Grüss Type Inequalities for Complex Functions Defined on Unit Circle with Applications for Unitary Operators in Hilbert Spaces

Some Grüss type inequalities for the Riemann-Stieltjes integral of continuous complex valued integrands defined on the complex unit circle C(0, 1) and various subclasses of integrators are given. Natural applications for functions of unitary operators in Hilbert spaces are provided.


Introduction
In [10], in order to extend the Grüss inequality to Riemann-Stieltjes integral, the author introduced the following µ Cebyšev functional The following result that provides sharp bounds for the µ Cebyšev functional de-…ned above was obtained in [10].The constant 1  2 is sharp.c) Assume that f; g are Riemann integrable functions on [a; b] and f satis…es the condition (1.2).If u : [a; b] !C is Lipschitzian with the constant L; then we have the inequality jT (f; g; u)j (1.5) The constant 1  2 is best possible in (1.5).For some recent inequalities for Riemann-Stieltjes integral see [1]- [5], [6]- [13] and [16].
For continuous functions f; g : C (0; 1) !C, where C (0; 1) is the unit circle from C centered in 0 and u : In this paper we establish some bounds for the magnitude of S C (f; g; u; a; b) when the integrands f; g : C (0; 1) !C satisfy some Hölder's type conditions on the circle C (0; 1) while the integrator u is of bonded variation.
It is also shown that this functional can be naturally connected with continuous functions of unitary operators on Hilbert spaces to obtain some Grüss type inequalities for two functions of such operators.
We recall here some basic facts on unitary operators and spectral families that will be used in the sequel.
We say that the bounded linear operator U : H ! H on the Hilbert space H is It is well known that (see for instance [15, p. 275-p. 276]), if U is a unitary operator, then there exists a family of projections fE g 2[0;2 ] , called the spectral family of U with the following properties: a) E E for 0 where the integral is of Riemann-Stieltjes type.Moreover, if fF g 2[0;2 ] is a family of projections satisfying the requirements a)-d) above for the operator U; then F = E for all 2 [0; 2 ] : Also, for every continuous complex valued function f : C (0; 1) !C on the complex unit circle C (0; 1), we have where the integral is taken in the Riemann-Stieltjes sense.
In particular, we have the equalities

Inequalities for Riemann-Stieltjes Integral
We say that the complex function f : C (0; 1) !C satis…es an H-r-Hölder's type condition on the circle C (0; 1) ; where H > 0 and r 2 (0; 1] are given, if (2.1) jf (z) f (w)j H jz wj r for any w; z 2 C (0; 1) : If r = 1 and L = H then we call it of L-Lipschitz type.
Consider the power function f : Cn f0g !C, f (z) = z m where m is a nonzero integer.Then, obviously, for any z; w belonging to the unit circle C (0; 1) we have the inequality jf (z) f (w)j jmj jz wj which shows that f is Lipschitzian with the constant L = jmj on the circle C (0; 1) : For a 6 = 1; 0 real numbers, consider the function f : jf a (z) f a (w)j = jaj jz wj j1 azj j1 awj for any z; w 2 C (0; 1) : If z = e it with t 2 [0; 2 ] ; then we have For other examples of Lipschitzian functions that can be constructed for power series on Banach algebras see [14].
The following result holds: jS C (f; g; u; a; b)j HKB r;q (a; b) Proof.We have the following identity exists and the following inequality holds (2.8) Applying this property twice, we have jS C (f; g; u; a; b)j (2.9) Utilising the properties of f and g we have In this case we get the inequality and the bound in (2.19) is proved.
The case of Lipschitzian integrators is of importance and can be stated as follows: The case of monotonic nondecreasing integrators is as follows: Theorem 4. Assume that f : C (0; 1) !C is of H-r-Hölder's type and g : Utilising this property and the identity (2.7) we have  Proof.We have to calculate If we insert these values in the right hand side of (2.27) we can get some expressions containing only Riemann integrals.However they are complicated and will not be presented here.

Applications for Functions of Unitary Operators
We have the following vector inequality for functions of unitary operators.
Theorem 5. Assume that f : C (0; 1) !C is of H-r-Hölder's type and g : C (0; 1) !C is of K-q-Hölder's type.If the operator U : H ! H on the Hilbert space H is unitary, then jhx; yi hf (U ) g (U ) x; yi hf (U ) x; yi hg (U ) x; yij (3.1) By the Cauchy-Buniakovski-Schwarz inequality for sequences of real numbers we also have that

g
(t) du (t) ; where f; g are continuous on [a; b] and u is of bounded variation on [a; b] with u (b) 6 = u (a) :

