So by using the

triangular inequality and (1), we get

Applying again the

triangular inequality and using the fact that mj j = [m.

1] = p and p [member of] [0, 2], using

triangular inequality and [absolute value of z] [less than or equal to] 1, we have

In order to prove the integral version of the above Theorem, it is necessary to have the integral inequality, corresponding to (A), be true for every pair of points x and y in the orbit of u, instead of just adjacent points in the orbit, because the

triangular inequality is not available.

138, is used only the triangular inequality of the norm and the fact that the real valued function [parallel] f(t) [parallel] is continuous on R, the present proof is then identical.

Then, since in the proof for Banach space valued functions, only the triangular inequality of the norm is used, it follows that the present proof is similar to the proof of Theorem 6.

Since in the case of Banach space valued functions, the proof use only the triangular inequality of the norm, we get that the proof of our Theorem 3.

By

triangular inequality, p(z, y) [less than or equal to] p(z, [T.

1) with x = a and x = b, adding the results, using the triangular inequality and then dividing by 2.

Summing over i from 0 to n - 1 and using the generalized triangular inequality, we deduce the desired estimation (3.