Let be defined on an interval of length at least 2 and suppose that exists there. If and for all show that on the interval.
The only way I can see to do this is by contradiction. Suppose that there is a point at which Start by using the mean value theorem, and the fact that , to show that if then , and therefore .
Now let with . Then
. . . . . . .
A similar calculation shows that if then Subtract that inequality from the previous one to get
Now the interval has length at least 2, so we can choose r and s so that . Notice that , and similarly Therefore It follows from the inequality in the previous paragraph that (because ). But that is impossible if and
That contradiction shows that for all . A similar argument shows that (or you can save work by applying the above argument to the function ).