Theorem 1 .g
Let f : [a; b] !R, g : [a; b] !C be continuous functions on [a; b] and u : [a; b] !C with u (a) 6 = u (b) : Assume also that there exists the real constants ; such that (1.2) f (t) for each t 2 [a; b] : a) If u is of bounded variation on [a; b] ; then we have the inequality jT (f; g; u)j (denotes the total variation of u in [a; b] : The constant 1 2 is sharp, in the sense that it cannot be replaced by a smaller quantity.b) If u : [a; b] !R is monotonic nondecreasing on [a; b] ; then one has the inequality: (s) du (s) du (t) : [a; b] [0; 2 ] !C is a function of bounded variation on [a; b] with u (a) 6 = u (b) ; we can de…ne the following functional as well (1.6) S C (f; g; u; a; b) := 1 u (b) u (a) Z b a f e it g e it du (t) 1 u (b) u (a) Z b a f e it du (t) 1 u (b) u (a) Z b a g e it du (t) :

f e i 2 d kE xk 2 = Z 2 0f e i 2 d 2 0exp e i dE and U n = Z 2 0 2 0 2 0cos e i dE and the hyperbolic functions by
hE x; xi ; for any x; y 2 H: Examples of such functions of unitary operators are exp (U ) = Z e in dE for n an integer.We can also de…ne the trigonometric functions for a unitary operator U by sin (U ) = Z sin e i dE and cos (U ) = Z z exp ( z)] and cosh (z) := 1 2 [exp z + exp ( z)] ; z 2 C.

1 2 r+q 1 it f e is g e it g e it du (s) dt M 2 2eUtilising the well known inequality jsin xj jxj for any x 2
(r + q + 1) (r + q + 2) (b a) r+q+2 :Proof.It is well known that if p : [a; b] !C is a Riemann integrable function and v : [a; b] !C is Lipschitzian with the constant M > 0, then the Riemann-Stieltjes integral R b a p (t) dv (t) exists and the following inequality holds (2.20) Z b a p (t) dv (t) M Z b a jp (t)j dt: Utilising this property and the identity (2.7) we have jS C (f; g; u; a; b)j (2.21) ju (b) u (a)j f e is g e it g e it dsdt: Utilising the properties of f and g we have f e it f e is g e it g e it HK e is e it r+q for any s; t 2 [a; b] ; which implies that Z is e it r+q dsdt = 2 r+q HK

Corollary 1 . 1 thenC 1 ; 1 ( 2 sin 2 b a 2 #
Assume that u : [a; b] [0; 2 ] !C is a M -Lipschitzian function with u (a) 6 = u (b) : If f and g are Lipschitzian with the constants L and N then (2.22) jS C (f; g; u; a; b)j 4M 2 N L ju (b) u (a)j t) dt = 1 cos (b a) and Z b a sin (a t) dt = cos (b a) and the inequality (2.22) then follows from (2.18).

2 : 2
23) jS C (f; g; u; a; b)j 2 p+q 1 HK [u (b) u (a)] 2 D r;q (a; b) where D r;q (a; b) max s;t2[a;b] 2 sin s t 2 r+q [u (b) u (a)] Proof.It is well known that if p : [a; b] !C is a continuous function and v : [a; b] !R is monotonic nondecreasing on [a; b] ; then the Riemann-Stieltjes integral R b a p (t) dv (t) exists and the following inequality holds ()j dv (t) :

Corollary 2 .
If f and g are Lipschitzian with the constants L and N and u : [a; b] [0; 2 ] !R is a monotonic nondecreasing function with u (a) < u (b) ; then jS C (f; g; u; a; b)j (2.27) LN [u (b) u (a)]

2 J 2 :Remark 3 .
t) du (s) du (t) : Since cos (s t) = cos s cos t + sin s sin t then J (a; b) Utilising (2.23) we deduce the desired result (2.27).Utilising the integration by parts formula for the Riemann-Stieltjes integral, we have Z b a cos sdu (s) = u (b) cos b u (a) cos a + Z b a u (s) sin sds and Z b a sin sdu (s) = u (b) sin b u (a) sin a Z b a u (s) cos sds: 1HK kxk 2 kyk In particular, if f and g are Lipschitzian with the constants L and N; then It is well known that, if P is a nonnegative selfadjoint operator on H; i.e., hP x; xi 0 for any x 2 H; then the following inequality is a generalization of the Schwarz inequality in H 2for any x; y 2 H: Remark 4. If U : H ! H is an unitary operator on the Hilbert space H; then for any integer m; n we have from (3.2) the power inequalities This function is Lipschitzian with the constant L a = jaj (1 jaj) 2 on the circle C (0; 1) : Now, if we take a; b 6 = 1; 0 and use the inequality (3.2) then we have Theorem 6.If f and g are Lipschitzian with the constants L and N and U : H ! H is an unitary operator on the Hilbert space H; then Since Re e it = cos t and Im e it = sin t; Remark 5.If U : H ! H is an unitary operator on the Hilbert space H; then for any integer m; n we have from (3.13) the power inequalities If we take n = 1 and m = n and take into account that U n x; x = (U n ) x; x = hx; U n xi = hU n x; xi for any x 2 H, then we get from (3.15) that Now, if we take a; b 6 = 1; 0 and use the inequality (3.13), then we get 2for any x; y 2 H:For a 6 = 1; 0 real numbers, consider the function f :C (0; 1) !C, f a (z) =1 1 az : 2 for any x; y 2 H: 2 for any x; y 2 H: 2 = jhRe (U ) x; xi + i hIm (U ) x; xij 2 = hRe (U ) x; xi 2 + hIm (U ) x; xi 2 we deduce from (3.14) the desired inequality (3.13). 2 i for any x 2 H: 2 i for any x 2 H: