JOURNAL OF COMPUTATIONAL ANALYSIS AND APPLICATIONS VOLUME 12, 2010

870 Pages·15.557 MB·English

Volume 12, Number 1-A January 2010
ISSN:1521-1398 PRINT,1572-9206 ONLINE

Journal of

Computational

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JOURNAL OF COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL.12, NO.1-A, 7-12,2010,COPYRIGHT 2010 EUDOXUS PRESS7, LLC
ON REPRESENTABILITY OF QUASI QUADRATIC
FUNCTIONALS BY SESQUILINEAR FUNCTIONALS
MEHMET A˙IKG(cid:214)Z AND ALIAKBAR GOSHABULAGHI
Abstract. In this paper, the representability of quasi-quadratic functionals
by sesquilinear ones over real and complex *-Banach algebras will be deter-
mined. Moreover, we will give some relative results for modules over general
*-rings.
1. Introduction
Let R be a *-ring with identity such that 2 is a unit in R and M be left R-
module. The mapping Q : M R is said to be quasi quadratic functional if for
!
any x;y M and (cid:11) R the parallelogram law
2 2
(1.1) Q(x+y)+Q(x y)=2Q(x)+2Q(y)
(cid:0)
and the homogeneity equation
(1.2) Q(ax)=aQ(x)a
(cid:3)
holds. A biadditive mapping S :M M R satisfying
(cid:2) !
(1.3) S(a x +a x ;y)=a S(x ;y)+a S(x ;y) (a ;a R; x ;x ;y M);
1 1 2 2 1 1 2 2 1 2 1 2
2 2
(1.4) S(x;a y +a y )=S(x;y )a +S(x;y )a (a ;a R; x;y ;y M)
1 1 2 2 1 (cid:3)1 2 (cid:3)2 1 2 2 1 2 2
is a sesquilinear functional.
Over *-algebra A an element with the property h = h is said to be hermitian.
(cid:3)
Anelementa Awillbecallednormalifa a=aa :Itisseenthatoveracomplex
(cid:3) (cid:3)
2
*-algebra A each element a A has a unique decomposition a = h + ik with
2
hermitian h and k: A *-algebra being a Banach algebra is said to be Banach *-
algebra. A Banach *-algebra is called hermitian if each hermitian element has real
spectrum. IfAbeahermitianBanach*-algebraandh Abeahermitianelement
2
then it is convenient to write h>0 if the spectrum of h be positive.
Now let M be a left R module over a *-ring R and Q : M R be a quasi
(cid:0) !
quadraticfunctional. Itisinterestingtoknowthatisthereasesquilinearfunctional
S such that Q(x)=S(x;x) for any x M?
2
In 1984 Vukman [10] possed the problem of representability of quasi quadratic
functionals by sesquilinear ones over complex *-algebras. The complete solution
was given in [2]. In our present work [1] we gave a solution to a general case,
modules over special *-rings including complex *-algebras. However here we will
giveasolutionoftheproblemofrepresentabilityofsomequasiquadraticfunctionals
Date:April12,2007.
2000 Mathematics Subject Classi(cid:133)cation. Primary 11E04,46C05,Secondary 11D09.
Key words and phrases. Quasi-quadratic functional, sesquilinear(bilinear) functional, *-
algebra,*-ring .
1
8
2 MEHMET A˙IKG(cid:214)Z AND ALIAKBAR GOSHABULAGHI
with sesquilinear functionals for ones over modules on general *-rings including
commutative *-rings with trivial involution.
2. Quasi-quadratic functionals over real Banach *-algebras
Lemma 1. Let X be a real vector space and Q : X R be additive functional.
!
Then the functional
1
m(x;y)= [Q(x+y) Q(x y)]
4 (cid:0) (cid:0)
is symmetric and biadditive.
Proof. See [10]: (cid:3)
Theorem 1. Let A be a commutative Banach algebra and X be a left A module.
(cid:0)
Moreover Q : X A be a mapping such that each invertible a A and x;y X
! 2 2
satis(cid:133)es:
1) Q(x+y)+Q(x y)=2Q(x)+2Q(y)
(cid:0)
2) Q(ax)=a2Q(x)
then for the functional S :X X A de(cid:133)ned by
(cid:2) !
1
S(x;y)= [Q(x+y) Q(x y)]
4 (cid:0) (cid:0)
the following conditions hold.
1) S is biadditive;
2) for any x;y X; S(x;x)=Q(x);
2
3) for any x;y X and a A;
2 2
S(ax;y)+S(x;ay)=2aS(x;y):
Proof. (1) and (2) are clear. To prove (3), let x;y X be (cid:133)xed and the mapping
2
f :A A be de(cid:133)ned in the way
!
f(a)=S(ax;y)+S(x;ay):
Since S is biadditive so f is additive. It is seen that for any invertible a A;
2
f(a) = a2f a 1 : For any a A the identity f(a) = af(e) holds(See[7]). So for
(cid:0)
2
any a A and x;y X;
2 (cid:0) (cid:1) 2
S(ax;y)+S(x;ay) = a[S(ex;y)+S(x;ey)]
= a[S(x;y)+S(x;y)]
= 2aS(x;y):
(cid:3)
