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Let $f:[0,1] \to [0,1]$ be continuous. Define the sequence $(x_n)$ by successive approximation as follows: $x_0 \in [0,1]$ and $x_{n+1} = f(x_n)$ for all non-negative integers n.

I wanted to prove that if $(x_{n+1}-x_n) \to 0$ as $n\to \infty$ then $(x_n)$ converges to a fixed point of $f$ , but I'm unsure about how rigorous my attempt is:

Consider the continuous function $g(x) = f(x) -x$ . Assume that $f$ has no fixed point, $\nexists \ c \in [0,1]$ such that $g(c)=0 \implies \epsilon' > 0$ such that $|g(x)| > \epsilon' \ \ \ \forall x\in[0,1]$

But we know that $\forall \epsilon > 0 \ \ \ \exists N , n\ge N \implies |x_{n+1}-x_n| = |f(x_n)-x_n | = |g(x_n)| < \epsilon$, which contradicts our assumption and implies that $(g(x_n)) \to c \ $ such that $g(c) = 0$ and therefore $(x_n) \to c \ $ as $n \to \infty$

Is my last logical jump rigorous, and if not, how would you go about fixing my argument?

Bernard
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CataSequence
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    With the definition $x_{n+1} = f(x_n)$, if $f(x) = 1-x$, and $x_0 = \frac{1}{4}$, then $x_1 =f(1/4) = 3/4$, $x_2 = f(3/4) = 1/4$, it is easy to see that $x_{2n} = \frac{1}{4}$ and $x_{2n+1} = \frac{3}{4}$ for every $n$. So the sequence $x_n$ is oscillating and not converging. You won't prove your statement that way, I'm afraid. – Thomas Nov 21 '21 at 19:29
  • @Thomas To be fair, that’s not what he wants to prove. He wants to prove that if the difference $x_{n+1}-x_n$ goes to zero then the sequence converges to a fixed point. And I think for that his reasoning is fine. Otherwise, a nice counterexample! – Maik Pickl Nov 21 '21 at 19:36
  • This is enough to prove $(x_n)_n$ is convergent because the limit will be directly a fixed point of $f$. By the way, a similar question was asked before, see https://math.stackexchange.com/questions/1721731/if-x-n1-fx-n-and-x-n1-x-n-to-0-then-x-n-converges – phdstudent Nov 21 '21 at 19:36
  • Okay, thanks everyone! I'll look into it @Elif – CataSequence Nov 21 '21 at 19:39
  • @MaikPickl it's not completely clear to me what should be proved, but I would assume it's the statement that a continuous function from $[0,1]$ to $[0,1]$ has a fixed point, and my understanding is that an approach using the iteration $x_n = f(x_{n-1})$ is intended. That won't work. – Thomas Nov 21 '21 at 19:40
  • @Thomas I can only assume that the second paragraph is what he wants to prove since it begins with „I wanted to prove…“. But you are right, the statement you mentioned is not true, as the counterexample you provided shows. – Maik Pickl Nov 21 '21 at 19:43
  • ...and if my suspicion about what should be proved is correct, you may look at this question: https://math.stackexchange.com/questions/2649268/prove-that-f0-1-to-0-1-has-a-fixed-point?rq=1 – Thomas Nov 21 '21 at 19:44
  • Hey @Thomas , I wanted to prove what Maik is saying but I appreciate the input greatly – CataSequence Nov 21 '21 at 20:07

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