Suppose $f\in C([a, b]→[a, b]),\:|f(x)−f(y)|≤\frac{1}{2}|x−y|$ for all $x,y∈[a, b]$. Show that there exists $ξ∈[a,b]$ such that $f(ξ)=ξ$

Asked Jun 21 '21 at 07:37

Active Jun 21 '21 at 10:06

Viewed 72 times

This is actually a Lipschitz function, so thought I should show that it's a Cauchy sequence. I assumed that $f(x_n)=x_{n+1}$. Putting $x=x_{n-1}$ and $y=x_{n-2}$, we get

$$|f(x_{n-1})-f(x_{n-2})| \le \frac 12 |x_{n-1} - x_{n-2}|$$

This gives us,

$$|x_n - x_{n-1}| \le \frac 12 |x_{n-1} - x_{n-2}|$$

Which ultimately gives that

$$|x_n - x_{n-1}| \le \frac{1}{2^{n-2}} |x_2-x_1|$$

Now, I don't know how to proceed from this.....

edited Jun 21 '21 at 10:06

Gary

31,845

asked Jun 21 '21 at 07:37

Demonic99

This is fine! Cauchy's criterion follows easily from what you already have. – Michał Miśkiewicz Jun 21 '21 at 07:40
2

$|x_n-x_m| \leq |x_n-x_{n-1}|+|x_{n-1}-x_{n-2}|+...+|x_{m+1}-x_m|$ for $m <n$. – Kavi Rama Murthy Jun 21 '21 at 07:40
Why don't just use Intermediate Value Property? $f(a)-a \ge 0 \ge f(b)-b$... – Paresseux Nguyen Jun 21 '21 at 07:41
1

@ParesseuxNguyen I would assume that this is an exercise designed to teach the Banach fixed point theorem on a very simple example. – Michał Miśkiewicz Jun 21 '21 at 07:44
@MichałMiśkiewicz Oh, I see – Paresseux Nguyen Jun 21 '21 at 07:45

Suppose $f\in C([a, b]→[a, b]),\:|f(x)−f(y)|≤\frac{1}{2}|x−y|$ for all $x,y∈[a, b]$. Show that there exists $ξ∈[a,b]$ such that $f(ξ)=ξ$

0 Answers0