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
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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
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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
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I don't know where to start.

Part 4
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