Proving f is a Constant Function: A Mathematical Investigation

[~Sam~]

Homework Statement

Suppose that |f (x) - f (y)|  (x-y)^2 for all real numbers x and y: Prove
that f is a constant function.

Homework Equations

No relevant equations..

The Attempt at a Solution

I'm really stuck..i'm thinking you're suppose to use mathematical induction.?

 

[Dick]

Can you show that |f(x)-f(y)|<=(x-y)^2 for all x and y implies that f is differentiable, and that it's derivative equals zero everywhere?

 

[lurflurf]

Dick's approach is popular, but I never liked it. I do not think this problem merits the use of the derivative. I like to show inductively that
|f(x)-f(y)|<=(x-y)^2 implies
|f(x)-f(y)|<=(x-y)^2/2^n for any natural number n from which the result is obvious.
hint
|f(x)-f(y)|=|[f(x)-f((x+y)/2]+[f((x+y)/2)-f(y)]|<=|f(x)-f((x+y)/2|+|f((x+y)/2)-f(y)|

 

[Dick]

Dick's approach is popular, but I never liked it. I do not think this problem merits the use of the derivative. I like to show inductively that
|f(x)-f(y)|<=(x-y)^2 implies
|f(x)-f(y)|<=(x-y)^2/2^n for any natural number n from which the result is obvious.
hint
|f(x)-f(y)|=|[f(x)-f((x+y)/2]+[f((x+y)/2)-f(y)]|<=|f(x)-f((x+y)/2|+|f((x+y)/2)-f(y)|

That is a nice alternative approach. There's more than one way to skin a cat.