since the continuous linear operators L 1 ( ) and L 2 ( ) acting on R ( , , ; ) do not exist in sharp upper bounds. Putting = = 1 in the first part of Theorem 1, we can get the following result.Abstract and Applied Analysis 5 Corollary 7. Let ≧ 1. Then R( , ) ⊂ S * for 3 ≦ < 1, where 3 is the solution of the following equation: