Originally Posted by

**kathrynmath** 0<c<1 and |f(x)-f(y)<c|c-y|

a) Show f is continuous on all of R

It's much, much stronger than continuous. If you're allowed to use the limit definition of continuity just note that $\displaystyle \displaystyle 0\leqslant \lim_{x\to y}|f(x)-f(y)|\leqslant \lim_{x\to y}c|x-y|=0$

b)Pick some point y1 in R and construct the sequence (y1,f(y1),f(f(y1)),....)

In general if y_(n+1)=f(yn) show that the resulting sequence yn is a Cauchy sequence. hence we may let y=limyn

Note that by induction $\displaystyle d\left(f^{m+1}(y_1),f^{m}(y_1)\right)<c^m d\left(f(y_1),y_1\right)$. Thus, we see that

$\displaystyle \begin{aligned}d\left(f^{n+k}(y_1),f^n(y_1)\right) &\leqslant \sum_{j=0}^{k-1}d\left(f^{n+j+1}(y_1),f^{n+j}(y_1)\right)\\ &< \sum_{j=0}^{k-1}c^{n+j}d\left(f(y_1),y_1\right)\\ &< \frac{c^n}{1-c}d\left(f(y_1),y_1\right)\end{aligned}$

From where it evidently follows, letting $\displaystyle n\to\infty$ that $\displaystyle \left\{f^{n}(y_1)\right\}_{n\in\mathbb{N}}$ is Cauchy, and thus by the completeness of $\displaystyle \mathbb{R}$ it follows that $\displaystyle y=\lim f^{n}(y_1)$ exists.

c)Prove that y is a fixed point of f and that is unique in this regard.

Uniqueness is clear in general since if $\displaystyle x\ne y$ but $\displaystyle f(x)=x,f(y)=y$ then we see our conditions state that $\displaystyle d(x,y)=d\left(f(x),f(y)\right)<cd(x,y)<d(x,y)$

Now, to see that $\displaystyle y$ is a fixed point we merely note that since $\displaystyle \lim f^n(y_1)=y$ that $\displaystyle \lim f^{n+1}(y_1)=y$ and so by the continuity of $\displaystyle f$ we have that $\displaystyle f(y)=f\left( \lim f^n(y_1)\right)=\lim f^{n+1}(y_1)=y$

d) Finally prove that if x is any arbitrary point in R then the sequence (x,f(x),f(f(x)),...) converges to y defined in (b).

Merely note that $\displaystyle y_1$ was arbitrary to begin with.

*Remark:* There is a much more elegant way of proving that a contractive map from a complete metric space to itself has a fixed point.