# Cauchy Sequence Homework Helper

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

real-analysissequences-and-seriescontinuityfixed-point-theorems

Меган сказала, что, если тереть глаза, будет только хуже. Он даже представить себе не может, насколько хуже. Не в силах сдержать нетерпение, Беккер попытался позвонить снова, но по-прежнему безрезультатно. Больше ждать он не мог: глаза горели огнем, нужно было промыть их водой. Стратмор подождет минуту-другую.