By Rita A. Hibschweiler, Thomas H. MacGregor

This quantity is concentrated on Banach areas of capabilities analytic within the open unit disc, equivalent to the classical Hardy and Bergman areas, and weighted models of those areas. different areas into consideration the following contain the Bloch house, the households of Cauchy transforms and fractional Cauchy transforms, BMO, VMO, and the Fock house. a number of the paintings bargains with questions about capabilities in numerous advanced variables

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The proof is complete. 1. 1, if we assume that φ Φ for i = 1, . . 7), we get Φ−φ 2 Cn ( f , f ) 2 . 9) 4 which in turn can be considered a discrete version of the well-known Gr¨uss inequality. |Cn ( f , g)| The discrete Gr¨uss-type inequality given in [138] is embodied in the following theorem. 2. Let f = ( f1 , . . , fn ), g = (g1 , . . , gn ) be two n-tuples of real numbers and p = (p1 , . . 10) 1 n ∑ pi gi . 11) j=1 where Cn (p, f , g) = 1 n ∑ pi fi gi − Pn i=1 1 n ∑ pi fi Pn i=1 ˇ Gr¨uss-and Cebyˇ sev-type inequalities Proof.

1, if we assume that φ Φ for i = 1, . . 7), we get Φ−φ 2 Cn ( f , f ) 2 . 9) 4 which in turn can be considered a discrete version of the well-known Gr¨uss inequality. |Cn ( f , g)| The discrete Gr¨uss-type inequality given in [138] is embodied in the following theorem. 2. Let f = ( f1 , . . , fn ), g = (g1 , . . , gn ) be two n-tuples of real numbers and p = (p1 , . . 10) 1 n ∑ pi gi . 11) j=1 where Cn (p, f , g) = 1 n ∑ pi fi gi − Pn i=1 1 n ∑ pi fi Pn i=1 ˇ Gr¨uss-and Cebyˇ sev-type inequalities Proof.

2) By direct computation it is easy to observe that the following discrete Korkine’s type identity holds: Cn ( f , g) = 1 n n ∑ ∑ ( fi − f j ) (gi − g j ) . 3) It is easy to observe that Cn ( f , f ) = 1 n 2 ∑ fi − n i=1 1 n ∑ fi n i=1 2 . 4) 46 Analytic Inequalities: Recent Advances Furthermore, by using the Cauchy-Schwarz inequality for sums, we observe that Cn ( f , f ) 0. Similarly, Cn (g, g) 0. 3) and using the Cauchy-Schwarz inequality for dou- ble sums, we have 1 n n ∑ ∑ ( fi − f j ) (gi − g j ) 2n2 i=1 j=1 |Cn ( f , g)|2 = 1 n n ∑ ∑ ( fi − f j )2 2n2 i=1 j=1 2 1 n n ∑ ∑ (gi − g j )2 2n2 i=1 j=1 = Cn ( f , f )Cn (g, g).