A N N A L E S S O C I E T A T I S M A T H E M A T I C A E P O L O N A E S e r i e s I : C O M M E N T A T I O N E S M A T H E M A T I C A E X V I I ( 1 9 7 3 )
R O C Z N I K I P O L S K I E G O T O W A R Z Y S T W A M A T E M A T Y C Z N E G O S é r i a I : P R A C E M A T E M A T Y C Z N E X V I I ( 1 9 7 3 )
Ma r e k Ku c z m a (Katowice)
On some set classes occurring in the theory oî convex functions
Introduction. Let A c= RN be a convex open domain in the real A-space R N. A function/: A -> R is called convex whenever the inequality
(0.1)
holds for all x, y e A. A function f: RN -> R is called additive if it satisfies Cauchy’s functional equation
(0.2) f(æ + y) = / ( ж ) + / ( у ) for all x, ye R n .
In [3] the set classes have been introduced as follows.
A set Г c Rn belongs to iff every convex function f : A - > R , s T с A c R n , which is bounded above on T , is continuous.
A set T c Rn belongs to iff every additive function /: RN -> R which is bounded above on T , is continuous.
A set T c Rn belongs to iff every additive function / : RN -> R which is bounded (bilaterally) on T, is continuous.
As M. E. Kuczma [4] has shown, we have
(0.3) s / N = &N,
whereas the class <£N is actually larger (obviously ^ n)'
In the present paper we are going to investigate the effect of some simple operations performed on the sets from the classes stfN, or from their complements s/'N, C€'N.
In the sequel we shall often appeal to the following fact established in [3].
0-1. I f Те ^ NJ then the set T spans over the field Q of rationals the whole space RN.
1. Linear transforms. We start with the following simple result.
1.1. Let T c= R n , be R n . I f an additive function is bounded (bounded above) on T , then it is also bounded {bounded above) on T + 6.
Proof. Let / : BN -> В be additive with
\f(x)I < M for xe T.
Any y e T -\-b may be written as x-\-b, xe T, whence by (0.2) f(y) = f ( æ + b) = f ( x ) + f ( b )
and
\ f ( y ) \ < M + \ f { b ) |.
T h u s/is bounded опТ + ü». For the upper-boundedness the proof is similar.
Let us note that an analogue of Theorem 1 . 1 for convex functions is not valid. To see this, take N = 1, A = (0, oo),/(a?) = аГ1, T = (1 , 2).
Then / satisfies (0.1) and is bounded on T, but not on T — 1 = ( 0 ,1 ) . However, taking into account relation (0.3), we obtain from Theorem
1 . 1 the following corollary:
1.2. Any set T a BN and its arbitrary translation are always in the same class s/ N, ^N, s/'N, ^N.
By a linear transform we understand any function F : BN —> BN of the form
(1.1) F(x) = +
where 51 is a real Ж x N matrix. Transform (1.1) is said to be non-singular whenever the matrix 5f is non-singular (i.e. det 5f ф 0).
1.3. I f F : BN BN is a non-singular linear transform, then, for every T c- BN, the sets T and F(T) are in the same class s#N, N, s#'N, <€'N.
Proof. In view of Theorem 1.2 we may assume that b in (1.1) is zero. Then F satisfies
(1.2) F(x + y) = F{x) + F{y) for X , y e BN.
Suppose that Те stfN, butF(T)^ s t N. Then there exists a discontin
uous additive function /: BN -> В and a constant M such that
(1.3) f ( x ) ^ M ÎOTXeF(T).
Take g(x) = f[F(x)). By (0.2) and (1.2) g is additive and is discontinuous, since F is non-singular. By (1.3) we have
g(x) < M for xe T ,
which contradicts the relation Те The proof in the other cases is similar.
But this time it is not true that if an additive function is bounded on a set T, then the same function is bounded on F(T). In order to exhibit a suitable example we take Ж = 1 and let R be a Hamel basis of the set В of real numbers over the field Q of rationals such that l e U. Fix an
Theory of convex functions 129
irrational h0e H and consider an additive function /: R -> R such that / ( 1 ) = 0 , /(Л0) = 1 .
(Any function/: H R can he extended to an additive function /: R- > R;
cf. [1].) Put
(1.4) F(x) = h0x.
