Using the Mean Value Theorem, prove the following:

Let $f : [a,b] \rightarrow [a,b]$ be differentiable. Then, $f$ is a contraction mapping if and only if there exists $r \in (0,1)$ such that $\lvert f'(x)\rvert \leq r$ for all $x \in [a,b]$.

I've seen other threads asking about this question, but none of them answer my question: How is the Mean Value Theorem enough to prove this statement?

Here's a proof in one direction:
Suppose $f$ is a contraction mapping.
Thus there exists an $r \in (0,1)$ such that $\lvert f(x)-f(y) \rvert \leq r \lvert x-y \rvert$ for all $x,y \in \mathbb{R}$.
By the Mean Value Theorem, there exists a point $c\in [a,b]$ such that$f(x)-f(y)=f'(c)(x-y)$ so $\lvert f(x)-f(y) \rvert = \lvert f'(c) \rvert \lvert x-y \rvert$
and thus $\lvert f'(c) \rvert \lvert x-y \rvert \leq r \lvert x-y \rvert \implies \lvert f'(c) \rvert \leq r$.

But what I am supposed to obtain is:
$\lvert f'(c)\rvert \leq r$ for all $c \in [a,b]$

How do I prove that this is true for all points, instead of just one point with the Mean Value Theorem? Am I misunderstanding the theorem?

  • 1
    $\begingroup$ The mean value theorem is only needed for the opposite direction. For the direction you are trying to prove, the answer to the thread you linked shows how to prove it directly using the definition of the derivative. $\endgroup$
    – angryavian
    Commented Nov 22, 2022 at 4:32
  • 1
    $\begingroup$ Seems hopeless. Yet it didn't asked you to use MVT for both directions. $\endgroup$
    – xbh
    Commented Nov 22, 2022 at 4:34
  • $\begingroup$ Actually, in your case, $c$ can be found strictly between $x$ and $y$. So, fix $x$ and vary $y$, say, let $y\to x$. $\endgroup$ Commented Nov 22, 2022 at 4:40

1 Answer 1


My suggestion and the suggestion of many other commenters is that the mean value theorem is most cleanly applied in the direction where you assume $|f'(x)| \leq r$ for $r\in (0,1)$. For any two $x,y\in [a,b]$ there is $c\in (a,b)$ with $|f(y) - f(x)| = |f'(c)||x-y|$. Now you can use the bound on $f'$ to your advantage.

If $f$ is a contraction mapping, then you've already shown that there is $r\in (0,1)$ such that $$\left|\frac{f(x) - f(y)}{x-y}\right|\leq r$$ for all $x,y\in [a,b]$ with $x\neq y$. What does this mean for the derivative of $f$ in $[a,b]$?


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