1. ## Contractive Map

Can anyone help me with the following proof please:

Let A>reals and Φ: A>reals Є C1.
Then the smallest k with
| Φ(x) – Φ(y)| ≤ |x-y| for all x,y in A is given by
k = max |Φ’(ε)| with ε ЄA.

Show that the contraction constant k is given by k = max |Φ’(ε)| with ε ЄA.

Thanks to anyone who is able to help.

2. Hi,
Originally Posted by smiler
Can anyone help me with the following proof please:

Let A>reals and Φ: A>reals Є C1.
Then the smallest k with
| Φ(x) – Φ(y)| ≤ |x-y| for all x,y in A is given by
k = max |Φ’(ε)| with ε ЄA.

Show that the contraction constant k is given by k = max |Φ’(ε)| with ε ЄA.

Thanks to anyone who is able to help.
The inequality $|\phi(x)-\phi(y)|\leq k|x-y|$ can be written as $\left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k$ for all $x,y\in A,\, x\neq y$. Letting $x\to y$ you can derive that $k \geq \max_{\varepsilon\in A}|\phi'(\varepsilon)|$, can't you ? Then, to get $k=\max_{\varepsilon\in A}|\phi'(\varepsilon)|$, it remains to be shown that $k>\max_{\varepsilon\in A}|\phi'(\varepsilon)|$ is impossible (hint : use a proof by contradiction and the mean value theorem).

3. Originally Posted by flyingsquirrel
Hi,

The inequality $|\phi(x)-\phi(y)|\leq k|x-y|$ can be written as $\left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k$ for all $x,y\in A,\, x\neq y$. Letting $x\to y$ you can derive that $k \geq \max_{\varepsilon\in A}|\phi'(\varepsilon)|$, can't you ? Then, to get $k=\max_{\varepsilon\in A}|\phi'(\varepsilon)|$, it remains to be shown that $k>\max_{\varepsilon\in A}|\phi'(\varepsilon)|$ is impossible (hint : use a proof by contradiction and the mean value theorem).
thanks for your assistance, but no i dont know how to derive that because we havent been shown, nor do i know what the mean value theorem is. brief help on this would be great please
thanks, smiler

4. Originally Posted by smiler
thanks for your assistance, but no i dont know how to derive that
Did you try ? That's less difficult than it looks like. Starting from

$\left|\frac{\phi(x)-\phi(y)}{x-y}\right|\leq k$ for all $x,y\in A,\, x\neq y$,
if we let $x$ tend to $y$ we get that for all $y \in A$,
$\lim_{x\to y}\left|\frac{\phi(x)-\phi(y)}{x-y}\right| \leq \lim_{x\to y} k$

What is the LHS (think about the definition of the derivative) ? What is the RHS ?

because we havent been shown
That sounds like "I don't have the solution so I can't do it"...
nor do i know what the mean value theorem is.
I can't think of a proof that doesn't use this theorem.