Let $\phi: \mathbb{R} \to \mathbb{R}$ be a continuous function such that:
$$-x+\beta < \phi(x)<x, \quad \text{for} \quad x>\frac{\beta}{2}$$
$$x < \phi(x) < -x+\beta, \quad \text{for} \quad x<\frac{\beta}{2}.$$
Show that the sequence $x_{n+1} = \phi(x_n)$ converges to $\frac{\beta}{2}$, for all $x_0$.
I was able to prove that
$$ \bigg| \phi(x)-\frac{\beta}{2}\bigg| < \bigg| x-\frac{\beta}{2}\bigg| \quad \text(1)$$
for all $x \in \mathbb{R} - \{\frac{\beta}{2}\}$, using the given inequalities.
Also, I could prove that $\phi\big(\frac{\beta}{2}\big) = \frac{\beta}{2}$ is the only fixed point of $\phi$ using that $\phi$ is continuous and the given inequalities. From (1), we can conclude that:
$$\bigg| x_{n+1}-\frac{\beta}{2}\bigg| < \bigg| x_{n}-\frac{\beta}{2}\bigg|.$$
Therefore, the sequence $e_n = |x_n-\frac{\beta}{2}| $ is monotone and bounded so it converges.
However, i'm having problems with showing that $e_n \to 0$. I tried by contradiction but it is not clear for me how to do it.
I appreciate any ideas or hints. Thanks in advance!
The argument by contradiction should be fairly straightforward. Assume that $\lim_{n\to\infty} e_n = e \neq 0$. Continuity of $\phi$ then implies that $\phi(\frac{\beta}{2}+e) = \frac{\beta}{2}+e$ which cannot be due to the definition of $\phi$.