Then we have f(x) = 0 for xeQ, whereas f(h0x) = x for xeQ. Thus / is bounded on Q, but not on F (Q), although F given by (1.4) is non-singular.
In the one-dimensional case, if an additive function / : R R is bounded (bounded above) on a set T, then it is also bounded (bounded above) on the set F(T) whenever in the linear transform
F(x) = ax + b
the coefficient a is rational (positive rational). This follows from the fact that, if / is additive, then for arbitrary qcQ and xe R (and, in more di
mensional case, for x e RN) we have f(Qx) = Qf(x)
([1], p. 32). However, in more dimensional case an additive function*/
bounded on a set T may be unbounded on F(T) even if all the entries of the matrix % in (1.1) are rational. For instance, take N = 2 and write x = (xx, x2). Let
f(x) = f ( x x, x 2) = X x ~ X 2 ,
F(x) = IF1(x1, x2) , F2 ( ^ 1 7 ^2)) ^ $2) j T = {x = (x±, x2) : xx — x2}.
Then
F(T) = {x = (xx, x2): x2 = \ x x]
and we have f(x) = 0 for xe T, whereas f(x) is unbounded on F(T). In the present case the matrix 51 has the form
Finally let us note the following fact.
1.4. I f F : RN -> RN is a singular linear transform, then for every set T cz Rn we have F(T)e ^'N cz s#'N.
Proof. Again we may assume that b in (1.1) is zero. Then F(R^) is a linear subspace of RN and, since F is singular, F(RN) ф RN. By Theorem 0.1 we have F(RN) e (^'N, whence also F(T) <= F(RN) belongs to V'N.
9 — K o c z n i k i P T M — P r a c e M a t e m a t y c z n e X V I I
2. Unions. We feel intuitively that the sets from the classes
are “large”, whereas the sets from the classes *N are “small”. This intuition is strengthhened by a theorem of Ostrowski [7], [6], [5] stating that all sets T c B N with positive Ж-dimensional inner Lebesgue measure belong to c= and, on the other hand, by the fact that all denumer
able sets belong to c since they do not span (over Q) the whole space B N.
The results of the present section show, however, that this intuition is deceiving.
2.1. The union of two sets from the class Ч,'# may belong to even if one of them is denumerable.
Proof. Let Я be a fixed Hamel basis of the space B N over Q and let us split Я into two disjoint, non-empty subsets:
Й = Е х и Е 2, E xc \E2 = 0 , Е ХФ 0 , Я2 # 0 .
Let T x and T 2be linear spaces over Q spanned by E x and E 2, respec
tively. By Theorem 0.1 the sets T x and T 2belong to ^ ’N.
Put T — Tx vj T 2and let f : B N -» В be an additive function bounded above on T. Take an arbitrary he E . Then ghe T for all ge Q xon the other hand, we have f ( g h ) = gf { h) . Since / is bounded above on T, we must have f ( h ) = 0. Consequently / vanishes on E , and hence on the whole
B N; in particular, / is continuous. This proves that Те s/ N.
If one of the sets Я* is denumerable, then so is also the correspond
i n g ^ .
2.2. The real line В can be represented as a union of c disjoint, congruent sets from the class A x.
Proof. Let Я be a fixed Hamel basis of В over Q and let E * be the set of all finite linear combinations of the elements of Я with integral coefficients :
Я* = \x : x = h e S , К integral}.
As has been shown in [2], E * e s / x.
Next, let Û be the set of all finite linear combinations of the elements of Я with rational coefficients from the interval [0 ,1 ]:
Q = {*: x = К е И , Q n [0,1)}.
Then д = и <Я*+*).
xe£3
The cardinal number of Q is (like that of Я) c. The sets E*-\-x are congruent and, by Theorem 1.2, all belong to A - It remains to show that for x , y e Q , x Ф y, we have
(2.1) (E* + x ) n{ E* + y) = 0 .
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Supposing the contrary, let u , v e H * be such that u-\-x = v Jr y, i.e.
(2.2) x — y = u — v.
Since H* is an additive group, (2.2) implies that x — yeH*. But x — y
= y^juaha with hae И and /uae Q n( — 1, 1). Since, on the other hand, all pa must be integral, necessarily pa = 0, i.e. x — y = 0 . This contradicts the assumption that x Ф y. Consequently (2.1) holds.
