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Let $f : [a,b] rightarrow [a,b]$ be a continuous function s.t. $f'(x)$ is defined on $(a,b)$ and $leftlvert f'(x)rightrvert leqq t$ where $0<t<1$. 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.

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.

The Mean Value Theorem tells you that the sequence $x_n $ is Cauchy because $|f(y)-f(x)| leq t|y-x|$. The space $[0, 1]$ is complete, so since the sequence is Cauchy, it must converge.

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

$$Big|fracf(x) - f(y)x - yBig| leq t$$

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

$$ x_n = f(x_n-1)$$