1. differentiation prove..

there is a function which is differentiable continuously on [a,b] .
suppose
|f'(x)|<1
for all $x\epsilon [a,b]$
prove that there exists
0<=k<1
$
x_1\epsilon [a,b]
$

$
x_2\epsilon [a,b]
$

that this equation is true
$
|f(x_1)-f(x_2)|<=K|x_1-x_2|
$

2. If you take $f(x)=0.9x$ and $K<0.9$ the equation isn't true :S

3. i got a solution to this..but i cant understand it.
they make mean value theorem on f'(x) [x1,x2] so x2>=x1
$
\frac{f'(x_2)-f'(x_1)}{x_2-x_1}=f'(c)
$

$
|f'(x_2)-f'(x_1)|=|f'(c)||x_2-x_1|
$

then they say that by the second law of weirsstas there is a minimum and maximum
so there is l,L on [a,b] so
f'(L)>=f'(x)>=f'(l) for every x in [a,b]

i cant understand what is the second law of weirshtass
that gives two extreme points??

there is another things but i got stuck on this
??

4. Originally Posted by transgalactic
i got a solution to this..but i cant understand it.
they make mean value theorem on f'(x) [x1,x2] so x2>=x1
$
\frac{f'(x_2)-f'(x_1)}{x_2-x_1}=f'(c)
$

$
|f'(x_2)-f'(x_1)|=|f'(c)||x_2-x_1|
$

then they say that by the second law of weirsstas there is a minimum and maximum
so there is l,L on [a,b] so
f'(L)>=f'(x)>=f'(l) for every x in [a,b]

i cant understand what is the second law of weirshtass
that gives two extreme points??

there is another things but i got stuck on this
??
I correct yours.
By the means of the mean value theorem, we have
$f(s)-f(t)=f^{\prime}(\xi)(s-t)$ for all $s\leq t$and for some $\xi\in[s,t]\subset[a,b]$.
Taking absolute value of both sides, we have
$|f(s)-f(t)|=|f^{\prime}(\xi)||s-t|\leq\mu|s-t|$,
where
$\mu:=\max\{|f^{\prime}(\zeta)|:\zeta\in[a,b]\}$.
Note that $\mu\in(0,1)$.

5. what is the meaning of "I correct yours."

??

6. Originally Posted by transgalactic
what is the meaning of "I correct yours."

??
I corrected your answer, because you used the terms $f^{\prime}$ in the numerator, may be it was just a typo.

7. it wasnt a mistake the function is continuesly differentiable
so i can do this thing with f'(x)

8. Originally Posted by transgalactic
there is a function which is differentiable continuously on [a,b] .
suppose
|f'(x)|<1
for all $x\epsilon [a,b]$
prove that there exists
0<=k<1
$
x_1\epsilon [a,b]
$

$
x_2\epsilon [a,b]
$

that this equation is true
$
|f(x_1)-f(x_2)|<=K|x_1-x_2|
$
This question may have been posted and answered at other websites that provide maths help.

9. it have been answered there
i am trying to solve it.

10. Originally Posted by transgalactic