Solution of the Higher Order Cauchy Difference Equation on Free Groups

Transcription

1 Global Journal of Pure and Applied Mathematics. ISSN Volume 13, Number 8 (2017, pp Research India Publications Solution of the Higher Order Cauchy Difference Equation on Free Groups K. Thangavelu Department of Mathematics, Pachaiyappa s College, Chennai , India. M. Pradeep Department of Mathematics, Arignar Anna Government Arts College, Cheyyar , India. Abstract Let f : G H be a function, where (G,. is a group and (H,+ is an abelian group. In this paper, the following n th Order Cauchy difference of f : C (n f(x 1,x 2,...,x n+1 = n+1 n+1 n+1 f(c n+1 ( x i f(c n ( x i +...+( 1 n f(c 1 ( x i x 1,x 2,...,x n+1 G is studied. Where f(c r ( n x i is defined as function of combination r at a time from n objects. We give some special solutions of C (n f = 0 on free groups. AMS subject classification: Keywords: Cauchy difference equation, free groups. 1. Introduction It is well known from [1] that Jenson s functional equation f(x+ y + f(x y = 2f(x (1.1 with additional condition f(0 = 0, is equivalent to Cauchy s equation f(x+ y = f(x+ f(y

2 4142 K. Thangavelu and M. Pradeep on the real line. Let (G,. be a group, (H, + is an abelian group. Let e G and 0 H denote identity elements. The study of (1.1 was extended groups for f maps G into H in [2], where the general solution for a free group H with two generators and G = GL n (z, n 3 (see [3]. Since the functional equations involve Cauchy difference, which made it become much more interesting [4 7]. For a function f: G H, its cauchy difference C (m f, is defined by C (0 f = f, (1.2 C (1 f(x 1,x 2 = f(x 1 x 2 f(x 1 f(x 2 (1.3 C (m+1 f(x 1,x 2,...,x m+2 = C (m f(x 1,x 2,x 3...,x m+2 C (m f(x 1,x 3,...,x m+2 C (m f(x 2,x 3,...,x m+2 (1.4 The first order cauchy difference C (1 f will be abbreviated as Cf. In [9], by using the reduction formulas and relations, as given in [2, 3], the general solution of third order Cauchy difference equation was provided on free groups. In this paper, we consider the following functional equation: n+1 f(c n+1 ( n+1 x i f(c n ( n+1 x i ( 1 n f(c 1 ( x i = 0 (1.5 It follows from (1.4 that (1.5 is equivalent to the vanishing n th order cauchy difference equation C (n f = 0 The purpose of this paper is to determine the solutions of equation (1.5. The solution of equation (1.5 will be denoted by KerC (n (G, H = {f : G H fsatisfies(1.5} (1.6 Remark KerC (n (G, H is an abelian group under the pointwise addition of functions; 2. H om(g, H KerC (n (G, H.

3 Solution of the Higher Order Cauchy Difference Equation Properties of Solutions Lemma 2.1. Suppose that f KerC (n (G, H. Then f(e = 0, (2.1 Cf (x 1,x 2 = 0, when x 1 = e or x 2 = e (2.2 C (2 f(x 1,x 2,x 3 = 0, when x 1 = e or x 2 = e or x 3 = e ( C (n 1 f(x 1,...,x n = 0, when x 1 = e or x 2 = e or...or x n = e (2.4 C (n 1 f is a homomorphism with respect to each variable (2.5 f(x n = n f (x + nc 2 Cf (x, x + nc 3 C (2 f(x,x,x...+ nc n C (n 1 f(x,...,x(n times (2.6 for all x 1,...,x n G and n Z. Proof. Putting x 1 = e in (1.5 we get (2.1. Then from (2.1 we obtain (2.2 (2.4 Cf (x 1,e= f(x 1 e f(x 1 f(e = f(x 1 f(x 1 = 0 Similarly we can obtain Cf (e, x 2 = 0, C (2 f(e,x 2,x 3 = f(ex 2 x 3 f(ex 2 f(ex 3 f(x 2 x 3 + f(e+ f(x 2 + f(x 3 = 0, Similarly we can obtain C (2 f(x 1,e,x 3 = 0, C (2 f(x 1,x 2,e= 0,... C (n 1 f(e,...,x n = 0, C (n 1 f(x 1,e,...,x n = 0,... C (n 1 f(x 1,...,e= 0.

