Volume 8 (2007), Issue 2, Article 35, 9 pp.
A SIMULTANEOUS SYSTEM OF FUNCTIONAL INEQUALITIES AND MAPPINGS
WHICH ARE WEAKLY OF A CONSTANT SIGN
DOROTA KRASSOWSKA AND JANUSZ MATKOWSKI
FACULTY OF M ATHEMATICS , C OMPUTER S CIENCE AND E CONOMETRICS
U NIVERSITY OF Z IELONA G ÓRA
P ODGÓRNA 50, PL-65-246 Z IELONA G ÓRA , P OLAND
D.krassowska@wmie.uz.zgora.pl
I NSTITUTE OF M ATHEMATICS
S ILESIAN U NIVERSITY
BANKOWA 14, PL-40-007 K ATOWICE
P OLAND
j.matkowski@wmie.uz.zgora.pl
Received 17 October, 2006; accepted 07 June, 2007
Communicated by Zs. Pales
A BSTRACT. It is shown that, under some algebraic conditions on fixed reals α1 , α2 , . . . , αn+1
and vectors a1 , a2 , . . . , an+1 ∈ Rn , every continuous at a point function f : Rn → R satisfying
the simultaneous system of inequalities
f (x + ai ) ≤ αi + f (x),
x ∈ Rn , i = 1, 2, . . . , n + 1,
has to be of the form f (x) = p · x + f (0), x ∈ Rn , with uniquely determined p ∈ Rn . For
mappings with values in a Banach space which are weakly of a constant sign, a counterpart of
this result is given.
Key words and phrases: Linear functional inequalities, Kronecker’s theorem, Mappings of weakly constant sign .
2000 Mathematics Subject Classification. Primary 39B72; Secondary 26D05.
1. I NTRODUCTION
In this paper we consider the simultaneous system of functional inequalities
(1.1)
f (x + ai ) ≤ αi + f (x),
x ∈ Rn ,
i = 1, 2, . . . , n + 1,
where n ∈ N, α1 , α2 , . . . , αn+1 ∈ R, a1 , a2 , . . . , an+1 ∈ Rn are fixed and f : Rn → R is an
unknown function.
In Section 3, assuming Kronecker’s type conditions on the vectors a1 , a2 , . . . , an+1 and a inequality involving some determinants depending on these vectors and scalars α1 , α2 , . . . , αn+1 ,
we show (Theorem 3.1) that for every continuous at least at one point function f satisfying (1.1)
there exists a unique vector p ∈ Rn such that
f (x) = p · x + f (0),
267-06
x ∈ Rn .
2
D OROTA K RASSOWSKA AND JANUSZ M ATKOWSKI
This result seems to be a little surprising in the context of an obvious fact that, in general,
the accompanying simultaneous system of functional equations has a lot of very regular but
nonlinear solutions, even “depending on an arbitrary function” (cf. M. Kuczma [4], [2] or [3],
where the full construction of the solution in a special one dimensional case is given). Theorem
3.1 generalizes a suitable result of [2], where the one dimensional case
f (x + a) ≤ α + f (x),
f (x + b) ≤ β + f (x),
x ∈ R,
is considered. Let us mention that in the case when α = β = 0, Montel [5] considered the
accompanying simultaneous system of equations.
In Section 4 we define a mapping to be weakly of a constant sign. Using this notion and a
total system of linear functionals, we present a counterpart of Theorem 3.1 for functions with
values in a Banach space.
2. K RONECKER ’ S T HEOREM
AND A
L EMMA
The symbols N, Z, Q, R are reserved, respectively, for the set of natural, integer, rational and
real numbers.
We begin this section by recalling the following:
Theorem 2.1 (Kronecker (cf. [1, p. 382])). If the reals υ1 , υ2 , . . . , υn , 1 are linearly independent
over the field Q, the numbers α1 , α2 , . . . , αn ∈ R are arbitrary, and N and are positive, then
there are the integers p1 , p2 , . . . , pn , and m > N, such that |mυi − pi − αi | < for each
i = 1, 2, . . . , n.
