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Assume $f$ has a finite derivative and $|f'(x)| \leq y < 1$ for all $x \in (a,b)$
$f$ is continuous and $a \leq f(x) \leq b$ for all $x \in [a,b]$. Prove $f$ has a unique fixed point in $[a,b]$.

So far I have for every c in (a,b) |f'(c)| ≤ y => lim x->c |f(x) - f(c)|/|x-c| ≤ y => lim x->c |f(x) - f(c)| ≤ y lim x->c |x-c|
Would that be the definition of a contractive map in R?
Therefore by Banach Fixed Point Theorem, f has a unique fixed point.
Can I prove Banach's theorem using the mean value theorem?

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Doesn't the Fixed point theorem require the metric space to be complete? $(0,1)$ is not complete. –  anonymous Nov 24 '10 at 3:43
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But doesn't f live in [a,b] and [a,b] is complete? –  JimJones Nov 24 '10 at 3:50
    
Sorry Jim. I was thinking of $f'$ –  anonymous Nov 24 '10 at 3:54
    
Sorry to be pedantic, but it is $f(x)$ that lives in $[a,b]$. $\,f\,$ will live in $C[a,b]$. –  Henry Shearman Dec 30 '11 at 13:09

2 Answers 2

up vote 3 down vote accepted

$\lim_{x\to c} |f(x) - f(c)|\leq y\ \lim_{x\to c} |x-c|$

Would that be the definition of a contractive map in R?

No, this is just the statement that $f$ is continuous at $c$, because the right-hand side is $0$. The fixed point theorem will apply, but to show that $f$ is contractive you will want to use the mean value theorem. Suppose that $a\leq z\lt x\leq b$. By the mean value theorem, there is a $c$ in $(z,x)$ such that $f'(c)=\frac{f(x)-f(z)}{x-z}$. Apply absolute values, rearrange, and use the hypothesis on the derivative to conclude that $f$ is contractive.

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Yuval already gave a good alternative approach, but I added this to address the approach in the question. –  Jonas Meyer Nov 24 '10 at 4:39

To show that there is some fixed point, consider $g(x) = f(x)-x$. Then $g(a)\geq 0$ and $g(b) \leq 0$. To show that it is unique, use the mean-value theorem for the two purported fixed points.

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I understand how to show the uniqueness, but I'm not sure how you use this to show that the fixed point exists –  JimJones Nov 24 '10 at 4:21
    
@JimJones: If $g(a)\geq 0$ and $g(b)\leq0$, what must happen somewhere between $a$ and $b$? –  Jonas Meyer Nov 24 '10 at 4:38
    
How do we know g(a)≥0 and g(b)≤0? –  JimJones Nov 24 '10 at 4:54
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@JimJones: $g(a)=f(a)-a$ and $g(b)=f(b)-b$; use the inequality you have for the values of $f$. –  Jonas Meyer Nov 24 '10 at 4:57

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