4 4144 K. Thangavelu and M. Pradeep Furthermore, by the definition of C (n 1 f,wehave C (n 1 f(x 1,x 2 x 3,...,x n+1 = f(x 1...x n+1 f(x 1...x n f(x 1...x n 2 x n+1... f(x 1 x 4...x n+1 f(x 2 x 3...x n+1 + +( 1 n [ f(x 1 + f(x 2 x 3 + f(x f(x n+1 ] and C (n 1 f(x 1,x 2,x 4,...,x n+1 + C (n 1 f(x 1,x 3,x 4,...,x n+1 = f(x 1 x 2 x 4...x n+1 f(x 1 x 2 x 4...x n f(x 1 x 2 x 4...x n 1 x n+1 f(x 1 x 4...x n+1... f(x 2 x 4...x n ( 1 n [ f(x 1 + f(x 2 + f(x f(x n+1 ] +f(x 1 x 3 x 4...x n+1 f(x 1 x 3 x 4...x n f(x 1 x 3 x 4...x n f(x 1 x 3 x 4...x n 1 x n+1... f(x 1 x 4...x n+1 f(x 3 x 4...x n ( 1 n [ f(x 1 + f(x 3 + f(x f(x n+1 ] One can easily check that C (n 1 f(x 1,x 2 x 3,...,x n+1 C (n 1 f(x 1,x 2,x 4,...,x n+1 C (n 1 f(x 1,x 3,x 4,...,x n+1 = C (n f(x 1,x 2,x 3,...,x n+1 = 0 Hence, the above relations imply the C (n 1 f (., x 2,...,x n is a homomorphism. Similarly, the fact is also true for C (n 1 f(x 1,...,x 3,...,x n,, C (n 1 f(x 1,...,.,x n and C (n 1 f(x 1,...,x n 1,.. This proves (2.5. We now consider (2.6. Actually, it is trivial for n = 0, 1, 2,...,n 1 by (2.1 and by the definition of Cf. Suppose that (2.6 holds for all natural numbers smaller than

5 Solution of the Higher Order Cauchy Difference Equation 4145 k n + 1, then f(x n = f(xxx...x(n times = n f(xx...x(n-1 times nc n 2 f(xx...x(n-2 times +nc n 3 f(xx...x(n-3 times...+ ( 1 n n f (x +C (n 1 f(x,x,...,x(n times = n f(x n 1 nc n 2 f(x n 2 + nc n 3 f(x n ( 1 n n f (x + C (n 1 f(x,x,...,x(n times [ = n (n 1 f (x + (n 1C 2 Cf (x, x + (n 1C 3 C (2 f(x,x,x (n 1C n 1 C n 2 f(x,x,...,x(n-1 times ] nc n 2 [(n 2 f (x + (n 2C 2 Cf (x, x + (n 2C 3 C (2 f(x,x,x (n 2C n 2 C n 3 f(x,x,...,x(n-2 times ] +nc n 3 [(n 3 f (x + (n 3C 2 Cf (x, x + (n 3C 3 C (2 f(x,x,x (n 3C n 3 C n 4 f(x,x,...,x(n-3 times ]...+ ( 1 n n f (x + C (n 1 f(x,x,...,x(n times = n f (x + nc 2 Cf (x, x + nc 3 C (2 f(x,x,x nc n C (n 1 f(x,...,x(n times where the definition of C (n 1 f and (2.5 are used in the second equation. This gives (2.6 for all n 0. On the other hand, for any fixed integer n>0, by (1.4 and (2.1, we have f(x n = n f (x + nc 2 Cf (x, x + nc 3 C (2 f(x,x,x nc n C (n 1 f(x,...,x(n times from (2.5 and the above claim for n>0. This confirms (2.6 for n<0. Remark 2.2. For any function f : G H, the following statements are pairwise equivalent: (i The function f KerC (n (G, H ; (ii C (n 1 f (., x 2,...,x n is a homomorphism; (iii C (n 1 f(x 1,.,x 3,...,x n is a homomorphism; (iv C (n 1 f(x 1,...,.,x n is a homomorphism;

