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The following problem is from Ch. 22 of Spivak's Calculus

  1. Suppose that $f$ is continuous and that the sequence

$$x,f(x),f(f(x)),f(f(f(x))),...\tag{1}$$

converges to $l$. Prove that $l$ is a "fixed point" for $f$, ie $f(l)=l$.

I'd like to go through the terse solution manual solution in more intermediate steps to make sure that my understanding of them is correct.

Here the solution manual solution verbatim

Let us denote

$$f(f(f\text{...}f(x)\text{...}))$$

by $f^k(x)$. Then by Theorem 1,

$$f(l)=f(\lim\limits_{k\to\infty} f^k(x))=\lim\limits_{k\to\infty} f(f^k(x))$$

$$=\lim\limits_{k\to\infty} f^{k+1}(x)=l$$

Let me try to go through this with more intermediate steps.

This solution uses the following theorem

Ch. 22, Theorem 1 Let $f$ be a function defined in an open interval containing $c$, except perhaps at $c$ itself, with

$$\lim\limits_{x\to c} f(x)=l$$

Suppose that $\{a_n\}$ is a sequence such that

$$(1)\text{ each }a_n\text{ is in the domain of }f,$$

$$(2)\text{ each }a_n\neq c$$

$$(3) \lim\limits_{n\to\infty} a_n=c$$

Then the sequence $\{f(a_n)\}$ satisfies

$$\lim\limits_{n\to\infty} f(a_n)=l$$

Conversely, if this is true for every sequence $\{a_n\}$ satisfying the above conditions then $\lim\limits_{x\to c} f(x)=l$.

By assumption, $f$ is continuous everywhere. Thus, it is defined everywhere (in particular in an open interval containing $l$), and

$$\lim\limits_{x\to l} f(x)=f(l)$$

This is the antecedent in Theorem 1.

Our sequence is

$$a_n=f^{(n-1)}(x)$$

Does this sequence fulfill the three criteria in Theorem 1?

  • By assumption, $\lim\limits_{n\to\infty} f^n(x)=l$ (this is criterion $(3)$)
  • Each $a_n$ is in the domain of $f$, which is all numbers. (criterion $(1)$)
  • As for criterion $(2)$, $a_n$ may be equal to $l$. But since we know that $f$ is defined at $l$, this is okay.

Thus, by Theorem 1, we can infer that the sequence $\{f(a_n)\}$ satisfies

$$\lim\limits_{n\to\infty} f(a_n)=\lim\limits_{n\to\infty} f(f^{(n-1)}(x))=\lim\limits_{n\to\infty} f^{(n)}(x)=l$$

Thus, if we evaluate $f$ at both sides of the equation

$$\lim\limits_{n\to\infty} f^{(n)}(x)=l$$

we get

$$f(l)=f(\lim\limits_{n\to\infty} f^{(n)}(x))\tag{2}$$

$$=\lim\limits_{n\to\infty} f(f^{(n)}(x))\tag{3}$$

$$=\lim\limits_{n\to\infty} f^{(n+1)}(x)=l$$

Thus

$$f(l)=l$$

Questions

  • Is this understanding of the intermediate steps of the solution manual solution correct?
  • What is it that allows us to go from $(2)$ to $(3)$?
Mittens
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xoux
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    $(2) \implies (3)$ because $f$ is continuous. – Martin R Oct 21 '22 at 20:21
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    That is what I thought. I had in mind that $\lim\limits_{x\to a} f(g(x))=f(\lim\limits_{x\to a} g(x)$ if $f$ is continuous at $\lim\limits_{x\to a} g(x)$. So in this case, we have $g(n)=f^{(n)}(x)$, and because $f$ is continuous everywhere it is continuous at $\lim\limits_{n\to\infty} g(n)$ and thus $\lim\limits_{n\to\infty} f(g(n))=f(\lim\limits_{n\to\infty} g(n))$ – xoux Oct 21 '22 at 20:29
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    Let $a_n\to a.$ Then by continuity $f(a_n)\to f(a).$ Define $a_n=f^{(n)}(x).$ Assume $a_n\to a.$ Then $a\leftarrow a_{n+1}=f^{(n+1)}(x)=f(a_n)\to f(a).$Therefore $f(a)=a.$ – Ryszard Szwarc Oct 22 '22 at 09:28
  • "As for criterion (2), $a_n$ may be equal to $l$. But since we know that $f$ is defined at $l$, this is okay." - while this is true, it means you are no longer applying Theorem 1, as that theorem requires $a_n \ne l$. Instead you are applying a distinct, though related, result. – Paul Sinclair Oct 23 '22 at 01:50

1 Answers1

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Your understanding is correct but I think we just need to take a step back by first extracting ourselves from this problem and convincing ourselves of a simple fact to demonstrate: either $(a_n)_{n\geq0 }$ a sequence of real numbers converging to $l$. Then the sequence $(b_n)_{n\geq0}$ defined by $b_n:=a_{n+1}$ for all $n\geq 0$ is such that $(b_n)_{n\geq0}$ also converges to $l$. So we come back to our problem:

As you put it, here, $a_n:=f^{(n)}(x)$ for a fixed $x$. $b_n=a_{n+1}=f(a_n)$. By hypothesis, $a_n\to l$ when $n\to +\infty$. And, according to our preliminary remark, the same for $(b_n)_n\color{green}{(1)}$. What allows us to go from (2) to (3) is that $f$ is continuous in $l$: $\lim_{n \to +\infty} f(a_n)=f(l)$ , that's to say $\lim_{n \to +\infty} b_n=f(l)\color{green}{(2)}$. From $\color{green}{(1)}$ and $\color{green}{(2)}$, we conclude.

Stéphane Jaouen
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