THE SPACE OF HENSTOCK INTEGRABLE FUNCTIONS OF TWO VARIABLES 15

We consider the space of Henstock integrable functions of two variables. Equipped with the Alexiewicz norm the space is proved to be barrelled. We give a partial description of its dual. We show by an example that the dual can't be described in a manner analogous to the one-dimensional case, since in two variables there exist functions whose distributional partials are measures and which are not multipliers for Henstock integrable functions.

written for the value of the integral, if for every e > 0 there exists a positive I 0 R such that if {((x i,yi ),I i) i 1,2 n} (2) is a partition of I 0 (i.e., Ii's are nonoverlapping subintervals of I 0 whose union is I O) for which (xi,Yi) li A((xi 'yi)'6(xi,yi) ), (3) where A((a,b),r) stands for the disk centered at (a,b) of radius r, then n Z f(xi,Yi) %(Ii)-f f(x,y)dxdy < , (4) i=l I 0 where %(li) denotes the area of I..i We will write H for the class of Henstock integrable functions on I O. H is a linear space.If we replace %(Ii), for I i [ai,bi] x [ci,di], in (4) by g(ai,ci) g(ai,di) g(bi,ci) + g(bi,di), for a certain g I 0 +, then we obtain the definition of the Henstock integral of f with respect to g, written as fl 0 [dg.
Henstock integral in the plane is fully discussed in [7]. 2. DEFINITION.
We will call (6) the Alexiewicz norm on H.

PROPOSITION.
T e H if and only if there is a finite signed Borel measure on (O,l] The norm of T is equal to the norm of .
PROOF.Let C be the space of continuous real-valued functions on I 0.

Define
Thn if we assign H is mapped isomorphically into a dense subset of C O (since every polynomial is the indefinite Henstock integral of its second mixed partial).Thus, we can identify A function g:I 0 -]R is a multiplier for H if for every f e H we have also fg ell.
In the one-dimensional case the dual of the space of Henstock-integrable functions is given by the class of multipliers (see [6]).The multipliers are functions whose distributional derivatives are measures.The two-dimensional case is different.
In [4] Kurzweil defines g:l 0 + to be of strongly bounded variation if for every x,g(x,-) is of bounded variation, for every y,g(-,y) is of bounded variation, and n M(g) ci,di]  where sup is taken over all partitions {of 10) {Ii}i=I, (10) so that every g of strongly bounded variation is a continuous linear functional on H.
consisting of non-overlapping, nondegenerate closed intervals.Then he shows that functions of strongly bounded variation are multipliers for H, and for f e H, g of The connection between this result and Proposition 3 is not known.It is not known either if functions of strongly bounded variation and those equivalent to them are the only multipliers.

EXAMPLE.
There exists a function g I 0 R whose distributional partials are measures and which is not a multiplier.Define g(x,y) x-y for x _> y, (12) 0 otherwise Note that Krickeberg shows in [2] that g I 0 R has its distributional partials being measures if and only if it is of bounded variation in the sense of Tonelli.
For g, var g(.,y) l-y var g(x,.) and So g is of bounded variation in the sense of Tonelli.
Define for n > 2 fn is equal to 0 on the boundary of K n, nonnegative on Ln and fL fn(X y)dxdy (15) n 3 and f (x,y) 0 for every (x,y) e K such that Ix-yl < n-Then for f given by n n f (x,y) for (x,y) e K for some n > 2, O n n f(x,y) otherwise. ( we have f e H, yet fg % H. 7.

REMARK.
It is shown in [8] that the space of Henstock integrable function of one variable is barrelled.We will show it to be true also in two dimensions.

DEFINITION.
If E is a topological vector space then a set B E is a barrel if B is closed, convex, circled and radial at O. A locally convex space in which every barrel is a neighborhood of 0 is termed a barrelled space.It should be noted that each barrel in a space E which is of the second category in itself is necessarily a neighborhood of O.In particular, every Banach space is barrelled.
The importance of barrelled spaces lies in the following Barrel Theorem.

THEOREM.
Let be a barrelled space and F be a pointwise bounded family of continuous linear functions on into a locally convex space K. Then the family F is equi- continuous.Consequently, in this case, F is uniformly bounded on each bounded subset of E.
This theorem implies in particular that the Banach-Steinhaus Theorem holds for barrelled spaces.Let X be a subspace of S satisfying the following two conditions: (a) If F e X and J I0, and Fj(I) F(I n J) (18) for I I 0 then Fj X; (b) I O, F e S, and Fj X for every J I 0 such that if 1,2 are the vertical and the horizontal line segments through c then J n i ' J n 2 ' then F e X.Then X is barrelled.R 2 PROOF.In the proof we will denote for z I z 2 e by [Zl,Z2] an interval for which Zl,Z 2 are opposite vertices.Let B be a barrel in X.If B is not a neighborhood of zero, then it is nowhere dense.To show that, suppose that a barrel is not nowhere dense.There is an open set U such that U B. Since is convex and circled (U-U) (B-B) U U is a neighborhood of zero, and so is .
For every I I 0 write and Then B(1) is a barrel in X(1).Consequently, B(II) +...+ B(I n B(I).The space X(I) is a topological n direct sum of X(II) X(In )" If (I I) ,(In are neighborhoods of zero in )<(II) X(I (respectively) then B(1) is a neighborhood of zero in X(I).Thus, n if (I) is nowhere dense in X(I) then at least one of (Ii)'s, i 1 n, is nowhere dense in the corresponding X(li).
Therefore, if we divide I O into four subintervals by splitting the sides into halves, among so obtained intervals there is at least one, call it Ii, such that (Ii) is nowhere dense in X(II).Applying the same procedure to Ii, and then continuing it, we obtain a sequence of intervals I such that n nNIn {c}.where c is a certain point in I O, and B(In is nowhere dense in X(In for every n E N.
For every n E N, write parallel to its sides and going through c.We can assume that li's are numbered n so that (25) for every n and i.Notice that since B(I is nowhere dense in X(I for every n, n n there is at least one i such that B(Ii)" is nowhere dense in x(Ii).
n n Consider the four sequences {Ii} for i 1,2,3,4 If in each of them n n N there is only finitely many n N such that (I) is nowhere dense in x(Ii)n then after passing those finitely many indices we will get all four (li), I 1,2,3,4, n being neighborhoods of zero.This will force B(I to be a neighborhood of zero, n a contradiction.Therefore, among the four sequences {I i) there has to be one n n N which produces infinitely many B(Ii)'s which are nowhere dense in the corresponding n x(Ii)'s.n i 0 i 0 Let {In }n e N be that sequence, and let {Ink} k e N be its subsequence such i 0 i 0 i 0 that B(Ink) is nowhere dense in X(In for every k e N. Write Jk In k for k N, k and let Jk [C'Xk]- Let u 1 x 1.There exists a function G 1 e X(J 1) such that G 1 8 and llGll < I/2.Then since 8 is closed and lim G I G I (in X) there is a Proceeding by induction, if n e N, then we have a function Gn e X(Jk such that Gn { n and IIGnl < 1/2 n.Since is closed and c) + f(b,d)Notice that if F e S then there is a unique f e C, such that f(x,y) A defined as the closed convex hull of the sequence {F n in S. Every element of A is of the form + F= Z F n=l n n for some sequence of scalars { with Z < i.Take a u [Cl,Ul], u c, 1980 MATHEMATICS SUBJECT CLASSIFICATION.PRIMARY 46EI0.SECONDARY 26A39.
with DEFINITION.