Show a sequence of functional iterates converges uniformly Problem.
Consider $f(x) = 4^{-x}, x \in (-\infty, \infty)$. Define $f_1 = f$ and, recursively, $f_{n+1} = f \circ f_n$, thus $$f_n(x) = [\underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}}](x).$$ Prove that the sequence $\{f_n(x)\}_{n \in \mathbb{N}}$ converges uniformly on $(-\infty, \infty)$ and find its limit.

My Question.
Does this seem like the kind of problem where I can appeal to some famous theorem in order to obtain the desired result? Or must I resort to analyzing the functions directly? I am studying for my qualifying exam and this was a question from a previous administration of the exam. I have a solution from a peer, and his solution looks directly at the iterates of $f$ and is what I would consider a "brute force" type solution. So I am basically wondering if there is some deeper theory or meaning behind this problem that I can research?
Here's what I do know so far: the original $f$ has a fixed point at $f(\frac{1}{2}) = \frac{1}{2}$. Further, I surmise that the sequence of functions converges to the constant function $g(x) = \frac{1}{2}$. And, I know from analyzing $f'$ that from a certain $b$ onward, $|f'(x)| < 1$, meaning (I think) that from that $b$ onward, $f$ is a contraction and so, on closed restrictions of the domain of the form $[b, a]$, I could hypothetically apply the Banach fixed point theorem. But I don't know if any of these facts are helpful.
 A: You don't need too much machinery for your particular problem. But certainly the Banach fixed point theorem may be will used to provide a satisfactory solution to your problem. The following result, which you may know already, has many uses in analysis and numerical analysis:

Lemma: If $f:(a,b)\rightarrow\mathbb{R}$ is continuously differentiable,  $f(c)=c$ for some $a<c<b$,  and $|f'(c)|<1$, then there is a small closed interval $I$ containing $c$ such that  $f(I)\subset I$, and $f:I\rightarrow I$ is a contraction.

Proof: The assumptions imply that for $|f'(c)|<k<1$, there is $\delta'>0$ such that $|f'(x)|\leq k$ for all $x\in[c-\delta',c+\delta']$. Also,
$$f(x)=c+f'(c)(x-c)+(x-c)\frac{r(x-c)}{x-c}$$
where $\lim_{x\rightarrow c}\Big|\frac{r(x-c)}{x-c}\Big|=0$. Hence
$$|f(x)-c|\leq|x-c|\left(|f'(c)|+\Big|\frac{r(x-c)}{x-c}\Big|\right)$$
There is $0<\delta\leq\delta'$ such that if $|x-c|\leq\delta$, then $\Big|\frac{r(x-c)}{x-c}\Big|<k-|f'(c)|$. Consequently, for any $x\in I=[c-\delta,c+\delta]$
$$|f(x)-c|<k|x-c|<|x-c|$$
This shows that $f(I)\subset I$. The mean value theorem then shows that  $f:I\rightarrow I$ is a contraction: for $x,y\in I$
$|f(x)-f(y)|\leq |f'(x+\theta_{x,y}(x-y)||x-y|\leq k|x-y|$ (here $0<\theta_{x,y}<1$). $\Box$
Remark 1: The Lemma above shows that the functions $f_n$ on $I$ defined by $f_0=f$ and $f_n=f\circ f_{n-1}$, $n\geq1$, will converge uniformly to the constant function $f^*(x)\equiv c$, for
$$\sup_{x\in I}|f_n(x)-c|\leq k^n\operatorname{diam}(I)\xrightarrow{n\rightarrow\infty}0$$
where $k$ is as in the proof of the Lemma.
Remark 2: If there is $m\in\mathbb{M}$ such that $f_m(a, b)\subset I$, then the sequence $f_n$ converges uniformly to the constant function $f^*\equiv c$ on $(a, b)$.
Remark 3: Under the assumptions of the Lemma above, suppose there is an closed interval $J\subset(a, b)$ containing $c$ and such that $f((a, b))\subset J$. If $f:J\rightarrow J$ is a contraction, then $f_n$ converges uniformly to the constant function $f_*\equiv c$ on $(a, b)$.
The Lemma above on its own will not solve your problem (that of showing that $f_n$ is uniformly convergent), for it does not tell you what the good interval $I$ is and more importantly, there is no reason to think that $f_m((a, b)\subset I$ for some $m$ in general.
For your specific problem, properties of the function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=4^{-x}$ should be exploited: $f$ is smooth (infinitely continuously differentiable); $f$ has a unique fixed point at $x=\tfrac12$: if there were a second fixed point $c'$, then there would be a point $x_*$ between $1/2$ and $c'$ such that $0=f'(x_*)-1=-\log(4)f(x_*)-1$, which is not possible since $f>0$; $f$ strictly convex, and $f$ is  monotone decreasing. This last property is what we concentrate on. Notice that

*

*$f(\mathbb{R})\subset(0,\infty)$,

*$f([0,\infty))\subset[0,1]=[0,f_0(0)]$.

*The only fixed point of $f$, $1/2$, belongs to $[0,1]$.
Observe that
$$f([0,f_0(0)])=[f_1(0),f_0(0)]\subset[0,f_0(0)]$$
By induction, we have that
$$ \begin{align}
f([f_{2n+1}(0),f_{2n}(0)])&= [f_{2n+1}(0),f_{2n+2}(0)]\subset[f_{2n+1}(0),f_{2n}(0)]\\
&=f([f_{2n-1}(0),f_{2n}(0)])\subset[f_{2n-1}(0),f_{2n}(0)]
\end{align}$$
and $\frac{1}{2}\in[f_{2n+1}(0),f_{2n+2}(0)]\subset[f_{2n+1}(0),f_{2n}(0)]$ for all $n\in\mathbb{N}$.

Notice that for $n=0$, $f([\tfrac14,1])\subset[\tfrac14,1]$ and by convexity, for $\frac14\leq x\leq 1$
$$ -1<-\sqrt{2}\log(2)=f'(\tfrac14)\leq f'(x) \leq f'(1)=-\tfrac{\log4}4<0$$.
All this shows that $f_3(\mathbb{R})\subset[\frac14,1]$ and $f:[\tfrac14,1]\rightarrow[\tfrac14,1]$ is a contraction. Then, as in Remark 3, we have that $f_n$ converges uniformly to the constant function $f_*\equiv\frac12$.