As an immediate consequence of this theorem we obtain the following
Corollary 2.2. Let e1 , e2 , . . . , en be the standard base of the real linear space Rn . If reals
υ1 , υ2 , . . . , υn , 1 are linearly independent over the field Q, and v : = (υ1 , υ2 , . . . , υn ), then the
set
{mv + p1 e1 + p2 e2 + · · · + pn en : m ∈ N, p1 , p2 , . . . , pn ∈ Z}
n
is dense in R .
In sequel we need a more special result which guarantees that the set of all linear combinations of the elements of the set B = {a1 , a2 , . . . , an+1 } with natural coefficients is dense in
Rn .
Lemma 2.3. Let {a1 , a2 , . . . , an } be an arbitrary base of linear space Rn over R. Suppose
that υ1 , υ2 , . . . , υn ∈ R are negative and the system of numbers υ1 , υ2 , . . . , υn , 1 is linearly
independent over the field Q. If
an+1 = υ1 a1 + υ2 a2 + · · · + υn an ,
then the set
A :=
( n+1
X
)
m(i) ai : m(i) ∈ N, i = 1, 2, . . . , n + 1
i=1
is dense in Rn .
Proof. Let N > 0 be fixed. Take x ∈ Rn . Then there exists a unique system of numbers
x1 , x2 , . . . , xn ∈ R such that
x =x1 a1 + x2 a2 + · · · + xn an .
By Kronecker’s theorem, for every k ∈ N there exist mk >N and pik ∈ Z such that
1
|mk υi + pik − xi | = |mk υi − (−pik ) − xi | < , i = 1, 2, . . . , n,
k
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
http://jipam.vu.edu.au/
F UNCTIONAL I NEQUALITIES
3
whence
lim (mk υi + pik ) = xi ,
i = 1, 2, . . . , n.
k→∞
Since mk ∈ N, υi < 0 for i = 1, 2, . . . , n, it follows that pik ∈ N for k large enough (in the
opposite case we would have limk→∞ (mk υi + pik ) = −∞). It follows that, for k large enough,
(n+1)
(i)
setting mk
:= mk and mk := pik , we get
(i)
mk ∈ N,
i = 1, 2, . . . , n + 1.
Put
tk :=
n X
(n+1)
mk
(i)
υi + mk
ai ,
k ∈ N.
i=1
By the definition of an+1 we hence get
tk =
n+1
X
(i)
mk ai ,
k ∈ N,
i=1
whence tk ∈ A for k sufficiently large. Moreover
lim tk = lim
k→∞
n X
k→∞
=
=
n
X
i=1
n
X
(n+1)
mk
(i)
υi + mk
ai
i=1
(n+1)
(i)
lim mk
υi + mk ai
k→∞
xi ai = x.
i=1
This completes the proof of the density of A in Rn .
3. M AIN R ESULT
Theorem 3.1. Let n ∈ N, ai = (ai,1 , ai,2 , . . . , ai,n ) ∈ Rn , i = 1, 2, . . . , n, negative reals
υ1 , υ2 , . . . , υn and α1 , α2 , . . . , αn+1 ∈ R be fixed and such that:
(i) a1 , a2 , . . . , an form a base of the linear space Rn over R;
(ii) υ1 , υ2 , . . . , υn , 1 are linearly independent over Q,
(iii) an+1 = υ1 a1 + υ2 a2 + · · · + υn an ,
(iv)
α1 α2 · · · αn αn+1
a
1,1 a2,1 · · · an,1 an+1,1
(−1)n sgn W(n) ··· ··· ··· ···
···
a1,n a2,n · · · an,n an+1,n
≤ 0,
where
W(n)
:= J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
a1,1
a1,2
···
a1,n
a2,1
a2,2
···
a2,n
· · · an,1
· · · an,2
··· ···
· · · an,n
.