6 4146 K. Thangavelu and M. Pradeep (v C (n 1 f(x 1,...,x n 1,.is a homomorphism; Before presenting Proposition 2.4, we first introduce the following useful lemma, which was given in [8]. Lemma 2.3. (Lemma 2.4 in [8] The following identity is valid for any function f : G H and l N; f(x 1 x 2...x l = C (m 1 f(x i1,x i2,...,x im (2.7 m l 1 < <...<i m l Proposition 2.4. Suppose that f KerC (n (G, H. Then f(x n 1 1 xn 2 = 2...xn l 1 i l l [ n i f(x i + n i C 2 Cf (x i,x i + n i C 3 C (2 f(x i,x i,x i ] n i C n C (n 1 f(x i,...,x i (n times + Cf (x n,x n + n i1 n i2 n i3 C (2 f(x i1,x i2,x i3 1 < l < <...<i n l 1 < <i 3 l n i1 n i2...n in C (n 1 f(x i1,x i2,...,x in (2.8 for n i Z and all x i G, i = 1, 2,...l such that x j = x j+1, j = 1, 2,...,l 1. Proof. Replacing x i in (2.7 by x n i i,wehave f(x n 1 1 xn xn l l = m l 1 < <...<i m l The vanishing of C (m 1 f for m (n + 1 yields f(x n 1 1 xn xn l l = f(x n i i + 1 i l + 1 < <i 3 l Therefore, by (2.6 and (2.5, we have C (m 1 f(x n,x n,...,x n im i m 1 < l Cf (x n,x n C (2 f(x n,x n,x n i 3 i 3 1 < <...<i n l C (n 1 f(x n,x n,...,x n in i n f(x n i i = n i f(x i + n i C 2 Cf (x i,x i n i C n C (n 1 f(x i,x i,...,x i (n times

7 Solution of the Higher Order Cauchy Difference Equation 4147 C (2 f(x n,x n,x n i 3 i 3 = n i1 n i2 n i3 C (2 f(x i1,x i2,x i3 C (3 f(x n,x n,x n i 3 i 3,x n i 4 i 4 = n i1 n i2 n i3 n i4 C (3 f(x i1,x i2,x i3,x i4... C (n 1 f(x n,x n,...,x n in i n = n i1 n i2...n in C (n 1 f(x i1,x i2,...,x in which is (2.8. This completes proof. Remark 2.5. In particular, if l = 1, then Proposition 2.4 holds. 3. Solution on a free group In this section, we study the solutions on free group. We first solve (1.5 for the free group G on a single letter a. Theorem 3.1. Let G be the free group on one letter a. Then f KerC (n (G, H iff it is given by f(a n = n f (a + nc 2 Cf (a, a + nc 3 C (2 f(a,a,a nc n C (n 1 f(a,a,...,a(n times n Z (3.1 Proof. Necessity. It can be obtained from (2.6 in Lemma 2.3. Sufficiency. Taking (3.1 as the definition of f on G=< a>. By Remark 2.5, we only need to verify that C (n 1 f is a homomorphism with respect to each variable and thus f belongs to KerC (n (G, H. Let x = a m 1,y = a m 2,...,u= a m n be any four elements of G. Then it follows from (1.4 and (3.1 that C (n 1 f(x,y,...,u= C (n 1 f(a m 1,a m 2,...,a m n = f(a m 1+m m n f(a m 1+m m n 1 f(a m m n 2 +m n... f(a m m n + +( 1 n [ f(a m 1 + f(a m f(a m n ] = N 1 f(a+ N 1 C 2 Cf (a, a N 1 C n C (n 1 f(a,a,...,a(n times N 2 f(a N 2 C 2 Cf (a, a... N 2 C n C (n 1 f(a,a,...,a(n times N 3 f(a N 3 C 2 Cf (a, a... N 3 C n C (n 1 f(a,a,...,a(n times

