Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that its derivative $f'$ is a continuous function. Moreover, assume that for all $x \in \mathbb{R}$,

$$0 \leq |f'(x)| \leq \frac{1}{2}.$$

Define a sequence of real numbers $\{a_n\}_{n \in \mathbb{N}}$ by:

$$a_1 = 1,$$ $$a_{n+1} = f(a_n) \text{ for all } n \in \mathbb{N}.$$

Prove that there exists a positive real number $M$ such that for all $n \in \mathbb{N}$,

$$|a_n| \leq M.$$

Answer 1

The sequence $\{a_n\}$ is bounded. There exists a positive real number $M$ such that for all $n \in \mathbb{N}$, $|a_n| \leq M$. A possible value for $M$ is $1 + 2|f(1) - 1|$.

Answer 2

The condition $0 \leq |f'(x)| \leq \frac{1}{2}$ implies that $f$ is a Lipschitz function with a Lipschitz constant $L = \frac{1}{2}$. By the Mean Value Theorem, for any two distinct real numbers $u$ and $v$, there exists a $c$ between $u$ and $v$ such that $f(u) - f(v) = f'(c)(u - v)$. Taking the absolute value, we get $|f(u) - f(v)| = |f'(c)| |u - v|$. Since $|f'(c)| \leq \frac{1}{2}$, we have $|f(u) - f(v)| \leq \frac{1}{2} |u - v|$ for all $u, v \in \mathbb{R}$.

Applying this to consecutive terms of the sequence: $|a_{n+1} - a_n| = |f(a_n) - f(a_{n-1})| \leq \frac{1}{2} |a_n - a_{n-1}|$ for $n \geq 2$. Let $C = |a_2 - a_1| = |f(a_1) - a_1| = |f(1) - 1|$. Then, $|a_{k+1} - a_k| \leq C \left(\frac{1}{2}\right)^{k-1}$ for $k \geq 1$.

We can express $a_n$ as a telescoping sum: $a_n = a_1 + \sum_{k=1}^{n-1} (a_{k+1} - a_k)$ for $n \geq 2$. Using the triangle inequality: $|a_n| \leq |a_1| + \sum_{k=1}^{n-1} |a_{k+1} - a_k| \leq 1 + \sum_{k=1}^{n-1} C \left(\frac{1}{2}\right)^{k-1}$.

The sum is a finite geometric series: $\sum_{k=1}^{n-1} C \left(\frac{1}{2}\right)^{k-1} = C \sum_{j=0}^{n-2} \left(\frac{1}{2}\right)^j = C \cdot \frac{1 - (1/2)^{n-1}}{1 - 1/2} = 2C \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)$.

Thus, for $n \geq 2$: $|a_n| \leq 1 + 2C \left(1 - \left(\frac{1}{2}\right)^{n-1}\right)$. Since $1 - (1/2)^{n-1} < 1$, we have $|a_n| < 1 + 2C$. For $n=1$, $|a_1|=1$, which is also $\leq 1 + 2C$ as $C \geq 0$. Therefore, for all $n \in \mathbb{N}$, $|a_n| \leq 1 + 2C = 1 + 2|f(1) - 1|$. Let $M = 1 + 2|f(1) - 1|$. Since $f(1)$ is a real number, $M$ is a positive real number. This proves that the sequence $\{a_n\}$ is bounded.