these equlcontlnuity conditions for operators which satisfy the conclusion of Theorem 3.  We give order analogues of these conditions below and consider the relationship between them.If for each a e A, {xia is a sequence in X, we say that ia is u-convergent to 0 uniformly for a A if there exists a sequence tl0 such that IXla ttu for all a A; we write u ltm Xia 0 uniformly for a e A.
PROPOSITION 4. Let T X Y be order bounded.Consider 1 (i) (T1) is uniformly order bounded on order bounded subsets of X.
(11) if r llm xj O, then v llm Tlx J 0 uniformly for e for some v e Y, v > O, (111) r 11m Tlx 0 whenever r Jim x 0.
Let u >_ O, u e X.By the boundedness property, it suffices to show that if (Tkl) is a subsequence, Ixi <_ u and tl0, then r lim tlTklX I O. But, since u lim tixi O, this follows from (iii).Theorem 3 gives sufficient conditions for the equlcontlnulty condition (I) and, therefore, (ll) and {lli), to hold.
As in the classical case we can apply the Uniform Boundedness Principle given in Theorem 3 to obtain a Banach-Stelnhaus type result for order bounded operators ([1] 5.1).
There is an order version of the Banach-Steinhaus Theorem for linear functlonals due to Nakano glven in [5] IX. 1.1, and a general form of the Banach-Steinhaus Theorem for operators between vector lattices given in [4].The results in [4] treat a different class of operators than that considered below In Corollary 5 and the assumptions on the spaces are not as restrictive.COROLLARY 5.
Let X, Y be as in Theorem 3 and let T X-, Y be order I bounded.If 0 lim Tix Tx exists for each x e X, then T X Y Is an order bounded linear operator.PROOF.
For each x, (TlX) is order bounded since the sequence is order convergent.
Therefore, from Theorem 3, (T i) is uniformly order bounded on order bounded subsets of X.
If [u, v] is an order interval in X, then there is an order interval I-w, w] in Y such that Ti([u, v])c_ [-w, w] for all i.Hence, T{[u, v]) c I-w, w] and T is order bounded.
The sequence of operators (T i) in Example is pointwlse order convergent to the operator T X Y given by Tf fjej.IIowever, T is not order bounded j=l since (TFj) (ej/2), where Vj(t) sln//Jt.This shows that in general the /J)lTi(x)l + 2-iv (2.2)