http://jipam.vu.edu.au/
4
D OROTA K RASSOWSKA AND JANUSZ M ATKOWSKI
If a continuous at least at one point function f : Rn −→ R satisfies the simultaneous system
of functional inequalities
x ∈ Rn , i = 1, 2, . . . , n + 1,
f (x + ai ) ≤ αi + f (x),
(3.1)
then there exists a unique p ∈ Rn such that
x ∈ Rn .
f (x) = p · x + f (0),
Moreover,
where
(n)
pi
:= i
1 h (n) (n)
(n)
p = (n) p1 , p2 , . . . , pn ,
W
a1,1
a2,1 · · · an,1 ···
··· ···
· · · a1,i−1 a2,i−1 · · · an,i−1 α1
α2 · · ·
αn ,
i = 1, 2, . . . , n.
a1,i+1 a2,i+1 · · · an,i+1 ···
··· ···
··· a1,n
a2,n · · · an,n Proof. Suppose that the function f : Rn −→ R satisfies system (3.1). Using an easy induction
argument, one can show that f satisfies the inequality
! n+1
n+1
X
X
(3.2)
f
m(i) ai + x ≤
m(i) αi + f (x),
i=1
i=1
for all m(i) ∈ N, i = 1, 2, . . . , n + 1 and all x ∈ Rn . By Lemma 2.3, the set
( n+1
)
X
A :=
m(i) ai : m(i) ∈ N, i = 1, 2, . . . , n + 1
i=1
(i)
is dense in Rn . Thus there exist the sequences of positive integers (mk )k∈N for i = 1, 2, . . . , n+
1 such that
!
n+1
n+1
n+1
n+1
X
X
X
X
(i)
(i)
(i)
(i)
mk ai,1 ,
mk ai,2 , . . . ,
mk ai,n
lim
mk ai = lim
k→∞
k→∞
i=1
i=1
i=1
i=1
= (0, 0, . . . , 0),
and, obviously,
(i)
lim mk = ∞,
i = 1, 2, . . . , n + 1.
k→∞
(i)
It follows that, for each i = 1, 2, . . . , n, the limit limk→∞
n+1
X
i=1
(i)
lim
k→∞
n+1
P
!
mk
a
(n+1) i,j
mk
= lim
k→∞
i=1
mk
(n+1)
mk
exists, and
(i)
mk ai,j
(n+1)
= 0,
j = 1, 2, . . . , n.
mk
Consequently,
(i)
(3.3)
lim
k→∞
mk
(n+1)
(n)
=
mk
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
Wi
,
W(n)
i = 1, 2, . . . , n,
http://jipam.vu.edu.au/
F UNCTIONAL I NEQUALITIES
5
where
(n)
Wi
:= a1,1
a1,2
···
a1,n
· · · ai−1,1 −an+1,1 ai+1,1
· · · ai−1,2 −an+1,2 ai+1,2
··· ···
···
···
· · · ai−1,n −an+1,n ai+1,n
,
· · · an,1
· · · an,2
··· ···
· · · an,n
i = 1, 2, . . . , n.
Let x0 be a point of the continuity of f . From inequality (3.2), we get
n+1
P (i)
f
mk ai + x0
n+1
(i)
X
mk
f (x0 )
i=1
≤
α
+
.
i
(n+1)
(n+1)
(n+1)
mk
m
m
i=1
k
k
Letting here k → ∞ and applying (3.3), we obtain
0≤
(3.4)
n
(n)
X
W
i
i=1
W(n)
αi + αn+1 .
Setting
(n)
Wi
:= − a1,1
a1,2
···
a1,n
· · · ai−1,1 ai+1,1
· · · ai−1,2 ai+1,2
··· ···
···
· · · ai−1,n ai+1,n
,
· · · an,1 an+1,1
· · · an,2 an+1,2
··· ···
···
· · · an,n an+1,n
i = 1, 2, . . . , n,
we can write inequality (3.4) in the form
0 ≤ sgn W(n)
n
X
!
