# How to finish this proof

1. Mar 13, 2009

### transgalactic

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

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

so f(x) is continues and differentiable
so i use mvt on x1 and x2
$$f'(k)=\frac{f(x1)-f(x2)}{x1-x2}$$
and i combine that with
|f'(x)|<1

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

what now?

Last edited: Mar 13, 2009
2. Mar 13, 2009

### yyat

Show that $$|f'(x)|\le k$$ for all x in [a,b] and for some 0<k<1. Hint: A continuous function defined on a compact interval has a maximum.

3. Mar 13, 2009

### transgalactic

they sign L as the minimum of f'(x)
and they sign l as the maximum of f'(x)

then they say that k=max{|f(L)|,|f(l)|}

but its not true
i am given that 0=<k<1
and i dont know if i get the same values

why
k=max{|f(L)|,|f(l)|} equals 0=<k<1
??

Last edited: Mar 13, 2009
4. Mar 13, 2009

### yyat

I was refering to the http://en.wikipedia.org/wiki/Extreme_value_theorem" [Broken].

What does it tell you about the function |f'(x)| on the interval [a,b]?
You will find that is has a maximum M. Now use |f'(x)|<1 to show that M<1.

Last edited by a moderator: May 4, 2017
5. Mar 13, 2009

### transgalactic

why this maximum is K ??
and not 1 ??

6. Mar 13, 2009

### yyat

Use the following: If M is the maximum of |f'(x)| then there exists a point a with |f'(a)|=M (by definition), but |f'(x)|<1 for all x (including a), so what does that tell you about M?

7. Mar 13, 2009

### transgalactic

that M<1

8. Mar 13, 2009

### transgalactic

i have been told that l,L are the maximum and minimum points on [a,b] interval so
f'(l)<=f'(x)<=f'(L)

what is the meaning of
k=max{|f'(L)|,|f'(l)|}
why not just say that
k=f'(L)
we see that f'(L) is the maximum in the innequality
??

9. Mar 13, 2009

### transgalactic

ok here is the complete proof as i see it

if we take K as maximum of the sequence
we get
|f'(x1)-f'(x2)|=|f'(c)||x1-x2|<=K|x1-x2|

but how to show that k
is 0<=k<1
??

10. Mar 13, 2009

### yyat

Look at posts #6 and #7.

11. Mar 13, 2009

### transgalactic

ok i agree that k<1

but its -1<k<1 because of |f'(x)|<1

not
0<=k<1

12. Mar 13, 2009

### yyat

If k=max |f'(x)| then k>=0 because $$|f'(x)|\ge 0$$ for all x.
Modify your original proof where you used |f'(x)|<1 and use $$|f'(x)|\le k$$ instead to get a better estimate.

13. Mar 13, 2009

thanks