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 $RR$ .?

Question

3839 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-28T17:43:38

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

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

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

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