(n)
(−1)n−i+1 Wi αi + W(n) αn+1
,
i=1
whence, by the Laplace expansion of a determinant,
α1 α2
n
(n) a1,1 a2,1
0 ≤ (−1) sgn W ··· ···
a1,n a2,n
· · · αn αn+1
· · · an,1 an+1,1
··· ···
···
· · · an,n an+1,n
Hence, taking into account condition (iv), we infer that
α1 α2 · · · αn αn+1
a
1,1 a2,1 · · · an,1 an+1,1
··· ··· ··· ···
···
a1,n a2,n · · · an,n an+1,n
.
= 0.
Now, by Laplace’s expansion theorem,
n
P
αn+1 = (−1)n+1 i=1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
(n)
(−1)1−i Wi αi
W(n)
,
http://jipam.vu.edu.au/
6
D OROTA K RASSOWSKA AND JANUSZ M ATKOWSKI
whence by (3.2),
n+1
X
f
!
(i)
m ai + x
i=1
≤
n
X
n
P
(n)
(−1)1−i Wi αi
m(i) αi + m(n+1) (−1)n+1 i=1
+ f (x)
W(n)
i=1
" n
#
n
X
X
1
(n)
(i)
(n)
(n+1)
n+2−i
=
m W αi + m
(−1)
Wi αi + f (x),
W(n) i=1
i=1
for all m(i) ∈ N, and x ∈ Rn .
(n)
Applying Laplace’s theorem for the ith columne of W(n) and for the last columne of Wi ,
we have
n
X
(i)
m W
(n)
αi + m
n
X
(n+1)
i=1
(n)
(−1)n+2−i Wi αi
i=1
=
n
X
(i)
m αi
i=1
n
X
(n)
(−1)i+j ai,j Wij
j=1
n
X
+ m(n+1)
(−1)n−i ai,j αi
i=1
(n)
n
X
!!
(n)
(−1)n+j an+1,j Wij
,
j=1
(n)
(n)
where Wij is obtained from Wi by deleting the jth row and ith column, and Wij is obtained
(n)
from Wi
by deleting the jth row and the last column. Since
(n)
(n)
Wij = Wij ,
i, j = 1, 2, . . . , n,
applying Fubini’s theorem for sums, we have
n
X
(i)
m W
(n)
αi + m
(n+1)
i=1
n
X
(n)
(−1)n+2−i Wi αi
i=1
=
n X
n
X
(k)
k+j
m ak,j (−1)
(n)
Wkj αk
+m
(n+1)
j=1 k=1
(i)
m W
(n)
αi + m
(n+1)
i=1
(n)
(−1)j−i Wij αi .
i=1
Adding and subtracting the term
n
X
an+1,j
n
X
n
X
(n)
j−i
Wij αi
i=1 (−1)
Pn
in the sum over k gives
(n)
(−1)n+2−i Wi αi
i=1
=
n
X
n
X
j=1
k=1
+
!
m(k) ak,j + m(n+1) an+1,j
n
X
!
(n)
(−1)j−i Wij αi
i=1
n
X
n
X
j=1
k=1
m(k) ak,j
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
n
X
!!
(n)
(n)
(−1)j−i+1 Wij αi + (−1)k+j Wkj αk
.
i=1
http://jipam.vu.edu.au/
F UNCTIONAL I NEQUALITIES
Let us note that
n
n
X
X
j=1
m(k) ak,j
n
X
!!
(n)
(−1)j−i+1 Wij αi
+
(n)
(−1)k+j Wkj αk
= 0.
i=1
k=1
Indeed, we have
n
n
X
X
j=1
7
m(k) ak,j
n
X
!!
