[SOLVED] A mean value theorem problem

**nqramjets** · June 1st 2009, 12:01 PM

This one has been stumping me for quite some time know...
Given $f:[a,b]\rightarrow[a,b]$ continuous on $[a,b]$ and differentiable on $(a,b)$ such that there exists an $\alpha\in(0,1)$ so that $|f'(x)|<\alpha$ for all $x\in(a,b)$ , show that there exists a unique point $c\in[a,b]$ such that $f(c)=c$ .

I can show that if there is a point, then it is unique, but showing the point exists is challenging.
Hint: Pick $x_0\in[a,b]$ and let $x_{k+1}=f(x_k)$ .
I have used repeated applications of the mean value theorem to achieve the following inequality
$|x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|$
Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!

**TheEmptySet** · June 1st 2009, 12:06 PM

Originally Posted by nqramjets

This one has been stumping me for quite some time know...
Given $f:[a,b]\rightarrow[a,b]$ continuous on $[a,b]$ and differentiable on $(a,b)$ such that there exists an $\alpha\in(0,1)$ so that $|f'(x)|<\alpha$ for all $x\in(a,b)$ , show that there exists a unique point $c\in[a,b]$ such that $f(c)=c$ .

I can show that if there is a point, then it is unique, but showing the point exists is challenging.
Hint: Pick $x_0\in[a,b]$ and let $x_{k+1}=f(x_k)$ .
I have used repeated applications of the mean value theorem to achieve the following inequality
$|x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|$
Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!

ahh but it is

$\alpha \in (0,1)$ so as k increases

$\alpha ^k \to 0, \mbox{ an } k \to \infty$

Just pick k large enough such that $\alpha^k =\frac{\epsilon}{x_1-x_0}$

**HallsofIvy** · June 1st 2009, 01:14 PM

Originally Posted by nqramjets

This one has been stumping me for quite some time know...
Given $f:[a,b]\rightarrow[a,b]$ continuous on $[a,b]$ and differentiable on $(a,b)$ such that there exists an $\alpha\in(0,1)$ so that $|f'(x)|<\alpha$ for all $x\in(a,b)$ , show that there exists a unique point $c\in[a,b]$ such that $f(c)=c$ .

I can show that if there is a point, then it is unique, but showing the point exists is challenging.
Hint: Pick $x_0\in[a,b]$ and let $x_{k+1}=f(x_k)$ .
I have used repeated applications of the mean value theorem to achieve the following inequality
$|x_{k+1}-x_k|\leq\alpha^k|x_1-x_0|$
Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!

Am I missing something here? You want to prove f(c)= c so that is an "intermediate value" theorem, not a "mean value" theorem.

You need neither that the function is differentiable nor the bound on the derivative to prove "existence", perhaps to prove "uniqueness".

This is a well known proof:
Suppose f:[a,b]->[a,b] is continuous on [a, b]. If f(a)= a, then we are done. So we can assume f(a)> a. If f(b)= b, we are done. So we can assume f(b)< b. Let h(x)= f(x)- x. Then h(a)= f(a)-a> 0 and h(b)= f(b)- b< 0. But h(x) is also continuous so there must exist c such that h(c)= f(c)- c= 0. Therefore f(c)= c.

**nqramjets** · June 1st 2009, 07:53 PM

It turns out this really is Cauchy, but only because if you write out $|x_m-x_n|$ you get a partial geometric sum. The problem is that $|x_{k+1}-x_k|<\epsilon$ doesn't necessarily make $x_k$ Cauchy. In this though, it does. This is what was making me struggle.

**nqramjets** · June 1st 2009, 07:53 PM

Hey this is slick, thanks. And yes, to show uniqueness you do need MVT.

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