0

I was working on the following problem:

Problem: Let f be a continuous function defined on all $\mathbb{R}$ s.t $$|f(x)-f(y)| \leq c|x-y|$$ for c $\in (0,1)$ . Pick a point $y \in \mathbb{R}$ and construct the sequence $$(y,f(y),f(f(y)),...)$$ show that the sequence is cauchy.

My attempt: I showed that $$|y_n-y_{n+1}|< \frac{c^n}{c}|y_1-y_2|$$ therefore because $c^n \to 0$ $$|y_n-y_{n+1}| \to 0$$

My question: Why is this not enough, I generalized this to the case where $m \neq n+1$ just to make sure I got it right but, I dont understand why this is not enough for a proof. According to rudin:enter image description here

if all we need is for $m > N$ then why is my argument not good enough, since what I showed above was that there exists some $N$ such that when $n>N$ and $m=n+1 > N$ $$|y_n-y_m|=|y_n-y_{n+1}|< \epsilon$$

Thank you in advance!

Samael Manasseh
  • 693
  • You were told this is not enough? – Divide1918 Dec 11 '20 at 04:59
  • No, but I assumed it was not since every other proof online uses this only as a step to a more generalized argument. All the other proofs out there break $|y_m-y_n|$ into $|y_n-y_{n+1}|$ parts using the triangle inequality and then use geometric series. – Samael Manasseh Dec 11 '20 at 05:02
  • 1
    @SamaelManasseh The problem is you don't get to pick $m$ in the Cauchy definition. Whatever inequality you prove needs to be true for any $n,m$ satisfying $n,m>N$, regardless of how many indices they are apart (you only checked one index apart). – Glare Dec 11 '20 at 05:06
  • @Divide1918 Of course, this is not good enough. OP took $m=n+1$ which is not good enough. – Kavi Rama Murthy Dec 11 '20 at 05:07
  • Thank you all for your answers :) – Samael Manasseh Dec 11 '20 at 05:07
  • @KaviRamaMurthy True, although in OP's case, it's immediately clear that one can split $|y_n-y_m|$ into the sum of $|y_k-y_{k+1}|$ which forms a convergent geometric series. So the missing part is just a minor detail that might be omitted. – Divide1918 Dec 11 '20 at 05:14
  • Compare https://math.stackexchange.com/questions/75840/a-sequence-of-real-numbers-such-that-lim-n-to-inftyx-n-x-n1-0-but-it – Martin R Dec 11 '20 at 05:15

2 Answers2

1

The condition $|y_n - y_{n+1}|\to 0$ is not enough to deduce $y_n$ is Cauchy as can be seen by considering the following counterexample: $y_n = \log n\to\infty$, but $|y_n - y_{n+1}| = \log(1+\frac1n) \to 0$. So you need to use the explicit rate $|y_n - y_{n+1} | \le ac^n$ where $c<1$. This of course gives for $m>n$ $$ |y_n-y_m| \le |y_{n}-y_{n+1} | + \dots + |y_{m-1} -y_m| \le a \sum_{k=n+1}^m c^k \le a\sum_{k=n+1}^\infty c^k \to 0 $$

Calvin Khor
  • 34,903
1

To prove that $(y_n)$ is Cauchy you have to find $N$ such that $|y_n-y_m| <\epsilon$ for all $n$ and all $m$ greater than $N$. You cannot choose $m$ to be $n+1$.

Kavi Rama Murthy
  • 311,013