3. Products. Let p, q be positive integers such that p + q = N .
We shall consider the space BN as the product B N = Bp x B q,
and we shall represent the points x e B N in the form x — (xx, x 2) with xxe Bp, x2e Bq. Then we have the following fact ([1], p. 216):
3.1. I f f : BN -> В is additive, then
(3.1) f(x) = f { x x, x 2) = f1{x1) + f2{x2),
where the functions f x: Bp -> В and f 2: Rq В are additive.
Now we shall prove the following result.
3.2. Let Tx с= Bp, T2 c Bq and write T = T j X T 2 c BN. Then T belongs to \to if and only if T x e s / p and T2 e [Tx e <ёр and T2e #®].
Proof. Suppose that e.g. T xe stf'p . Then there exists a discontinuous additive function f x : Bp В bounded above on Tx. Then the function / : BN -> В defined by the formula
f(x) = f ( x x, x 2) = f x(xx)
is a discontinuous additive function bounded above on T. This proves that Те sf'N.
Now suppose that Те stf'N and let f: BN -> В be a discontinuous additive function bounded above on T. By Theorem 3.1 / may be written in form (3.1). At least one of the functions occurring on the right-hand side of (3.1), say f x, must be discontinuous. Keeping x2 fixed and letting xx range over Tx we see that f x is bounded above on Tx. Consequently Txe
If a set T c BN is not a product of lower-dimensional sets, the sit
uation becomes more complicated. For arbitrary xx e Bp write (3.2) T[xx] = {x2e B q: (xx,X2) e T ) .
Further, let Tx and T2 be the projections of the set T onto the spaces Bp and Bq, respectively.
3.3. Suppose that T a BN belongs to ^'N and that its projection T2 onto B9 is bounded. I f there exists an xxe Bp such that T[^x]e (cf. (3.2)), then the projection T x of T onto Bp belongs to %?'p.
P roof. By hypothesis there exists a discontinuous additive function / : BN В bounded on T. This function may be written in form (3.1).
For a fixed xX€ Bp such that T[ xx] € ^ q we have /a(®a) = / ( » i, ® a )-/i(® i),
and thus / 2 is bounded on T[ x x\, and consequently it is continuous. The
refore it is bounded on T2, since the latter set is bounded. Now, varying xlt x2 in such a manner that xx runs over the whole T x and (xx, x 2) e T we see from (3.1) that f x remains bounded. Moreover, f x must be discon
tinuous, since / is discontinuous and f2 is continuous. Thus f x is a discon
tinuous additive function bounded on T x, which proves that T xe ^'p . In the above theorem the assumption that T2 is bounded is essential what may be seen from the following example.
Take N — 2, p = g = 1, and let f x: В В be a discontinuous ad
ditive function. Put
T = {(a?i, x2) e B 2: xxe B , f x(xx) < x2 < f x(xx) + l } .
Then for every xxe В the set T[ xx] — {fi{^i),fi{^i) + l) belongs to &х in virtue of the theorem of Ostrowski [7]. Similarly the set Tx = В belongs to ^fx. On the other hand, the function f: В2 -> В given by
f ( Xx, X2) = X2- f x{Xx)
is a discontinuous additive function and is bounded on T : 0 < f ( X x, X2) < l for (xx, X2) e T , and consequently
In this example we have clearly T2 = B.
Let d be the Euclidean distance in BQ. In Theorem 3.3 the requi
rement of the boundedness of T2 may be replaced by the condition that the function
(3.3) D{xx) = sup d(x2, T [ x 1}) X2eB^
is bounded on T x. (This condition is certainly fulfilled if the set T \xf\
is dense in B9 for every xxe T x.) More exactly, we have the following result.
3.4. Suppose that T cz BN belongs to and that function (3.3) is bounded on a set TQ <= T x:
(3.4) D( xx) ^ M for ххеТ0,
where Tx is the projection of T onto Bp. I f there exists an xxe Bp such that then the set T0 belongs to (6,p .
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Proof. The proof runs similarly as that of Theorem 3.3, only we must change the argument showing that is bounded on T0, since now f2 need not be bounded on T2. Let К c Rq be a closed disc with radius M.
By (3.4) for every x^ Tq we may find an x2e К such that (aq, x2)e T.
