f(z)-∫(0)=(x-0)f(2),5∈(0,z) (1) f(1)-∫()=(1-z)∫(m),T∈(z,1) (2) 注意到f(0)=0,f(1)=1,由(1),(2)有 b f(r atb l-∫(z)_a+b f"(4)f(5) (3)1-z ∫(m)f(m7) (4) b (3)+(4),得1 f"(4)(a+b)f(m)(a+b) b =a+b ∫"(4)f(7)f ( ) − f (0) = ( − 0) f ( ), (0, ) f (1) − f ( ) = (1− ) f (), ( ,1) (1) (2) 注意到 f (0) = 0, f (1) = 1, 由(1), (2)有 ( )( ) ( )( ) 1 f a b b f a b a + + + = f ( ) a b a + = (3) ( ) (4) 1 ( ) 1 f f − − = f () a b b + = (3)+ (4),得 ( ) ( ) f f = . ( ) ( ) a b f b f a = + +