Mathematica Pannonica 7

(1)

Mathematica Pannonica 7

/2 (1996), 223 { 232

BOUNDED SOLUTIONS OF

SCHILLING'S PROBLEM

Janusz ^Morawiec

Instytut Matematyki, Uniwersytet Slaski, ul. Bankowa 14, PL{40{007 Katowice, Poland

Received: August 1995 MSC 1991: 39 B12, 39 B 22

Keywords: Schilling's problem, bounded and continuous functions.

Abstract: Let ⁿ be a positive integer, ^qⁿ be the unique ^x ² (¹³¹²) with

x n+1

;3^x+ 1 = 0 and ^q ² (0^qⁿ]. We found a set ^Aⁿ^q of reals with the following property (P): Every solution ^f :^R!^Rof the functional equation

f(^{q x}) = 14^q^f(^x^;1) +^f(^x+ 1) + 2^f(^x)]

which vanishes outside of ^;^1;q^q ^1;q^q ] and is bounded in a neighbourhood of a point of that set vanishes everywhere. It is also observed that for^q²(0¹³] the set^S¹ⁿ⁼¹^Aⁿ^q, which equals then

n 1

X

n=1

"(ⁿ)^qⁿ : ^"²^f;101^g^N^o is the largest one with property (P).

Following R. Schilling 9] we consider solutions f : ^R^! ^Rof the functional equation

(1) f(qx_{) = 14}q f⁽x^;1) +f(x+ 1) + 2f(x) such that

(2) f(x) = 0 for ^jx^j> Q

where q is a xed number from the open interval (01) and

(2)

Q= q 1^;q:

In what follows any solution f : ^R^! ^Rof (1) satisfying (2) will be called a solution of Schilling's problem.

(3) If 3q1^;^p³2 +^p³4

then according to 7] the zero function is the only solution of Schilling's problem which is bounded in a neighbourhood of a point of the set (4) ⁿ"^Xⁿ

i⁼¹qⁱ : n² ^N^f0+^1g "²^f;11^g^o:

This generalizes in particular 1 Th. 1]. It is the aim of the present paper to obtain such a result with the set (4) replaced by a larger one.

However, we are not able to enlarge (4) for allq's satisfying (3) but, on the other hand, for q ¹³ we succeeded in nding even the largest set to be put in the place of (4) (cf. Cor. 1).

Given a positive integernand q ²(01) consider the setA_nq of all the real numbers of the form

(5)

"^X^L

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾⁺^M +"(^;1)^L ^X^M

m⁼¹q^m where " ² ^f;11^g, M, L are non-negative integers, K¹:::KL ²

2 f1:::n^g, and :^f1:::L^g^f1:::n^g^!^N. Evidently, the set (4) is a subset of clA_nq. Let us observe also that for l¹l² ² ^f1:::L^g, k(6)_K¹ ²^f1:::Kl¹^g, k²² ^f1:::Kl²^g, if (l¹k¹)⁶= (l²k²) then

l¹

X

m⁼k¹(l¹m) + ^X^L

j⁼l¹⁺¹ Kj

X

m⁼¹(jm)⁶= ^K^X^l²

m⁼k²(l²m) + ^X^L

j⁼l²⁺¹ Kj

X

m⁼¹(jm): The proof of the following fact is left to the reader (cf. also 6 Th. 21(a), (d)]).

Remark 1.

If q ²(0 ¹³] then cl¹

n⁼¹A_nq =ⁿ^X¹

n⁼¹"(n)qⁿ : "²^f;101^g^N^o and if q ²¹³1) then

(3)

n 1

X

n⁼¹"(n)qⁿ : "²^f;101^g^N^o= ^;QQ]:

For every positive integer n let qn denote the unique x ² (¹³¹²) with(7) xⁿ⁺¹^;3x+ 1 = 0

and observe that ifq ²(0 ¹²) then

qqn i qⁿ⁺¹^;3q+ 1 0: Our main result reads.

Theorem 1.

If n is a positive integer and q ² (0qn] then the zero function is the only solution of Schilling's problem which is bounded in a neighbourhood of a point of the set clA_nq.

The proof of this theorem is based on four lemmas. However, we start with the following simple remarks.

Remark 2.

If f is a solution of Schilling's problem then so is the function g:^R^!^Rdened by the formulag(x) =f(^;x).

Remark 3.

Assume f is a solution of Schilling's problem.

If q ⁶= ¹⁴ then f(^;Q) =f(Q) = 0. If q < ¹² then f(0) = 0.

Lemma 1.

^Assume q²(0¹²). Iff is a solution of Schilling's problem then

(8) f(q^N⁺^Mx+"^X^M

m⁼¹q^m) =1 2

M 1 2q

N⁺Mf(x)

for all x ² (Q^;11^;Q) (for all x ² Q^;11^;Q] if q ⁶= ¹⁴), for all

"²^f;11^g, and for all non-negative integers M and N.

Forx ²(Q^;11^;Q) this was proved in 7] as Lemma 2. In the case of the closed interval Q^;11^;Q] and q ⁶= ¹⁴ we argue similarly as in the proof of 7 Lemma 2] using also 7 Remarks 1 and 2(i)].

Lemma 2.

Let n²^N,q ²(0qn] and y =q^N⁺^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk⁾x+

+^X^L

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾

whereN is a non-negative integer, Lis a positive integer, K¹:::KL²

2f1:::n^g, : ^f1:::L^g^f1:::n^g^! ^N, and x ² 01^;Q) (x ²

201^;Q] if q ⁶= ¹⁴).

If L is even then y²(01^;Q).

(4)

If L is odd then y ²Q^;10] (y ²(Q^;10] if q < ¹³).

Proof.

Since q qn< ¹² we have

(9) Q <1:

Moreover, asqn is a solution of (7), (10) ^Xⁿ

i⁼¹qⁱ ^Xⁿ

i⁼¹q_in= 1^; qn

1^;qn 1^; q

1^;q ^{= 1}^;Q and

(11) if q < ¹_{3 then} ^Xⁿ

i⁼¹qⁱ < Q <1^;Q:

Observe also that

(12)

y =q⁽^LK^L⁾q^N⁺^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk^);⁽^LK^L⁾x+ +^L^X^;1

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm^);⁽^LK^L⁾+ + (^;1)^{L K}^X^L^;1

k⁼¹ q^P^KL^m⁼^;1^k ⁽^Lm⁾ + (^;1)^L

(13)

y=q^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk⁾^;q^Nx^;1^;

;

K¹

X

k⁼²q^P^K^m¹⁼^k⁽¹^m⁾⁺^P^Lj⁼²^P^Kj^m⁼¹⁽^jm⁾+ +^X^L

l⁼²(^;1)^{l K}^X^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾ (14) q^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk⁾^;q^Nx^;1<0

and

(15) ^X^K^L

k⁼¹q^P^KL^m⁼^k⁽^Lm⁾ ^X^K^L

k⁼¹q^P^KL^m⁼^k¹ ^Xⁿ

k⁼¹q^k:

Suppose rst L is even. Applying (13), (14), (6), (15), (9) and (10) we obtain

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y <^X^L

l⁼²(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾

L^X^;2 l⁼²

K_l

X

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾^;

;

KX_L^;1

k⁼¹ q^P^KL^m⁼^;1^k ⁽^L^;1^m⁾⁺^P^KL^m⁼¹⁽^Lm⁾+^X^K^L

k⁼¹q^P^KL^m⁼^k⁽^Lm⁾

1

X

i⁼¹q⁽^L^;1^K^L^;1⁾⁺^P^KL^m⁼¹⁽^Lm⁾⁺ⁱ^;

;q⁽^L^;1^K^L^;1⁾⁺^P^KL^m⁼¹⁽^Lm⁾ +^Xⁿ

k⁼¹q^k =

=q⁽^L^;1^K^L^;1⁾⁺^P^KL^m⁼¹⁽^Lm⁾(Q^;1) +^Xⁿ

k⁼¹q^k <^Xⁿ

k⁼¹q^k 1^;Q whereas (12), (6) and (9) give

y q⁽^LK^L⁾^;^X¹

i⁼¹qⁱ+ 1=q⁽^LK^L⁾(^;Q+ 1)>0:

