f is differentiable continuously(which means that f(x) and f'(x) are continues and differentiable) on [a,b]

suppose that |f'(x)|<1 for x in [a,b]

prove that there is 0<=k<1 so there is x1,x2 in [a,b]

|f(x1)-f(x2)|<=k|x1-x2|

??

i started from the data that i was given and i know that f(x) is differentiable

so

$\displaystyle

f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}

$

i use

|f'(x)|<1

so

$\displaystyle

|f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1

$

so f(x) is continues and differentiable

so i use mvt on x1 and x2

$\displaystyle

f'(k)=\frac{f(x1)-f(x2)}{x1-x2}

$

and i combine that with

|f'(x)|<1

$\displaystyle

\frac{|f(x1)-f(x2)|}{|x1-x2|}<1

$

so

$\displaystyle

|f(x1)-f(x2)| <|x1-x2|

$

what now?