# Edelstein's Version of the Banach Fixed Point Theorem

The statement of the theorem is

If $$X$$ is a complete, compact metric space and $$f:X\to X$$ is continous and satisfies $$d(f(x),f(y))\lt d(x,y)$$ for $$x\neq y$$ then the recursive sequence $$f^{(n)}(x)$$ is convergent.

Now, in the context of the answer, I have understood that $$f$$ has a unique fixed point but I can't understand what makes the recursive sequence convergent in the first place at all?

My thinking :

Let $$\{a_n\}$$ be the recursive sequence where $$a_1=x$$ and $$a_{n+1}=f(a_n) ,\forall n\in \mathbb{N}$$. Then since the space $$X$$ is compact , there is a convergent subsequence $$\{a_{r_n}\}$$ .

Let $$a_{r_n} \to l$$ as $$n\to \infty$$.

Then $$a_{r_n+1}=f(a_{r_n})\to f(l)$$ as $$n\to \infty$$ by the continuity of $$f$$.

Similarly $$a_{r_n+2}=f(a_{r_n+1})\to f(f(l))$$ as $$n\to \infty$$

Similarly, for $$k\in \mathbb{N}$$

$$a_{r_n+k}\to f^{(k)}(l)$$ as $$n\to \infty$$

As a sidenote, before you ask the question why didn't I ask there in the comments , I want to make it clear that I asked it there but didn't get any reply. So I thought of posting this as a separate question.

Thanks for your time and attention.

• The space is compact and metric, so every infinite sequence has a convergent subsequence. The contraction-like property lets you extend it to the full sequence rather than just a subsequence. Does this help? Commented Aug 6, 2020 at 16:46
• @Clayton May be you are correct but I am not sure how to use your hint , sir. Commented Aug 6, 2020 at 17:40
• It is spelled Caccioppoli, with double p. ;-) Commented Aug 7, 2020 at 0:30
• A proof of this result is given as Theorem 3.5 in kconrad.math.uconn.edu/blurbs/analysis/contraction.pdf.
– KCd
Commented Aug 7, 2020 at 4:53
• @KCd Thank you. Commented Aug 7, 2020 at 6:36

In the aforementioned link in the OP, it is proven that $$f$$ has a unique fixed point, say $$w$$.

To show that for any $$x\in X$$, $$f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$$, we show that any subsequence of $$\{f^{(n)}(x)\}$$ admits a subsequence that converges to $$w$$.

Following the notation of the link, define the function $$Q(x):=d(f(x),x)$$. Since $$f$$ continuous, so is $$Q$$; moreover, unless $$x$$ is a fixed point of $$f$$, we have that $$Q(f(x))=d(f(f(x)),f(x))

If $$Q(f^{(n)}(x))=0$$ for some $$n_0$$, then $$f^{(m)}(x)=f^{n_0}(x)$$ for all $$m\geq m_0$$ and so, $$f^{(n)}(x)\xrightarrow{n\rightarrow\infty}f^{(n_0)}(x)=w$$ since $$f^{(n_0)}(x)=f(f^{(n_0-1)}(x))=f^{(n_0-1)}(x)$$.

Suppose $$x$$ such that $$Q(f^{(n)}(x))>0$$ for all $$n$$. Then, \begin{align} Q(f^{(n)}(x)) and so, $$Q(f^{(n)}(x))$$ converges. On the other hand, as $$X$$ is compact, any subsequence $$\{f^{(n')}(x)\}$$ of $$\{f^{(n)}(x)\}$$ admits a convergent subsequence $$\{f^{(n_k)}(x)\}$$. Say, $$f^{(n_k)}(x)\xrightarrow{k\rightarrow\infty}y\in X$$

For any $$n$$, there is a unique $$k$$ such that $$n_k\leq n; hence $$Q(f^{(n_{k+1})}(x)) and so, by the continuity of $$Q$$ \begin{align} \lim_nQ(f^{(n)}(x))=Q(y).\tag{1}\label{one} \end{align} By $$\eqref{zero}$$, $$Q(f^{(n)}(x))>Q(y),\quad \forall n\in\mathbb{N}$$

We claim that $$y$$ is a fixed point. Otherwise, $$Q(f(y)). However, $$Q(f(y))=\lim_k Q(f(f^{(n_k)}(x))\geq Q(y)$$ which is a contradiction; hence $$y$$ is a fixed point, and by uniqueness $$y=w$$.

We have shown that any subsequence of $$\{f^{(n)}(x)\}$$ admits a subsequence that converges to the unique fixed point $$w$$ of $$f$$. From this, we conclude that in fact $$f^{(n)}(x)\xrightarrow{n\rightarrow\infty}w$$.

Edit: This is to address a comment from the OP:

Lemma: Suppose $$(X,d)$$ is a metric space, $$a\in X$$ and $$\{a_n:n\in\mathbb{N}\}\subset X$$. The sequence $$a_n$$ converges to $$a$$ iff any subsequence $$a_{n'}$$ of $$a_n$$ admits a subsequence $$a_{n''}$$ that converges to $$a$$.

Here is a short proof:

($$\Longrightarrow$$) Obvious.

($$\Longleftarrow$$) Suppose $$a_n$$ does not converge to $$a$$. Then, there is $$\varepsilon>0$$ such that for any $$k\in\mathbb{N}$$, there is $$n_k\in \mathbb{N}$$ such that $$d(a_{n_k},a)\geq \varepsilon$$. Without loss of generality, we may assume that $$n_k. Then $$\{a_{n_k}:k\in\mathbb{N}\}$$ is a subsequence of $$\{a_n:n\in\mathbb{N}\}$$, and no subsequence of $$\{a_{n_k}\}$$ converges to $$a$$ (for $$d(a_{n_k},a)\geq\varepsilon$$ for all $$k$$).

• Thanks sir. I have some questions/confusions. Please clarify these before I accept the answer. $(1)$ Why did you prove the result via subseqence of a subsequence ? I mean you could have started " Since the space $X$ is compact, there is a convergent subsequence of $\{f^{(n)}(x)\}$ ". and then rest of the work. $(2)$ Also in the last paragraph , are you trying to say "If for a sequence $\{a_n\}$ , if it happens that any subsequence of it admits a subsequence that converges at a particular point $a$ (say) , then the original sequence $\{a_n\}$ converges to $a$. ? Commented Aug 7, 2020 at 6:31
• @user710290: (1) originally, I had taken a convergent subsequence of $f^{(n)}(x)$. But I quickly realized that by considering subsequences from the beginning avoids repetition of arguments. Thus, I change the strategy slightly and based it on the fact that (2) if a sequence $A=\{a_n\}$ (in a metric space) is such that for some $a$ any subsequence $a_{n'}$ of $A$ admits a convergent subsequence $a_{n''}$ with limit $a$, then necessarily $a_n$ converges to $a$. If that fact I not clear to you, I can write a short proof. Commented Aug 7, 2020 at 17:53
• Great , really appreciate your help. Everything's clear now . Thanks again. Commented Aug 7, 2020 at 20:18