Can Nearby Contractions Imply Nearby Fixed Points?

[mattmns]

Here is the question from our book:
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Let $(X,d)$ be a complete metric space, and let $f:X\to X$ and $g:X\to X$ be two strict contractions on $X$ with contraction coefficients $c$ and $c'$ respectively. From the Contraction Mapping Theorem we know that $f$ has some fixed point $x_0$, and $g$ has some fixed point $y_0$. Suppose we know that there is an $\epsilon > 0$ such that $d(f(x),g(x)) \leq \epsilon$ for all $x \in X$. Show that $d(x_0,y_0) \leq \epsilon/(1-\max(c,c'))$. Thus nearby contractions have nearby fixed points.
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Below is my proof. I believe everything I have is correct. My question is why do we not conclude that $d(x_0,y_0) < \epsilon/(1-\min(c,c'))$ as I believe this is a stronger result that follows (with very slight modification) from my proof below.

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Proof.

Let $m = \max(c,c')$.

So we have $d(f(x),f(y)) < md(x,y)$ and $d(g(x),g(y)) < md(x,y)$ for all $x,y \in X$.

Since $x_0$ is a fixed point of $f$, and $y_0$ is a fixed point of $g$, we know that $f(x_0) = x_0$, and $g(y_0) = y_0$. Hence we get the following

$$\begin{align*}<br /> d(x_0,y_0) & = d(f(x_0),g(y_0)) \\<br /> & \leq d(f(x_0),f(y_0)) + d(f(y_0),g(y_0)) \\<br /> & \leq md(x_0,y_0) + \epsilon<br /> \end{align*}$$

So $d(x_0,y_0) \leq md(x_0,y_0) + \epsilon$.

Hence, $d(x_0,y_0)(1 - m) \leq \epsilon$.

Therefore $$d(x_0,y_0) \leq \dfrac{\epsilon}{1 - \max(c,c')}$$.

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I can't see anything wrong, but there is probably some subtlety that I am missing. Any ideas? Thanks!Definitions if needed.

Strict contraction: Let $f:X\to X$ be a map, then $f$ is a strict contraction if $d(f(x),f(y) < d(x,y)$.

Fixed point: x is a fixed point of a function f if f(x) = x.

Contraction mapping theorem. Let $(X,d)$ be a metric space, and let $f:X\to X$ be a strict contraction. Then f can have at most one fixed point. Moreover, if we assume X is complete and non-empty then f has exactly one fixed point.

 

[Hurkyl]

I don't see a problem.

 

[AKG]

The proof looks fine, and yes, with a slight modification of your proof you can derive the stronger result that replaces 'max' with 'min'.

 

[mattmns]

Thanks! (10 char)[/color]