Suppose now L is odd. If L = 1 then using the denition of y, (15) and (10) we see that

y ^;^X^K¹

k⁼¹q^P^K^m¹⁼^k⁽¹^m⁾ ^;^Xⁿ

k⁼¹q^k Q^;1

with the last inequality being strict ifq < ¹³ (cf. (11)). If L 3 then on account of the denition of y, (6), (15), (9) and (10) we have

y ^;^L^X^;2

l⁼¹ K_l

X

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾+ +^K^X^L^;1

k⁼¹ q^P^KL^m⁼^;1^k ⁽^L^;1^m⁾⁺^P^KL^m⁼¹⁽^Lm⁾ ^;^X^K^L

k⁼¹q^P^KL^m⁼^k⁽^Lm⁾

; 1

X

i⁼¹q⁽^L^;1^K^L^;1⁾⁺^P^KL^m⁼¹⁽^Lm⁾⁺ⁱ+

(6)

+q⁽^L^;1^K^L^;1⁾⁺^P^KL^m⁼¹⁽^Lm⁾^;^X^K^L

k⁼¹q^k

>^;^Xⁿ

k⁼¹q^k Q^;1:

Finally, ifL is odd then taking into account (12) and (6) we obtain yq⁽^LK^L⁾x+^X¹

i⁼¹qⁱ^;1q⁽^LK^L⁾(1^;Q) +Q^;1] = 0:

Lemma 3.

Assume n ² ^N and q ² (0qn]. If f is a solution of Schilling's problem then for every x²01^;Q), for every non-negative integers MLand N, for every K¹:::KL ² ^f1:::n^g, and for every : ^f1:::L^g^f1:::n^g^!^Nwe have

(16)

f(q^N⁺^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk⁾⁺^Mx+ +^X^L

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾⁺^M+ + (^;1)^{L M}^X

m⁼¹q^m) =

=1 2

PLl⁼¹Kl⁺M 1 2q

N⁺^P_Ll⁼¹^P^Kl_k⁼¹⁽lk⁾⁺Mf(x):

Proof.

According to Lemma 1, (16) holds for L = 0. Assume L is a positive integer.

Consider rst the case M = 0.

Let L= 1. Equality (16) takes then the form (17) f(q^N⁺^P^K^k⁼¹¹ ⁽¹^k⁾x^;^X^K¹

k⁼¹q^P^K^m¹⁼^k⁽¹^m⁾) =

=1 2

K¹ 1 2q

N⁺^P^K_k⁼¹¹ ⁽¹k⁾f(x)

and making use of Lemma 1 we see that if K¹ = 0 then (17) holds for all x ² (Q^;11^; Q) (for all x ² Q^; 11^;Q] if q ⁶= ¹⁴) and for every non-negative integer N. Fix now a K¹ ² ^f0:::n^;1^g and suppose that (17) is satised for every non-negative integerN, for every

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: ^f1^g^f1:::n^g^!^N, and for allx²01^;Q) (for allx²01^;Q] if q⁶= ¹⁴). Let N ²^N^f0^g, :^f1^g^f1:::n^g^!^Nand x²01^;Q) (x ²01^;Q] ifq ⁶= ¹⁴). Putting

z=q^N⁺^P^K^k⁼¹¹ ⁽¹^k⁾x^;^X^K¹

k⁼¹q^P^K^m¹⁼^k⁽¹^m⁾ ^;1 we have

(18) zx^;1

and, according to Lemma 2, y :=qz ² Q^;10] (and y ² (Q^;10] if q < ¹³). This jointly with the denition of z, Lemma 1, (1), (17), (2), Remark 3 and (17) gives

f(q^N⁺^P^K^k⁼¹¹⁺¹⁽¹^k⁾x^;^K^X¹⁺¹

k⁼¹ q^P^K^m¹⁼⁺¹^k ⁽¹^m⁾) =

= f(q⁽¹^K¹^+1);1y) == 1 2q

⁽¹K¹^+1);1f(y) =

= 1 2q

⁽¹K¹^+1);1 1

4qf(z^;1) +f(z+ 1) + 2f(z)] =

= 1 2q

⁽¹K¹⁺¹⁾1

2f(z+1) =

= 1 2q

⁽¹K¹⁺¹⁾1 2

1 2

K¹ 1 2q

N⁺^P^K_k⁼¹¹ ⁽¹k⁾f(x) =

= 1 2

K¹⁺¹ 1 2q

N⁺^P^K_k⁼¹¹⁺¹⁽¹k⁾f(x):

