Functions involving real values If $f$ is a real valued differentiable function satisfying 

$|f(x)-f(y)|&lt;(x-y)^2$ for all real $x$ and $y$ 

and $f(0)=0$ 

then $f(1)$ equals:
Hint: First you don't need the assumption differentiability. You can prove that $f' =0$ (use definition of differentiability). This shows $f$ is constant, and then $f(1) = f(0)=0$
We can simpify it as f(x)-f(y)/x-y= x-y
If we take that x approaches to y it implies that derivative of f is zero 
Since f'(x)= f(x)-f(y)/x-y
Now, if f'(x) is zero then we can say f is a constant function since derivative of a constant function is zero. 
It implies f(0)=f(1)=0