Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.