Hence (17) holds for every K¹ ² ^f1:::n^g, for every non-negative integerN, for every :^f1^g^f1:::n^g^!^N, and for allx ²01^;Q) (for all x ² 01 ^;Q] if q ⁶= ¹⁴). Consequently, taking into account Remark 2 we have also

(19) f(q^N⁺^P^K^k⁼¹¹ ⁽¹^k⁾x+^X^K¹

k⁼¹q^P^K^m¹⁼^k⁽¹^m⁾) =

=1 2

K¹ 1 2q

N⁺^P^K_k⁼¹¹ ⁽¹k⁾f(x)

for every K¹ ² ^f1:::n^g, for every non-negative integerN, : ^f1^g

f1:::n^g ^! ^N, and for all x ² (Q^;10] (for all x ² Q^;10] if q ⁶= ¹⁴).

(8)

Fixnow a positiveintegerLand suppose that (16) holds withM =

= 0 for every K¹:::KL ² ^f1:::n^g, for every non-negative integer N, : ^f1:::L^g^f1:::n^g ^! ^N, and for all x ² 01^;Q) (for all x ²01^;Q] if q ⁶= ¹⁴). Dening y as in Lemma 2 and making use of Lemma 2, (17) and (19) with x replaced byy, and (16) withM = 0 we obtain

f(q^N⁺^P^L^l⁼¹⁺¹^P^Kl^k⁼¹⁽^lk⁾x+ +^L^X⁺¹

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^L^j⁼⁺¹^l⁺¹^P^Kj^m⁼¹⁽^jm⁾) =

=f(q^P^KL^k⁼¹⁺¹⁽^L⁺¹^k⁾y+ (^;1)^L⁺¹^K^X^L⁺¹

k⁼¹ q^P^KL^m⁼⁺¹^k ⁽^L⁺¹^m⁾) =

=1 2

KL⁺¹ 1 2q

PKLk⁼¹⁺¹⁽L⁺¹k⁾f(y) =

=1 2

KL⁺¹ 1 2q

PKLk⁼¹⁺¹⁽L⁺¹k⁾1 2

PLl⁼¹Kl

1

2q ^N

+

PLl⁼¹^PKlk⁼¹⁽lk⁾

f(x) =

=1 2

PLl⁼¹⁺¹Kl 1 2q

N⁺^P^L_l⁼¹⁺¹^P^Kl_k⁼¹⁽lk⁾f(x): This ends the proof of (16) in the case where M = 0.

IfM is a positive integer then dening once moreyas in Lemma2 and making use of this lemma, (8) with N = 0 and x replaced by y, and (16) with M = 0 we get

f(q^N⁺^P^Ll⁼¹^P^Kl^k⁼¹⁽^lk⁾⁺^Mx+ +^X^L

l⁼¹(^;1)^l^X^K^l

k⁼¹q^P^Kl^m⁼^k⁽^lm⁾⁺^P^Lj⁼^l⁺¹^P^Kj^m⁼¹⁽^jm⁾⁺^M + (^;1)^{L M}^X

m⁼¹q^m) =

=fq^My+ (^;1)^{L M}^X

m⁼¹q^m=1 2

M 1 2q

Mf(y) =

=1 2

PLl⁼¹K_l⁺M 1 2q

N⁺^P_Ll⁼¹^P^Kl_k⁼¹⁽lk⁾⁺Mf(x):

(9)

The fourth lemma is just 7 Lemma 1].

Lemma 4.

Assume q ² (0¹²). If a solution of Schilling's problem vanishes either on the interval (^;q0) or on the interval (0q) then it vanishes everywhere.

Proof of Theorem 1.

