Do you want to help me with my homework? The exercise is as follows:

Consider a Lipschitz function, $h:\mathbb{R}\rightarrow\mathbb{R}$, satisfying for every $x, y$: $$\left| h(x)-h(y) \right| \leq \alpha \left| x-y \right|$$ with $0 \lt \alpha \lt 1$

  1. Show that there exists at most one $x$, with $h(x)=x$.
  2. Prove that $h$ is uniformly continuous.
  3. Take some $x_1\in\mathbb{R}$, and define inductively the sequence $(x_n)$ as $$ x_{n+1}=h(x_n), \quad n= 1, 2, \cdots$$ Show that for every $x_1$ the sequence $(x_n)$ is a Cauchy sequence.
  4. Take some $x_1$. Define $x= \lim x_n$. Show that $x=\lim{x_n}$ satisfies $h(x)=x$
  5. Show (using the first part), that the limits of the sequences $(x_n)$ for all choices of $x_1$ are all the same.

My work until now:

Part 1

Suppose $h(x_1)=x_1, h(x_2)=x_2$, and $x_1 \not = x_2$. Then, following the condition, $$\left| h(x_1)-h(x_2) \right|= \left| x_1-x_2 \right| \le \alpha \left|x_1-x_2\right|.$$ This means that $\frac{\left|x_1-x_2\right|}{\left|x_1-x_2\right|}=1 \le \alpha$. But $\alpha <1$ was given, so $x_1 = x_2$. There exists at most one $x$ with $h(x)=x$

Part 2

Let $\epsilon>0$ be given and choose $\delta=\frac{\epsilon}{\alpha}$. Then, for any $x, y$ with $\left|x-y\right|<\delta = \frac{\epsilon}{\alpha}$, I have $$\left| h(x)-h(y) \right| \le \alpha \left|x-y\right| \lt \alpha\left(\frac{\epsilon}{\alpha}\right)=\epsilon,$$ which shows that $f$ is uniformly continuous.

Part 3

I don't know where to start.

Part 4

  • 1
    $\begingroup$ You sometimes write $f(x)$ and sometimes $h(x)$. You mean the same function in both cases, right? $\endgroup$ – Martin Sleziak Oct 21 '12 at 13:51
  • $\begingroup$ yes, ty, fixed that $\endgroup$ – Applied mathematician Oct 21 '12 at 14:45
  • 1
    $\begingroup$ What is the point of defacing your question in this way more than half a year after posting it? $\endgroup$ – Martin May 29 '13 at 10:36

The only difficult part is Part 3.

Let $p$ be a positive integer. Then $|x_{k+1}-x_k|\leq\alpha|x_k-x_{k-1}|$ and therefore $$\eqalign{(1-\alpha)|x_p-x_0|&\leq(1-\alpha)\sum_{k=1}^p |x_k-x_{k-1}|\leq \sum_{k=1}^p |x_k-x_{k-1}|-\sum_{k=1}^p |x_{k+1}-x_k|\cr &\leq|x_1-x_0|=:\delta\ .\cr}$$ It follows that $$|x_p-x_0|\leq{\delta \over1-\alpha}\qquad(p\geq1)\ ;$$ whence we obtain that $$|x_{n+p}-x_n|\leq {\delta\over1-\alpha} \ \alpha^n\qquad(n\geq0,\ p\geq1)\ .$$ Assume now that an $\epsilon>0$ is given. There is an $n_0$ such that $${\delta\over1-\alpha}\ \alpha^n<\epsilon\qquad(n>n_0)\ .$$ It follows that $|x_m-x_n|<\epsilon$ as soon as $m>n>n_0$.


(3.) Suppose $m>n$, and consider $x_0$ instead of $x_1$. Then $|x_n-x_m| =|f^{(\circ n)}(x_0)-f^{(\circ n)}(x_{m-n}) | < \alpha^n\cdot |x_0 - x_{m-n}|$ may help. Note also that $|x_0-x_2|\le |x_0-x_1|+|x_1-x_2|<(1+\alpha)|x_0-x_1|$, and $|x_0-x_3| \le |x_0-x_1|+|x_1-x_3|<(1+\alpha(1+\alpha))|x_0-x_1|$, so $$|x_0-x_d|<(1+\alpha+\alpha^2+..\alpha^{d-1})|x_0-x_1|< \frac1{1-\alpha}|x_0-x_1|$$

(4.) Use that $h$ is continuous, hence preserves limit.

(5.) Follows from these, using (1.).

  • $\begingroup$ what do you mean with $f^{(\circ n)}$ ?? $\endgroup$ – Applied mathematician Oct 21 '12 at 15:42

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