A N N A L ES S O C IE T A T IS M A T H E M A T IC A E P O LO N A E Series I : C O M M E N TA TIO N E S M A T H E M A T IC A E X X I I I (1983) R O C Z N IK I P O L S K IE G O T O W A R Z Y S T W A M A TEM A TY C ZN EG O
Séria I : P R A C E M A T E M A T Y C Z N E X X I I I (1983)
Z. K ominek (Katowice)
Some remarks on the set classes A% and B ÿ
Introduction. The problem of the continuity of convex (in the Jensen’s sense) and additive functions is very important in the theory of such functions. Let A a B N be a convex open domain in the real Ж-space B N.
A function /: A->B is called convex iff f( b (a + y)) < i (f {x )+ f(y ))
holds for all x, y e A. A function/: RN->R is called additive iff it satisfies Cauchy’s functional equation
f{ x -f y) = /(a?) + f{y ) for all x ,y e RN.
In [5] the set classes
A qand
Bqwere introduced as follows:
A set T c B N belongs to the class A q iff every convex function /: A->R, T c A, such that the restriction f\T is continuous, is continuous in A.
A set T c B N belongs to the class B q iff every additive function /: such that the restriction f\T is continuous, is continuous.
Some informations on these classes may be found in [2 ], [3], and [5].
1. Let p, q be positive integers such that p-\-q = N. We shall con
sider the space B N as the product
B N = R p x B ? .
We have the following fact ([1], p. 216) : Every additive function/: B N^ B bas a unique representation
/(») =/i(® i)+/2W i
where x = {xx, x2), xx e B p, x2 eB ? and f v f 2 are additive functions.
For a set T c B N we denote by Tp and Tq the projections of the set T on the spaces B p and BA, respectively.
T
h e o r e m1. 1/ T
cB n belongs to the class A q
{ Bq),then Tp e Af, (Tp
e B c ) and Tq e Aqc (Tq e Щ ).
Roczniki
PTM — Prace Matematyczne t. X X III50 Z. K o m in e k
P roo f. Without loss of generality we may assume that 0 g T. Suppose that Tp does not belong to the class Af, (for example). There exists a dis
continuous convex function f : Ap^-B, Ap a Tp, such that the restriction f\T is continuous. Putting
g (x): — f{ x i, 0) for x e Ap x B Q,
we note that g is convex and discontinuous function. On the other hand, the restriction g\TpXTq continuous and therefore it is continuous in Ap x B!\
because Tp x T q ^> T and T belongs to the class A
q. The proof in the case of the class B
qis similar.
T heorem 2. Let T — Tp x Tq. I f Tp belongs to the class B
qand Tq belongs to B% then T belongs to B
q.
P ro o f. If/ : B N ^-B is an additive and f\T is continuous, then by (1) and from the fact that Tp and Tq belong to the classes B% and B
q, resp.?
f x and /2 are continuous. Thus / is continuous, and hence T belongs to B q . T heorem 3. Let T = T\x ... x T x . I f T\,i = 1, . .. ,A , belong to the class A
q, then T belongs to the class A
q.
P ro o f. Induction. Let, for every subsystem {A1, ..., Ak) <= [T\, ...
. .. , T f } , 2 < h < Ж, the set A 1 x ... x A k belong to the class A q and let {Zl, ..., Zk+l) c. {T\, . . . r T x } be an arbitrary subsystem such that f\z ix...xzk+l continuous, where/: A->B, A => Z1 x ... x Z k+1, is a convex function. Take a1, bl e ZÀ, a1 < bl, i = 1, . .. , h + 1 . Then for each xk+1 e Zk+1 the function f { ' , x k+1)\z \x...xzk is continuous and therefore it is bounded on the interval [(a1, . .. , ak), lb1, . .., bk)]. We shall show that / is upper bounded on the interval [(a1, . .. , ak, ak+l), lb1, ..., bk, 6&+1)]. By induction’s assumption, / is bounded on every dimensional face on this interval.