Suppose f is a solution of Schilling's problem bounded in a neighbourhood of a point x⁰ ² clA_nq. We may (and we do) assume that x⁰ is of the form (5), where " ² ^f;11^g, ML are non-negative integers, K¹:::KL ² ^f1:::n^g, and : ^f1:::L^g

f1:::n^g^! ^N. Moreover, according to Remark 2, we may (and we do) assume"= 1.

If x²01^;Q) is xed then the left-hand side of (16) is bounded with respect to N whereas the right-hand side is bounded if(x) = 0.

This shows thatf vanishes on 01^;Q). Hence and from (10) it follows that f vanishes, in particular, on 0q) which jointly with Lemma 4 proves that f vanishes everywhere.

To formulate a corollary accept the following denition.

Denition 1.

Let q ² (01) and x ² ^;QQ]. We say that x ² Bq

(resp. x ² Cq) if and only if the zero function is the only solution of Schilling's problem which is bounded in a neighbourhood of x (resp.

continuous at x).

We will use also the following result of W. Forg{Rob cf. 6 The- orems 20, 21, 23{26 and 28] and Remark 1.

If q ²(01) and f is a solution of Schilling's problem then suppf ⁿ^X¹

n⁼¹"(n)qⁿ : "²^f;101^g^N^o

and for every q² (0 ¹³]the Schilling's problem has a nonzero solution.

Corollary 1.

If q²(0¹³] then Bq =Cq =ⁿ^X¹

n⁼¹"(n)qⁿ : "²^f;101^g^N^o:

Proof.

Obviously Bq Cq, whereas the above quoted result of W.

Forg{Rob gives Cq ⁿ

1

X

n⁼¹"(n)qⁿ : "²^f;101^g^N^o: Moreover, applying Remark 1 and Th. 1 we obtain that

(10)

(

1

X

n⁼¹"(n)qⁿ : "²^f;101^g^N

)

Bq:

Applying Lemma 3 (formula (16) with x = 0 and Remark 2) and Remark 3 we obtain also the following result.

Theorem 2.

If n is a positive integer and q²(0qn]then any solution of Schilling's problem vanishes on the set A_nq.

The reader interested in further results on Schilling's problem is referred to 2] by K. Baron, A. Simon and P. Volkmann, 3] by K. Baron and P. Volkmann, 4] by J. M. Borwein and R. Girgensohn, 5] by G. Derfel and R. Schilling, 6] by W. Forg{Rob and 8].

Acknowledgement.

This research was supported by the Silesian Uni- versity Mathematics Department (Iterative Functional Equations pro- gram).

References

1] BARON, K.: On a problem of R. Schilling,Berichte der Mathematisch-statis- tischen Sektion in der Forschungsgesellschaft Joanneum{Graz, Bericht Nr.

286(1988).

2] BARON, K., SIMON, A. et VOLKMANN, P.: Solutions d'une equation fonctionnelle dans l'espace des distributions temperees, Comptes Rendus des Seances de l'Academie des Sciences. Serie I. Mathematiques. (Paris) ³¹⁹ (1994), 1249{1252.

3] BARON, K. et VOLKMANN, P.: Unicite pour une equation fonctionnelle, Rocznik Naukowo{Dydaktyczny WSP w Krakowie. Prace Matematyczne ¹³ (1993), 53{56.

4] BORWEIN, J. M. and GIRGENSOHN, R.: Functional equations and distri- bution functions,Results in Mathematics ²⁶(1994), 229{237.

5] DERFEL, G. and SCHILLING, R.: Spatially chaotic congurations and functional equations with rescaling, Manuscript.

6] FORG-ROB, W.: On a problem of R. Schilling,Mathematica Pannonica ⁵/1 (1994), I 29{65, II: 145{168.

7] MORAWIEC, J.: On bounded solutions of a problem of R. Schilling, Annales Mathematicae Silesianae ⁸(1994), 97{101.

8] MORAWIEC, J.: On continuous solutions of a problem of R. Schilling,Results in Mathematics ²⁷(1995), 381-386.

9] SCHILLING, R.: Spatially chaotic structures, in: H. Thomas (editor), Non- linear Dynamics in Solids, Springer, Berlin, 1992, 213-241.