Question

Let `RR` denoted the set of real numbers. Find all functions `f:RR->RR`,satisfying `abs(f(x) - f(y)) = 2 abs(x-y)` for all `x,y` belongs to `RR`.?

Answer 1

`f(x) = pm 2 x+ C_0`

Explanation:

If `abs(f(x)-f(y))=2abs(x-y)` then `f(x)` is Lipschitz continuous. So the function `f(x)` is differentiable. Then following,

`abs(f(x)-f(y))/(abs(x-y))=2` or
`abs((f(x)-f(y))/(x-y))=2` now

`lim_(x->y)abs((f(x)-f(y))/(x-y))=abs(lim_(x->y)(f(x)-f(y))/(x-y)) = abs(f'(y))=2`

so

`f(x) = pm 2 x+ C_0`