The function/2, being continuous, is bounded on K, and hence the function /1(^1) = f {00! , x2) - f2{x2)
is bounded on T Q. As in the proof of Theorem 3.3 we infer hence that t
0*K-
Theorems 3.3 and 3.4 say that, under some additional assumptions (the boundedness of T2 or of D(x-f)), if a set T c RN is not in then either all sections Tfaq] are not in or all sections
T[X2] = {x1e R p : (хг, X2)e T}
are not in Now, one could expect that if a set T c RN is in then all (or, in some sense, almost all) sections T[x{\ and Т[ж2] must be in resp. <ip . But this is not the case. As we shall show in the next section, there exist sets T e <ê2 such that all sections T[x{\, T [ x 2] contain at most one point each. In other words, there are functions (even invertible func
tions) f: R - > R whose graphs belong to c62.
4. Graphs. In the present section we shall consider functions /:
A -* R, where А c R. By I f we shall denote the graph of /, i.e.
If = {{x,y): y = f ( x ) , X € A}.
We are going to investigate the sets I f from the point of view whether they belong or not to the classes j/ 2, 2.
At first we have the following simple result.
4.1. I f f: R -> R has the property that g(x) = f ( x ) —f ( 0) is linear, then If e ^2 •
Proof. I g is a proper linear subspace of R 2 (over rationals) and hence I a€ by Theorem 0.1. Consequently also I f, being a translation of I g, belongs to ^2-
However, the situation described in Theorem 4.1 is rather exceptional, for we have the following theorem.
4.2. I f f: A —y R, where A is an interval, is continuous and non
linear, then If€ <s/2.
Proof. In virtue of Theorem 1.2 we may assume that 0 is an inner point of A, /(0) = 0, and / is not linear in any neighbourhood of 0. We consider the function
(4.1)
F(x, t) =f{t)+f{x-t)
for x > 0, xe A, and 0 < t < x. Suppose that for xe [0, a] <= Л the function F( x , t ) = F(x) does not depend on t. Setting x — t — s we have
(4.2) F(x) = F ( t + s) = f ( t ) +f ( s) ,
and since (4.2) with 5 = 0 implies F{t) = f ( t ) , we obtain finally (4.3) f(t + 8) = f ( t ) + f ( s )
for 0 < t < a, 0 < s < a, 0 < t + s < «. But (4.3) together with the conti
nuity of/im plies t h a t/is linear in [0, a] ([1], p. 48). Consequently function (4.1) must depend on t.
Thus there exists in the interior of A a point x0 > 0 such that Ш = sup F{x0,t) > mî F{x0,t) = m.
<е[0,а:0] И 0 , Я 0]
Let T0 and t0 be those points from the interval [0, x0] at which the supre- mum resp. infimum is attained:
F{x0, T0) = M , F( x0, t0) = m.
There exists an xx > x0 such that [ж0, n?x] с Л and
(4.4) F ( x , T0) > M —i d, F(x, t0) < m + \d for xe |>0, a?x], where d = Ж — m.
Now, take an arbitrary point {x, y) e [x0, aq] x [ m+ $d, M — $<?].
Relation (4.4) implies that there exists a t between t0 and T0 such that F( x, t ) — y. With s = x — t we may write
x = t + s, у = /(* )+ /(« ), i.e. (x, y)e Consequently
(4.5) [a?0, aq] x [m+^d, M — |d ] c I f + I f .
It follows from (4.5) that the set has a non-empty interior, which in turns implies (cf., e.g., [5], [3]) that I fe This completes the proof.
References
[1] J. A c z é l, Lectures on fu n c tio n a l equations and their applications, New York — London 1966.
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[3] — and M. K u c z m a , On the boundedness and continuity o f convex fun ctio n s and additive fu n c tio n s, A equationes Math. 4 (1970), p. 157-162.
[4] Marcin E. K u c z m a , On discontinuous additive fun ctio n s, Fund. Math. 66 (1970)*, p. 383-392.
Theory of convex functions 13Ô
[5] S. M a r c u s, Généralisation, a u x fonctions de plusieurs variables, des théorèmes de A lexander OstrowsM et de M asuo R ulcuhara concernant les fonctions convexes (J), J. Math. Soc. Japan 11 (1959), p. 171-176.
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