8 4148 K. Thangavelu and M. Pradeep N 4 f(a N 4 C 2 Cf (a, a... N 4 C n C (n 1 f(a,a,...,a(n times +( 1 n [m 1 f(a+ m 1 C 2 Cf (a, a m 1 C n C (n 1 f(a,a,...,a(n times + m 2 f(a+ m 2 C 2 Cf (a, a m n C n C (n 1 f(a,a,...,a(n times + + m n f(a+ m n C 2 Cf (a, a m n C n ] C (n 1 f(a,a,...,a(n times Where N 1 = m 1 + m 2 + +m n,n 2 = m 1 + m 2 + +m n 1,N 3 = m m n 2 + m n,n 4 = m 2 + +m n,. By a tedious calculation, we have C (n 1 f(a m 1,a m 2,...,a m n = m 1 m 2...m n C (n 1 f(a,a,...,a(n times which leads to the result that C (n 1 f is a homomorphism with respect to each variable. At the end of this section, for the free group on an alphabet A with A 2, we discuss some special solution of (1.5. An element x A can be written in the form x = a n 1 1 an an l l, where a i A,n i Z (3.2 For each fixed a A and fixed pair of distinct a,b A, define the functions W 1, W 2, W 3 : W 1 (x; a = a i =a n i (3.3 W 2 (x; a,b = W 3 (x; a,b = i<j,a i =a,a j =b i>j,a i =a,a j =b n i n j (3.4 n i n j (3.5 along with (3.2. Referring to [2, 3], the above functions are well defined. Furthermore, they satisfy the following relations: W 1 (xy; a = W 1 (x; a + W 1 (y; a (3.6 W 2 (x; a,b = W 3 (x; b, a. (3.7 Proposition 3.2. For any fixed a A and fixed pair of distinct a,b in A, the following assertions hold:

9 Solution of the Higher Order Cauchy Difference Equation 4149 (i W 1 (.; a belongs to KerC (n (A,Z; (ii W 2 (.; a belongs to KerC (n (A,Z; (iii W 3 (.; a belongs to KerC (n (A,Z. Proof. Claim (i follows from the fact that x W(x; a is a morphism from A to Z by (3.6. Now we consider assertion (ii. Let x 1,x 2,...,x n+1 in the free group be written as x 1 = a r 1 1 ar ar l l, x 2 = b s 1 1 bs bs p p, x 3 = c t 1 1 c t c t q q, x n 1 = d m 1 1 dm d m v v, x n = g n 1 1 gn gn k k, x n+1 = e u 1 1 eu eu h h, Let R = r i r j ; S = s i s j ; T = t i t j i<j,a i =a,a j =b M = R s = i<j,d i =a,d j =b a i =a,b j =b R m = S t = S m = a i =a,d j =b b i =a,c j =b i<j,b i =a,b j =b i<j,c i =a,c j =b m i m j ; N = n i n j ; U = u i u j r i s j ; R t = a i =a,c j =b i<j,g i =a,g j =b r i t j i<j,e i =a,e j =b r i m j ; R n = r i n j ; R u = r i u j s i t j a i =a,g j =b a i =a,e j =b s i m j ; S n = s i n j ; S u = s i u j b i =a,d j =b M n = b i =a,g j =b b i =a,e j =b m i n j ; M u = m i u j ; N u = n i u j d i =a,g j =b d i =a,e j =b g i =a,e j =b

10 4150 K. Thangavelu and M. Pradeep Then W 2 (x 1 x 2...x n+1 ; a,b = R + S + T N + U + R s + R t R n + R u +S t S n + S u M n + M u + N u W 2 (x 1 x 2...x n ; a,b = R + S + T M + N + R s + R t R m + R n +S t S m + S n M n W 2 (x 1...x n 1 x n+1 ; a,b = R + S + T M + U + R s + R t R m + R u +S t S m + S u M u W 2 (x 1 x 2 ; a,b = R + S + R s W 2 (x 1 x 3 ; a,b = R + T + R t W 2 (x 1 x n+1 ; a,b = R + U + R u W 2 (x 2 x 3 ; a,b = S + T + S t W 2 (x 2 x n+1 ; a,b = S + U + S u W 2 (x n x n+1 ; a,b = N + U + N u W 2 (x 1 ; a,b = R W 2 (x 2 ; a,b = S W 2 (x n ; a,b = N W 2 (x n+1 ; a,b = U Hence, we have W 2 (x 1 x 2...x n+1 ; a,b W 2 (x 1 x 2...x n ; a,b W 2 (x 1...x n 1 x n+1 ; a,b... W 2 (x 2...x n x n+1 ; a,b + +( 1 n [ W 2 (x 1 ; a,b + W 2 (x 2 ; a,b W 2 (x n 1 ; a,b +W 2 (x n ; a,b + W 2 (x n+1 ; a,b ] = 0. This concludes assertion (ii. Claims (iii follows from (3.7 directly.

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