View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector A Generalization of the Cosine-Sine Functional Equation on Groups J. K. Chung (J. K. Jong) Department of Mathematics South China Polytechnic Institute Canton, P.O.B. 51504, People’s Republic of China and PI. Kannappan and C. T. Ng* Department of Pure Mathematics University of Waterloo Waterloo, Ontario, Canada N2L 3Gl This paper is dedicated to Professor J, AczCl on his sixtieth birthday. Submitted by Riclxml A. Brualdi ABSTRACT The functional equation f(ry) = f(x)g(y)+ g(x)f(y)+ h(x)h(y) is solved where f, g, h are complex functions defined on a group. 1. INTRODUCTION Our aim is to obtain the general solution of the functional equation f(xy) = f(.r>g(y)+g(x>f(y)+h(x)h(y) (FE) where f, g, h : G + C, G is a group, and Q: is the complex field. This and all the subsequent equations are supposed to hold on the whole of G. No assumption is imposed on the group G; in particular, we do not require G to be abelian. The familiar cos(x + y) and sin(r + y) admit such expansion as in (FE). The main idea is to derive from (FE) similar identities for g(xy) and h(ry), and to solve these identities as a system of equations. *Research supported by NSERC grant of Canada. LlNEAR ALGEBRA AND ITS APPLICATlONS 66:259-277(1985) 259 (” Elsevier Science Publihing Co., Inc., 1985 52 Vanderbilt Ave.. New York, NY 10017 00243795/&s/$3.30 260 J. K. CHUNG, PL. KANNAPPAN, AND C. T. NG In the special case when G is abelian and h = 0, the general solution of (FE) follows easily from the earlier work in [4]. The case when the domain is an interval and the functions possess some regularity properties has been treated by several authors [l, p.197; 2, Lemma lo]. Here we make use of only the algebraic structures of G and C. All the results presented in this paper are valid when Q: is replaced by a field of characteristic different from two which is algebraically closed. Notation and terminoldgy A morphism A:G+C is said to be additive if A(xy)=A(x)+A(y); a map E : G + Q= is exponential provided E( xy ) = E( r )E( y ). An exponential map is either identically zero or nowhere zero. The capital letters A and E along with subscripts are used exclusively for additive and exponential maps respectively. For a map f: G -+ C, the notation f # 0 means that f is not identically zero on G; “f is nonzero” means f # 0. If x : G + Q= (i = l,%...,n), by span { fi, +. ) f, > we mean the linear space generated by f,,...,f, ouer C. Note that whenever (f, g, h) is a solution of (FE), so is (Ef, Eg, Eh). 2. PRELIMINARY RESULTS We prepare some lemmas covering some special cases of (FE) to be used in the main result given in the next section. LEMMA 1. Let n 2 1 be a fixed integer, and let E,, E,,. , E,, be exponential. Let f, g, h E span{ E,, E,, . , E, }, and suppose they satisfy (FE). Then there exist complex constants (a ,):= I, (b, )I’=,, (c, ):‘=I satisfying the relation . bl ‘2 . aI a2 . Cl c2 . so that f = C~=,a,Ei, g = C:‘=,b,E,, h = Cr=IciEi. Conversely, if f, g, and h are such combinutions with coeffacients satisfying the above relation, then they satisfy (FE). Furthermore, if any one off, g, h happens to be zero, the corresponding coefficients can be chosen to be zero. Proof. Let SC {1,2,...,n} be such that (E,),,, form a basis for span{ E,, E,, . , E, }. COSINE-SINE FUNCTIONAL EQUATION 261 We can write f=Xy=laiEi, g =XyElbiEi, and h = C:=lciEi, so that a i = bj = ci = 0 for all i @ S, in a unique manner. Putting f= Xi EsaiEi, g = CiEsbiEi, and h =IrESciEI in (FE), we use the independence of { Ei 1i E S} to conclude that (FE) holds if, and only if, the constants (ai)i,S,(bi)i,s,(Ci)iESSatiSfy uibi + biuj + cicj = Sijui for all i, j E S. With ui = b, = ci = 0 for i e S, this is equivalent to the