# Some homework questions about a Lipschitz function (cauchy sequence)

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

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You sometimes write $f(x)$ and sometimes $h(x)$. You mean the same function in both cases, right? – Martin Sleziak Oct 21 '12 at 13:51
yes, ty, fixed that – Joyeuse Saint Valentin Oct 21 '12 at 14:45

(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.).

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 what do you mean with $f^{(\circ n)}$ ?? – Joyeuse Saint Valentin Oct 21 '12 at 15:42

## 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.

## Part3

In this part, when I talk about f(x), I mean of course h(x).

I need to find an N, so that for a given $\epsilon>0$, and for all $n>m \ge N$: $|x_n-x_m|<\epsilon$. This n will be depend on $\epsilon$.

$|x_{n+1} - x_n|=|f(x_n) - f(x_{n-1})| \le \alpha |x_n - x_{n-1}|$

With indunction one can show that $|x_{n+1} - x_n| ≤ \alpha^n|x_2 - x_1|$

Base Case: $|x_2-x_1| = |f(x_1)-f(x_2)| \le \alpha|x_1-x_2$, for which the hypothesis is correct Induction Step: Assume that $|x_{n+1}-x_n| \le \alpha^{n}|x_2-x_1|$. Then $|x_{n+2} - x_{n+1}| = |f(x_n+1)-f(x_n)| \le \alpha |x_{n+1}-x_n| \le \alpha (\alpha^n|x_2-x_1|)= \alpha^{n+1}|x_2-x_1|$

So by induction, $|x_{n+1} - x_n| ≤ \alpha^n|x_2 - x_1|$

Now, by the triangle inequality:

$|x_n - x_m| ≤ |x_n - x_{n-1}| + |x_{n-1} - x_{n-2}| + ... + |x_{m+1} - x_m|$

$≤ \alpha^{n-1}|x_2 - x_1| + \alpha^{n-2}|x_2 - x_1| + ... + \alpha^m|x_2 - x_1|$

$= (\alpha^{n-m-1} + \alpha^{n-m-2} + ... +1)(\alpha^m|x_2 - x_1|)$

now $0 < \alpha < 1$, so $\sum_{1}^{\infty} {\alpha^k} = \frac{1}{1-\alpha}$

$(1 + \alpha + ....+ \alpha^{n-m-1})$ is just a partial sum of the infinite sum, so it is less than the convergent limit, so

$(\alpha^{n-m-1} + \alpha^{n-m-2} + ... +1)(\alpha^m|x_2 - x_1|)< (\alpha^m|x_2 - x_1|)/(1 - \alpha)$

if we choose N so large that $\alpha^N < ε(1 - \alpha)/|x_2 - x_1|$,

then $|x_n - x_m| ≤ (\alpha^m|x_2 - x_1|)/(1 - \alpha) ≤ (\alpha^N)(|x_2 - x_1|/(1 - \alpha)) < (ε(1 - \alpha)/|x_2 - x_1|)(|x_2 - x_1|/(1 - \alpha)) = ε$

Hence, $(x_n)$ is Cauchy.

## Part 4

$x= \text{lim}(x_n)$, then $|f(x)-x| \le |f(x)-f(x_n)| + |f(x_n)-x|$

$\le \alpha|x-x_n|+|x_{n+1}-x| <|x-x_n|+|x-x_{n+1}|$

For sufficiently large n, we can ensure $|x -x_n|, |x-x_{n+1}| < \frac{\epsilon}{2}$

So $|f(x)-x|<\epsilon$ for any $\epsilon>0$, so $f(x)-x=0$, which implies $f(x)=x$

## Part 5

Suppose $f(z)=z$ is another fixed point of f. Then $|x-z|=|f(x)-f(z)|\le \alpha|x-z|$. Since $0<\alpha \lt 1$, $0\le (1-\alpha)|x-z| \le 0$, which implies $|x-z|=0$, so $x=z$

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