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

I managed the "if part", but I really doubt the other way. We can have a curve with only 1 point of inflection, with derivative $1$, and the derivative is less then $1$ every other point on the curve. Won't this violate the "only if" ? Thanks.

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JSCB
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  • Related: http://math.stackexchange.com/q/60917/, http://math.stackexchange.com/q/92044/. To be explicit, you could take your example to be $\sin$ on $[-1,1]$. – Jonas Meyer Aug 13 '14 at 05:32

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Jason, your inclination is tempting, but you won't have a contraction mapping. Just use the definition of the derivative, starting with the definition of a contraction map, rather than the Mean Value Theorem.

Ted Shifrin
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  • Why doesn't my counter-example works? The contraction only requires abs(slope) between 2 distinct points to be less than 1, but haven't restrict the tangent case... – JSCB Aug 13 '14 at 05:01
  • @ᴊᴀsᴏɴ: "Contraction" means there exists $r\in(0,1)$ such that $|f(x)-f(y)|\leq r|x-y|$ for all $x$ and $y$, right? – Jonas Meyer Aug 13 '14 at 05:07
  • Yes, so my example is contraction? – JSCB Aug 13 '14 at 05:12
  • @ᴊᴀsᴏɴ: No. Your example satisfies $|f(x)-f(y)|<|x-y|$ for all $x\neq y$, but is not a contraction. Using the def of contraction and the definition of derivative (take a limit), you will get a bound on the derivative of a contraction. – Jonas Meyer Aug 13 '14 at 05:25
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You have to show that if $ f$ is a contraction with modulus $ r \in [0,1)$, then $ \lvert f^{\prime}(x) \rvert<r$ for all $x \in [a,b]$
By definition of derivative, we have:

$$ \lvert f^{\prime}(x) \rvert = \lim_{h \to 0}\left\lvert \dfrac{f(x+h)-f(x)}{h} \right\rvert.$$

But $ \lvert f(x+h)-f(x) \rvert \leq r \lvert h \rvert$ since $f$ is a contraction. Hence we've got

$$ \lvert f^{\prime}(x) \rvert = \lim_{h \to 0}\dfrac{\lvert f(x+h)-f(x)\rvert}{\lvert h \rvert} \leq r$$

for all $ x$

José Carlos Santos
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Alessandro
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