(n)
(−1)j−i+1 Wij αi
+
(n)
(−1)k+j Wkj αk
i=1
k=1
=
n
n X
X

=
X
m(k) ak,j 
j=1 k=1
n
X

i∈{1,2,...,n}−k

X
m(k) 
k=1
and
(n)
(−1)j−i+1 Wij αi 
αi
i∈{1,2,...,n}−k
n
X
n
X
!
(n)
(−1)j−i+1 ak,j Wij

j=1
(n)
(−1)j−i+1 ak,j Wij
j=1
is a determinant with two equal columns. Thus we have shown that f satisfies the inequality
! n+1
!!
!
n
n+1
n
X
X
X
X
1
(n)
(−1)j−i Wij αi
m(i) aij
+ f (x),
f
m(i) ai + x ≤
n
W i=1
i=1
i=1
j=1
for all m(i) ∈ N, and x ∈ Rn , which can be written in the form
(3.5)
t ∈ A, x ∈ Rn .
f (t + x) ≤ p · t + f (x),
Now take an arbitrary x ∈ Rn . By the density of A there is a sequence (tn ) such that
tn ∈ A (n ∈ N),
lim tn = x0 − x.
n→∞
From (3.5) we have
f (tn + x) ≤ p · tn + f (x),
n ∈ N.
Letting here n → ∞, and making use of the continuity of f at x0 , we obtain
x ∈ Rn .
f (x0 ) ≤ p·(x0 − x) + f (x),
To prove the converse inequality, note that replacing x by x − t in (3.5) we get
t ∈ A, x ∈ Rn .
f (x) ≤ p · t + f (x − t),
Taking a sequence (tn ) such that
tn ∈ A (n ∈ N),
lim tn = x − x0 ,
n→∞
(which, by the density of A, exists) we hence get
f (x) ≤ p · tn + f (x − tn ),
n ∈ N.
Letting here n → ∞, and again making use of the continuity of f at x0 , we obtain the inequality
f (x) ≤ p·(x − x0 ) + f (x0 ),
x ∈ Rn .
Thus
f (x) = p · x + (f (x0 ) − p · x0 ),
Since f (0) = f (x0 ) − p · x0 , the proof is completed.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
x ∈ Rn .
http://jipam.vu.edu.au/
8
D OROTA K RASSOWSKA AND JANUSZ M ATKOWSKI
4. A PPLICATION
FOR
M APPINGS W HICH
ARE
W EAKLY OF A C ONSTANT S IGN
We start this section with the following.
Definition 4.1. Let X be an arbitrary nonempty set, Y - an arbitrary real linear topological
space, Y ∗ - the conjugate space of Y , and T ⊂ Y ∗ .
(i) We say that a mapping G : X → Y is T -weakly of a constant sign if for each functional
φ ∈ T either φ ◦ G is nonpositive or φ ◦ G is nonnegative.
(ii) The mappings G1 : X → Y and G2 : X → Y are said to be T -weakly of the same sign
if (φ ◦ G1 ) · (φ ◦ G2 ) ≥ 0 for every functional φ ∈ T .
Now applying Theorem 3.1 we prove:
Theorem 4.1. Let Y be a real Banach space and {a1 , a2 , . . . , an } ⊂ Rn , ai = (ai,1 , ai,2 , . . . , ai,n ),
i = 1, 2, . . . , n be a base of the real linear space Rn . Suppose that reals υ1 , υ2 , . . . , υn are negative and υ1 , υ2 , . . . , υn , 1 are linearly independent over the field Q. Let an+1 = υ1 a1 + υ2 a2 +
· · · + υn an .
If a mapping F : Rn → Y is continuous at least at one point and there exist a total system
T ⊂ Y ∗ and the vectors y1 , y2 , . . . , yn+1 ∈ Y such that
(i) the mappings
Rn 3 x → F (x + ai ) − F (x) − yi ,
(*)
i ∈ {1, 2, . . . , n + 1},
are T -weakly of a constant and the same sign;
(ii) for every φ ∈ T,
φ(y1 ) φ(y2 ) · · · φ(yn ) φ(yn+1 )
a2,1 · · · an,1
an+1,1
n
(n) a1,1
(−1) sgn W
···
··· ···
···
···
a1,n
a2,n · · · an,n
an+1,n
≤ 0,
then there exists a unique linear continuous mapping L : Rn → Y such that
x ∈ Rn .
