I am working on Real Analysis question. I have been struggling to come up with the right idea solving the following problem.
Let $f(x)=\frac{1}{4^x}$ be a function defined on $\mathbb{R}$. Let $f_1=f$ and $f_{n+1}=f\circ f_{n}$ where $f_n$ is the composition of $f$ ($n$ times) . Show that the sequence $\{f_n(x)\}$ converges uniformly on $\mathbb{R}$ and find its limit.
Does anyone have a good idea to tackle this problem?
Limiting object: Consider the limiting object, $f_n(x)\to g(x).$ Since $f$ is differentiable, so are compositions with itself. Therefore, $g$ is differentiable. In the limit, $f_n\to g$ and $f_{n+1}\to g$ so it amounts to solving $g(x)=f(g(x)),$ or equivalently, $g(x)=\left(\frac{1}{4}\right)^{g(x)}.$ Differentiating, \begin{align*} g'(x)&=\left(\frac{1}{4}\right)^{g(x)}\ln\left(\frac{1}{4}\right)g'(x) \\ &=-\ln(4)\left(\frac{1}{4}\right)^{g(x)}g'(x). \end{align*} If $g'(x)\neq 0,$ then $1=-\ln(4)\left(\frac{1}{4}\right)^{g(x)}$. The right hand side is clearly negative, so we have a contradiction. Therefore, $g'(x)\equiv 0.$ Hence, the limiting object, if it exists, is a constant function $g(x)\equiv c^*.$ The existence and uniqueness of the number $c^*$ is clear if you try to solve $c=\left(\frac{1}{4}\right)^c.$ Obviously $c^*>0$ since a negative cannot equal a positive. To that end, consider $h(c)=c-\frac{1}{4}^c.$ We have $h(0)=-1<0,$ $h(1)=\frac34>0$ so existence follows from the intermediate value theorem. Moreover, $h'(c)=1-(-\left(\frac{1}{4}\right)^c\ln(4))=1+\left(\frac{1}{4}\right)^c\ln(4)>0$, showing that $c^*\in(0,1)$ is unique. Above proves that if a limit exists, it is unique.