Let $L(x)=a\,x+b$, What values of $a$ and $b$ makes $L^{[n]}(x)$ converge as $n\rightarrow \infty$.$\forall x\in R$?

Question

Let $L(x)=a\,x+b$, What values of $a\neq0$ and $b$ such that

$L^{[n]}(x)$ converge as $n\rightarrow \infty$, where $L^{[n]}(x)=L(L(...L(x))$ ($n$ times)

$\forall x\in R$?

Answer 1

As @Alephnull pointed out, for fixed $x\in\mathbb{R}$, if $L^{[n]}(x)$ converges when $n\to\infty$ then it must converge to $\frac{b}{1-a}$. The key idea here is to write $$ L(x)-\frac{b}{1-a}=a\left(x-\frac{b}{1-a}\right), $$ for it leads to $$ L^{[n]}(x)-\frac{b}{1-a}=a\left(L^{[n-1]}(x)-\frac{b}{1-a}\right)=\cdots=a^n\left(x-\frac{b}{1-a}\right). $$ This tells us that

$\bullet$ If $|a|>1$ then $L^{[n]}(x)$ will NOT converge except for $x=\frac{b}{1-a}$.

$\bullet$ If $a=-1$ then (for fixed $x$) $L^{[n]}(x)$ is sign-oscillating and therefore won't converge, except again for the trivial case $x=\frac{b}{1-a}.$

$\bullet$ Finally, if $|a|<1$ then we can estimate $$ \left|L^{[n]}(x)-\frac{b}{1-a}\right|\leq|a|^n\left|x-\frac{b}{1-a}\right|, $$ which (for fixed $x$) converges to $0$ as $n\to\infty$.

To sum it up: $L^{[n]}(x)$ converges (and it converges to $\frac{b}{1-a}$) as $n\to\infty$ for every $x\in\mathbb{R}$ if and only if $|a|<1$.