# Convergence to a fixed point [duplicate]

Let $$f : [a,b] \rightarrow [a,b]$$ be a continuous function s.t. $$f'(x)$$ is defined on $$(a,b)$$ and $$\left\lvert f'(x)\right\rvert \leqq t$$ where $$0. Prove that for any point $$x_0$$ in $$[a,b]$$ the sequence defined by $$x_n=f(x_{n-1}), n>0$$ converges to one unique fixed point.

Attempt: Frankly, I have struggled to make a real attempt due to the fact that I can't find notes relating to this.

Obviously, I am assuming that there exists $$x$$ in$$[a,b]$$ s.t. $$f(x)=x$$ but how do I relate the sequence to this $$x$$?

I'm strictly not allowed to assume Banach's theorem in this question, nor the Cauchy sequence because they come up on the second part of the course. I rather have to PROVE this.

## marked as duplicate by rtybase, Eevee Trainer, mrtaurho, Vinyl_cape_jawa, José Carlos SantosMar 7 at 13:30

• Another related question. – rtybase Mar 6 at 22:36

The Mean Value Theorem tells you that the sequence $$\{x_n \}$$ is Cauchy because $$|f(y)-f(x)| \leq t|y-x|$$, so for $$m \lt n, |x_m-x_n| \leq |x_1-x_2|\sum_{k=m}^nt^k \leq 2M \frac{t^m}{1-t}$$ (where $$M = \max \{f(x)~|~x \in [a, b]\}$$. The space $$[0, 1]$$ is complete, so since the sequence is Cauchy (choose $$m$$ sufficiently large that $$\frac{2Mt^m}{1-t} \lt \epsilon$$), it must converge.

• MVT does not immediately tell you that $(x_n)$ is Cauchy. You have to provide more details. – Kavi Rama Murthy May 10 at 23:10
• @KaviRamaMurthy Edited to add details. – Robert Shore May 11 at 0:20

To prove this from scratch note that by MVT $$|x_n-x_m| \leq |x_{m+1}-x_m|+|x_{m+2}-x_{m+1}|$$ $$+\cdots+|x_{n}-x_{n-1}|\leq |x_{m+1}-x_m| (1+t+t^{2}+\cdots+t^{n+m-1})$$ for $$n >m$$. Also $$|x_{m+1}-x_m| \leq t^{m-1} |x_2-x_1|$$. Using the convergence of the geometric series $$\sum t^{n}$$ conclude that $$\{x_n\}$$ is Cauchy. If $$x =\lim x_n$$ then the definition of $$x_n$$'s and continuity of $$f$$ tells you that $$f(x)=x$$. Uniqueness follows easily by MVT.

• Thank you! This is very helpful :-) – Gracious Mar 7 at 9:20

This can be proved using the Banach Fixed Point Theorem.

Intuitively, the BFPT tells us that if there is some function $$F$$ such that the distance between any two points $$x$$ and $$y$$ (when scaled by a constant $$q$$) is always larger than the distance between the corresponding images ($$F(x)$$ and $$F(y)$$), then the sequence $$x_n = F(x_{n-1})$$ will converge to a unique fixed point. For a more rigorous treatment of the BFPT statement: https://en.wikipedia.org/wiki/Banach_fixed-point_theorem.

Since in this case you know that $$|f'(x)| \leq t$$ This implies that

$$\Big|\frac{f(x) - f(y)}{x - y}\Big| \leq t$$

for all possible points $$x,y \in [a,b]$$. Think about why this is the case (Hint: Use the mean value theorem).

This implies that

$$| f(x) - f(y)| < t |x-y|$$

which is what the BFPT requires. From here, we can just apply the BFPT to state that there is a fixed point in $$[a,b]$$ for the sequence

$$x_n = f(x_{n-1})$$

• Should probably expressly mention that the space $[0, 1]$ is complete, since that is a necessary condition for the Banach fixed point theorem to hold. For an easy counterexample demonstrating that you need completeness, consider $f(x) = x/2$ on $(0, 1]$. – Robert Shore Mar 6 at 22:50
• @RobertShore Yes, you are right, thanks! – Sean Lee Mar 6 at 22:52