Theorem 2. Let A be a commutative Banach algebra with identity and X be a
unitary left A module and let Q:X A be a such mapping that for any invertible
(cid:0) !
a A and x;y X;
2 2
i) Q(x+y)+Q(x y)=2Q(x)+2Q(y);
(cid:0)
ii) Q(ax)=a2Q(x);
iii) Let Q : X A be a functional such that a Q(ax+y) be continuous for
! !
any (cid:133)xed x;y X:
2
9
ON QUASI QUADRATIC FUNCTIONALS BY SESQUILINEAR FUNCTIONALS 3
Then then the functional S :X X A de(cid:133)ned by
(cid:2) !
1
S(x;y)= [Q(x+y) Q(x y)]
4 (cid:0) (cid:0)
is a bilinear form and for any x X; S(x;x)=Q(x):
2
Proof. From Theorem 1, it su¢ cies to prove
S(ax;y)=aS(x;y); S(x;ay)=aS(x;y):
For (cid:133)xed x;y X; de(cid:133)ne f :A A such as
2 !
f(a)=S(ax;y) S(x;ay);
(cid:0)
f is additive since S is biadditive. From (iii) f is continuous and for any invertible
a A; f(a) = a2f a 1 holds hence from [10]; Corollary 1 (3), for any a A;
(cid:0)
2 (cid:0) 2
f(a)=0 (See [6];[7]), that yields;
(cid:0) (cid:1)
S(ax;y)=S(x;ay);
and from Theorem 1, (3), the proof completes. (cid:3)
Theorem3. LetX beavectorspaceoveracomplex*-algebraAhavinganidentity
element 1. For any quasi-quadratic functional Q:X A de(cid:133)ne S :X X : A;
! (cid:2) !
1 i
S(x;y)= [Q(x+y) Q(x y)]+ [Q(x+iy) Q(x iy)]:
4 (cid:0) (cid:0) 4 (cid:0) (cid:0)
The following conditions hold.
(a) S is biadditive;
(b) For x;y X and a A;
2 2
S(ax;y)=aS(x;y); S(x;ay)=S(x;y)a ;
(cid:3)
(c) For any x X; Q(x)=S(x;x):
2
Proof. See [10]: (cid:3)
3. Modules over *-rings
Theorem 4. Let R be a *-ring with identity and there exist a R satisfying
0
2
(3.1) a +a =0; a a =1
0 (cid:3)0 0 (cid:3)0
(3.2) a a=aa ; (a R):
0 0
2
then for any quasi quadratic functional Q:M R; the functional S :M M R
! (cid:2) !
de(cid:133)ned by
1 a
S(x;y)= (Q(x+y) Q(x y))+ 0 (Q(x+a y) Q(x a y))
0 0
4 (cid:0) (cid:0) 4 (cid:0) (cid:0)
is the unique sesquilinear functional satisfying Q(x)=S(x;x) for any x M:
2
Proof. From, Theorem 1 of [1], S is sesquilinear and for any x M; S(x;x) =
2
Q(x):Theonlyfactremainstoproveisuniqueness. LetT beanothersesquilinear
functional satisfying the condition of theorem. Now for x;y M;
2
S(x;x)+S(x;y)+S(y;x)+S(y;y) = S(x+y;x+y)=Q(x+y)
= T (x+y;x+y)
= T (x;x)+T (x;y)+T (y;x)+T (y;y)
10
4 MEHMET A˙IKG(cid:214)Z AND ALIAKBAR GOSHABULAGHI
So
(3.3) S(x;y)+S(y;x)=T (x;y)+T (y;x)
Now by replacing x with a x we conclude
0
S(x;y) S(y;x)=T (x;y) T (y;x)
(cid:0) (cid:0)
This with (3:3) yields S =T: (cid:3)
Proposition 1. Let R be a *-ring with identity then R R is a *-ring with the
(cid:8)
addition (a;b)+(c;d)=(a+c;b+d); product (a;b):(c;d)=(ac bd;ad+bc) and
(cid:0)
involution (a;b)(cid:3) =(a(cid:3); b(cid:3)):
(cid:0)
Proof. Since
(c;d)(cid:3):(a;b)(cid:3) = (c(cid:3); d(cid:3)):(a(cid:3); b(cid:3))=(c(cid:3)a(cid:3) d(cid:3)b(cid:3); c(cid:3)b(cid:3) d(cid:3)a(cid:3))
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
= (ac bd;ad+bc)(cid:3)
(cid:0)
= ((a;b):(c;d))(cid:3):
Reminder of proof is clear. (cid:3)
Proposition 2. Let M be a left R-module then M M is a left module over R R
(cid:8) (cid:8)
with the module action de(cid:133)ned by
(a;b):(x;y)=(ax by;ay+bx) (a;b R; x;y M):
(cid:0) 2 2
Proof. For any a;b;c;d R; x;y M we have;
2 2
(a;b):[(c;d):(x;y)] = (a;b):(cx dy;cy+dx)
(cid:0)
= (acx ady bcy bdx;acy+adx+bcx bdy)
(cid:0) (cid:0) (cid:0) (cid:0)
= ((ac bd)x (ad+bc)y;(ac bd)y+(ad+bc)x)
(cid:0) (cid:0) (cid:0)
= (ac bd;ad+bc):(x;y)
(cid:0)
= ((a;b)(c;d)):(x;y):
One can easily check other module action properties. (cid:3)
Proposition 3. i) R=R 0 R R;
(cid:24) (cid:8) (cid:18) (cid:8)
ii) (1;0) is the identity of R R and a =(0;1) satis(cid:133)es (3:1);(3:2):
0
(cid:8)
Theorem 5. Let M be a left module over *-ring R with identity such that 2 be a
unit in R: Moreover, let q : M R be a functional such that for any a;b R;
! 2
x;y M;
2
(3.4) q(ax by)+q(ay+bx)=a(q(x)+q(y))a +b(q(x)+q(y))b ;
(cid:3) (cid:3)
(cid:0)
and
(3.5) aq(x)b =bq(x)a
(cid:3) (cid:3)
then the functional S :M M R de(cid:133)ned by
(cid:2) !
1
S(x;y)= (q(x+y) q(x y));
4 (cid:0) (cid:0)
is sesquilinear and S(x;x)=q(x) for any x M:
2