Suppose that $\displaystyle f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous, bounded, strictly increasing function.

Questions:

(a) Show that there is a point $\displaystyle a_{1}\epsilon\mathbb{R}$ such that $\displaystyle f(a_{1})>a_{1}$

(b) For each $\displaystyle n\epsilon\mathbb{N}$, define $\displaystyle a_{n+1}=f(a_{n})$. Explain why $\displaystyle a_{2}>a_{1}$. Then explain why $\displaystyle a_{n+1}>a_{n}$ for all $\displaystyle n\epsilon\mathbb{N}$

(c) Explain why the sequence $\displaystyle {a_{n}}$ is bounded above.

(d) Explain why the sequence $\displaystyle {a_{n}}$ converges to some number, L.

(e) Explain why f(L) = L.

My work so far:

(a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point $\displaystyle x\epsilon\mathbb{R}$ such that x < f(x). I know this reason is not enough, still working on the proper proof.

(b) Now $\displaystyle a_{2}=a_{1+1}=f(a_{1})>a_{1}$ as defined by part (a).

Let $\displaystyle S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}$

$\displaystyle 1 \epsilon S$ as $\displaystyle a_{1+1}=a_{2} > a_{1}$

let $\displaystyle k \epsilon S$, then $\displaystyle a_{k+1} > a_{k}$

Now $\displaystyle a_{k+1+1} = f(a_{k+1}) > a_{k+1} $, thus proved $\displaystyle k+1 \epsilon S$

Therefore S is inductive, and proved $\displaystyle a_{n+1}>a_{n}$

(c) Because f is continuous and bounded.

(d) Because f is monotonic, implies $\displaystyle {a_{n}}$ is monotonic, thus $\displaystyle {a_{n}}$ is bounded. All bounded monotonic sequence converges, thus the sequence converges to L.

(e) $\displaystyle {a_{n}}$converges to L, then $\displaystyle f({a_{n}})$ converges to L because f is continuous.

I know I made some mistakes up there, please check, thank you.

KK