Let $f : \mathbb{R} \to \mathbb{R}$ be a $C^2$ increasing function. If $a \in \mathbb{R}$ is a fixed point of $f$ (so that $f(a) = a$) and $f'(a)=1, f''(a)>0$, show that $\exists \delta > 0$ such that $(a,a+\delta)\subset W^c$, where $W = \left\{x : \lim_{n\to\infty}f^n(x) = a\right\}.$
Note $f^n(x) := \underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(x)$
I tried to show that if $a<x^*<a+\delta$ then $\lim_{n\to\infty} f^n(x^*)\neq a$ but I don't really know how to continue.
Any ideas?
Since $f''$ is continuous, there exists $\delta>0$ such that $f''(x)>f''(a)/2>0$ whenever $|x-a|<\delta$. This implies $f'$ is increasing on the interval $(a-\delta,a+\delta)$; this is a standard exercise using the mean value theorem. Now fix $x\in(a,a+\delta)$. Applying the mean value theorem again, there exists $x^*\in(a,x)$ such that $$\frac{f(x)-f(a)}{x-a}=f'(x^*)>f'(a) = 1.$$ Since $f(a)=a$, one can rearrange the above inequality to deduce $f(x)>x$ for all $x\in(a,a+\delta)$.
To finish off the proof, assume for a contradiction that there exists $x\in(a,a+\delta)$ for which $f^n(x)\to a$ as $n\to\infty$. Then there exists $N$ such that $f^n(x)\in(a,a+\delta)$ for all $n\ge N$ (here we also use the fact that $f$ is increasing and $f'(a)>0$). But from our above work, this means $f^{n+1}(x)=f(f^n(x))>f^n(x)$, so this sequence is increasing, contradicting our assumption that its limit is $a$. This completes the proof.