# differentiation prove..

• Feb 28th 2009, 09:22 AM
transgalactic
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|
$
• Feb 28th 2009, 09:38 AM
Abu-Khalil
If you take $f(x)=0.9x$ and $K<0.9$ the equation isn't true :S
• Mar 3rd 2009, 10:30 AM
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
??
• Mar 3rd 2009, 10:55 AM
bkarpuz
Quote:

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 $\mu\in(0,1)$.
• Mar 3rd 2009, 11:05 AM
transgalactic
what is the meaning of "I correct yours."

??
• Mar 3rd 2009, 11:12 AM
bkarpuz
Quote:

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.
• Mar 3rd 2009, 11:19 AM
transgalactic
it wasnt a mistake the function is continuesly differentiable
so i can do this thing with f'(x)
• Mar 3rd 2009, 04:38 PM
mr fantastic
Quote:

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.
• Mar 3rd 2009, 11:47 PM
transgalactic