Q: f from the the reals to the reals satisfies |f(x)-f(y)|<=|x-y|^2 (*), for all x,y in the reals. Show f is constant.
A: My answer is this:
Suppose, without loss of generality, that y>x. Suppose also that f is strictly increasing, i.e. f(y)>f(x).
Then LHS of (*) is f(y)-f(x), and the RHS is (y-x)^2, with y-x>0,
Thus, (f(y)-f(x))/(y-x) <= y - x. Now, the limit of the LHS of inequality, as y tends to x, is f'(x), which is strictly postive, whereas the lim of the RHS as y tends to x is 0, so we have a contradiction, as the right hand side is always greater than the left hand side.
Similarly, we obtain a contradiction if we assume f is strictly decreasing, and thus I conclude f is constant.
Is this argument correct?