asserted constraint in the matrix form. n The next lemma holds for an arbitrary field, including the complex field. LEMMA 2. Let G be a group and F be a$eld, and let A:G+F be additive and A f 0. Ifchar( F) = 0, then 1, A, A’, . is independent over F. Zf char(F) = p # 0, then 1, A, A’, . , Ape’ is independent over F. Proof Suppose C:=,h i A’ = 0, where hi E F and n = char(F) - 1 in case char(F) # 0. Then Cr,,A,A’(x)= 0 for all x E 6; thus every element in the image A(G) is a zero of the polynomial q(z) = CyzoXiZi in F[.z]. If 9 # 0, then it has at most n distinct zeros. However, A(G), as a nontrivial additive subgroup of F, has more than n distinct elements. Thus 9 = 0 must follow, i.e. h,=X,= ... =X,,=O. n LEMMAS. Let n > 0 be fixed, and let E,, E,, . , E,, E, + 1 be exponen- tial and A be additive. Zff, g, h E span{ E,, E,, . , E,, E,+l, E,+,A} satisfy (FE), then there exist constants (a i):=Y:, ( bi)y,+f, ( ci):,if satisfying the relu- tion a1 a2 b, b2 ” ’ bn+2 a, u2 ... Unt2 : ICI c2 ... %I+2 U in+2 cn+2 1 1:n-c2 Ul 0 I I / I 0 0 ‘a,1 I r---------- ’ ant1 Un+2 0 ; Un+2 0 262 J. K. CHUNG, PL. KANNAPPAN, AND C. T. NC so that n+l f = C a,Ei + a,+tnE,+lA, i=l n+l g = C biEi + bn+zEn+,A, i=l n+l h = c ciEi + c,czE,+,A. i=l Conversely, iff, g, h are such linear combinations with coefficients satisfying the above relation, then they satisfy (FE). Furthermore if any one off, g, h happens to be zero, the corresponding coefficients can be chosen to be zero. Proof. If A=O,wecanspecifya,+,=b,+,=c,+,=O.If E,+l=O,we can specify a,,, = an+2 = b,+l = bn+2 = c,+l = c,+z = 0. In both cases Lemma 1 can be applied to support the current constraint. If both E,, 1 and A are nonzero, by Lemma 2 it follows that E,, 1 and E ,+lA are independent. Let S c {1,2,. , n} be such that (Ei)i,s and E n+l7 E ,,+iA form a basis for span{ E,, E, ,..., E,+l, E,+,A}. We can write f,g, h in the asserted form such that a i = bi = ci = 0 for all i @ S, in a unique manner. They satisfy (FE) if and only if the scalars satisfy the asserted relation. n LEMMA 4. Let f, h: G --z c be mappings. Zf they satisfy the functional equation f(xy) = f(x)f(y)+h(r)h(y) on G, (2.1) then h(xy) = f(x)h(y)+ h(x)f(y)+2ah(x)h(y) for some constant a, and f and h are of the form f = a,E, + a,E + a-&4, (2.2) h = clE, + c,E + c,EA with the constraint a1 0 0 0 a2 a3 . (2.2c) [ 0 a3 0 I COSINE-SINE FUNCTIONAL EQUATION 263 Conversely, if f and h are given by (2.2) where the coefficients satisfy (2.2c), then they form a solution of (2.1) [cf. Remurk 31. Proof. Let f, h satisfy (2.1). If h = 0, then f is exponential and they are of the form (2.2). In what follows, we suppose that h z 0. Computing f(xyz) first as f((xy)z)) and then as f(x( yz)), using (2.1) we get [hbd - f(x)h(y) - hb)f(y)lhb) = [hb4 - f(y)hb) - h(y)fb)l h(r). (2.3) We fix z = z0 with h(z,) + 0 and get hbd - fb)h(y) - hb)f(y) = hb)k(y), (2.4) where k(y): = h(z,)-‘[h(yz,)- f(y)h(z,)- h(y)f(zO)]. Putting this back in (2.3), we have h(x)k(y)h(z)= h(y)k(z)h(x); and as h # 0, it follows that k = 2ah for some constant oz.Thus (2.4) yields h(w) = fbNy)+ hb)f(y)+2~h(x)h(y). (2.5) Under (2.1) and (2.5), simple computation shows that (f + Xh)by) = (f + Wb)(f + Ah)(y) if, and only if, X satisfies A2 - 2ha - 1= 0. By fixing h to be such a constant we get f + Ah = E,. (2.6) Observe that X + 0. Substituting f = E, - Ah in (2.1), we get Ah(xy)=Xh(x)E,(y)+hE,(x)h(y)-(A2+l)h(x)h(y). (2.7) There are two possibilities. One is E, = 0, in which case (2.7) gives h = - A(A2 + l)-‘E,, and so f = - Ah = A2(A2+ l))‘E,. They are of the form (2.2). The other is E, f 0. Dividing (2.7) by E,(xy)= E,(x)E,(y) side by 264 J.K.CHWNG,PL.KANNAPPAN,ANDC.T.NG side, we obtain When A’+l=O, Xh/E,=A f o ll ows and so f and h are of the form (2.2) [l, p.