F (x) = L(x) + F (0),
(4.1)
Proof. Assume that F : Rn → Y satisfies assumptions (i) and (ii). By Definition 4.1, we have
either, for all φ ∈ T and i = 1, 2, . . . , n + 1,
φ (F (x + ai ) − F (x) − yi ) ≤ 0,
x ∈ Rn ,
or, for all φ ∈ T and i = 1, 2, . . . , n + 1,
φ (F (x + ai ) − F (x) − yi ) ≥ 0,
x ∈ Rn .
Without any loss of generality we may confine the consideration to the first case. From the
linearity of φ we obtain the simultaneous system of inequalities
φ (F (x + ai )) ≤ φ (F (x)) + φ (yi ) ,
x ∈ Rn ;
yi ∈ Y, i = 1, 2, . . . , n + 1;
φ ∈ T.
Let us fix φ ∈ T. Setting f := φ ◦ F and αi := φ (yi ), we hence get
f (x + ai ) ≤ αi + f (x),
x ∈ Rn ,
i = 1, 2, . . . , n + 1,
i.e. f satisfies system (3.1). Since all the required assumptions of Theorem 3.1 are satisfied,
there exists a vector p, such that
f (x) = p · x + f (0),
x ∈ Rn .
Hence, by the definition of f,
φ (F (x) − F (0)) = p · x,
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
x ∈ Rn .
http://jipam.vu.edu.au/
F UNCTIONAL I NEQUALITIES
9
It follows that for every φ ∈ T, there exists p : = pφ , such that
φ (F (x) − F (0)) = pφ ·x,
x ∈ Rn .
Setting L := F − F (0), we hence get
φ (L(x + y)) = φ (F (x + y) − F (0))
= pφ ·(x + y)
= pφ ·x + pφ ·y
=φ (F (x) − F (0)) + φ (F (y) − F (0))
= φ (L(x)) + φ (L(y)) ,
for all x, y ∈ Rn , whence
φ (L(x + y) − L(x) − L(y)) = 0,
x, y ∈ Rn ,
for every functional φ ∈ T. Since T is total set of functionals (cf. [6]),
L(x + y) − L(x) − L(y) = 0,
x, y ∈ Rn ,
that is, L is an additive mapping. Now the continuity of F at least at one point implies that
L := F − F (0) is continuous and, consequently, a linear map. The proof is completed.
R EFERENCES
[1] G.H. HARDY AND E.M. WRIGHT, An Introduction to the Theory of Numbers, Oxford, 1960.
[2] D. KRASSOWSKA AND J. MATKOWSKI, A pair of functional inequalities and a characterization
of Lp −norm, Ann. Polon. Math., 85 (2005), 1–11.
[3] D. KRASSOWSKA AND J. MATKOWSKI, A pair of functional inequalities of iterative type related
to a Cauchy functional equation, in Functional Equations, Inequalities and Applications, Kluwer
Academic Publishers, Dordrecht-Boston-London, 2003, 73–89.
[4] M. KUCZMA, An Introduction to the Theory of Functional Equations and Inequalities, Cauchy’s
Equation and Jensen’s Inequality, PWN - Polish Sci. Publ. and Silesian University, WarszawaKraków-Katowice, 1985.
[5] P. MONTEL, Sur les propriétés périodiques des fonctions, C. R. Acad. Sci. Paris, 251 (1960), 2111–
2112.
[6] A. WILANSKY, Functional Analysis, Blaisdel Publishing Company, New York-Toronto-London,
1964.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 35, 9 pp.
http://jipam.vu.edu.au/