JSTow we see that
i(F r[(o S ak+1), (bS bt+1)]+Fr[(<i\ a‘ +1), (b\ i t+I)])
= [(a1, - . ® * * 1)
and hence/ is upper bounded on the interval [(a1, . .., ak+1), (b1, . .. , 6fc+1)], because it is a convex function. Applying the theorem of Ostrowski [6]
which says that every convex function/: A-^B, A <= B N, upper bounded on the set T c A of a positive Lebesgue measure is continuous, we have that Z1 x ... x Z k+1 belongs to the class A q +1. This completes the proof.
2. Intuitively, the sets of the classes A q and B q are “large”. It may be interesting that there exists a set T с В such that T + T = В and T $ B q (see [3]) as well as there exists a set T belonging to B lc such that T\ {£„}, for some t0
gT, does not belong to B
q. For example, the set
T : = H u Q n (0 ,1 ],
51
where И is a Hamel basis of the space В (i.e. a basis of the space R over the field Q of all rationale), H a (0,1], 1 e Я , as' was proved in [2], belongs to jBlG and T\ {h}, where h e H \ {1} does not belong to B q (every set be
longing to B q contains a Hamel basis [3]).
T heorem 4. The set o f all cluster points o f a set belonging to the class B q belongs to the class B q .
P roo f. If it is not true, then there exists a discontinuous additive function f : R N^-B such that f\Td is continuous (here^TÆ denotes the set of all cluster points of the set T). But T \ T d is the set of isolate points, thus the restriction f\T is continuous, too. So / is continuous and we have a contradiction.
Hence and from the fact that every set belonging to A q contains a Hamel basis it follows
B e m ark 1. I f T belongs to B q , then the set of all cluster points of the set T contains a Hamel basis.
3
.It is easily seen that if I 7belongs to
Aq ( Bq),then every translation I 7-fa, a e B N, belongs to
Aq ( Bq).Let F : B N->BN be a linear transform
ation, i.e.
F (со) = 21r for x e B N,
where 21 is a two-dimensional matrix of range N. We say that F is non
singular whenever the matrix 21 is non-singular (i.*e. det 21 Ф 0 ).
T heorem 5. I f F is non-singular transformation, then the sets T and F (T) belong to the same classes A q or B q .
P roo f. Suppose that T e
Aq(T e
Bq)and F ( T ) $ A q (В (Т )ф В ^ ).
Then there exists a discontinuous convex (additive) function / defined on a convex open domain A c B N (defined on B N), F (T ) a A, such that the restriction f\F^T) is continuous. Taking
(2) g(x): = f[F (x )], x e F ~ 1(A )( x e B n ),
we observe that F ~ 1(A) is a convex open domain in B N (F~l (BN) = B N) and the restriction g\T is continuous. The function g is convex (additive) so g is continuous, because T e
Aq(T e
Bq).Hence, by (2), / is continuous in A and this contradiction finishes the proof.
Bern ark 2. The set o f all real numbers can be represented as a denumer
able union o f the sets which do not belong to
Bq.In fact, if f is an additive function such that f(B ) = Q and Tr :
= {x e B :f( x ) = r} for r e Q, we see that for every r, Tr does not belong to B la and u r , = « .
reQ
Bern ark 3. The set o f all reals can be represented, as a union o f c dis
joint, congruent sets from the class A lc .
Set classes A
qand B
q52 Z. K o m in e k
In [4 ] M. Kuczma proved that if Я is a Hamel basis,
H* = Jак x = J ? Àaha, ha e H, Xa integral, the sums are finitej, and
Q = {#: x = Xaha, ha e H, Xa eQ n [0 ,1 ], the sums are fin itej, then
В = U (Я* + x) • and (Я* + ж)п(-Н* + 2/) — 0 for x , y e Q , x Ф у .
xeO
