23

Given $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$ for all $x\in\mathbb{R}$. Show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ for all $n\in\mathbb{N}$.

I appreciate any help!

Edit 1: It is easy to see that $f$ is injective. So, if $f$ increases anywhere, $f$ will be an increasing function.

Edit 2: As said in the comments, $f$ must indeed be increasing, since $f(1)=f(e^0)=f(f(f(0)))=e^{f(0)}>f(0)$.

lone student
  • 1
JPMarciano
  • 1,055
  • 5
  • 15
  • 2
    Not an answer, but intuitively, suppose it is not true, then f(x) is bounded by a polynomial. But then applying f again to f (a polynomial) gives another polynomial, which still grows slower than exponentially. – Benjamin Wang Mar 05 '21 at 00:47
  • 2
    @BenjaminWang Agreed. But I found out some difficult trying to formalize this idea because not necessarily there is $x_0\in\mathbb{R}$ such that $f$ is dominated by $x^n$ for all $x\geq x_0$. – JPMarciano Mar 05 '21 at 01:22
  • 2
    Ah I see. There's the annoying case where the limit doesn't exist (?) – Benjamin Wang Mar 05 '21 at 01:57
  • 1
    @BenjaminWang Yes, exactly! – JPMarciano Mar 05 '21 at 01:59
  • 3
    Using your edit: $f$ must indeed be strictly increasing, since $f(1) = f(e^0) = f(f(f(0))) = e^{f(0)} > f(0)$. – hgmath Mar 06 '21 at 16:04
  • 1
    @TheSimpliFire what do you mean by this $f^{(n)}(x)$? If you meant the $n$-th derivative we would have to prove it exists firstly (which I don't think is necessarily true). – JPMarciano Mar 06 '21 at 16:50
  • Why in the edit 2, if $f(1) > f(0)$, you can deduce that $f(x)$ is increasing? – NN2 Mar 06 '21 at 17:02
  • @NN2 It is because $f$ is injective. As it is continuous, it must be increasing everywhere or decreasing everywhere. – JPMarciano Mar 06 '21 at 17:04
  • Ok, I see it now. – NN2 Mar 06 '21 at 17:06
  • 3
    Various properties of $f$ can be found here: https://math.stackexchange.com/q/65876/42969. – Martin R Mar 06 '21 at 17:16
  • 2
    If you can show $f(x)/x\rightarrow \infty$, then you can make the substitution $x=e^t$ to get $\lim_{x\rightarrow\infty} \frac{f(x)}{x^n} = \lim_{t\rightarrow\infty} \frac{e^{f(t)}}{e^{nt}} = \lim_{t\rightarrow\infty} e^{f(t)-nt} = \infty$. – Dark Malthorp Mar 06 '21 at 17:52
  • 3
    @MartinR $f(e^t)=f(f(f(t)))=e^{f(t)}$ – JPMarciano Mar 06 '21 at 18:29
  • @BenjaminWang Actually assuming not does not imply $f$ is bounded by a polynomial; all that follows is some sequence $x_n\to\infty$ such that $|f(x_n)|\le|p(x_n)|$ for some polynomial $p$. – David C. Ullrich Mar 10 '21 at 11:55

2 Answers2

16

We'll start with a weak estimate, and then use the functional equation to improve it iteratively.

Claim: There is a constant $d > 0$ such that $ f(x) \ge x+d$ for all $x \ge 0$.

Once we have that estimate (see below for the proof), we can can continue as follows: Setting $C = e^d > 1$ we get $$ f(x) = f(e^{\ln x}) = e^{f(\ln x)} \ge e^{\ln x + d} = xe^d = Cx $$ for $x \ge 1$, then $$ f(x) = e^{f(\ln x)} \ge e^{C\ln x} = x^C $$ for $x \ge e$, and finally $$ f(x) = e^{f(\ln x)} \ge e^{(\ln x)^C} $$ for $x \ge e^e$.

It follows that $$ \frac{f(e^x)}{e^{nx}} \ge \frac{e^{x^C}}{e^{nx}} = e^{x^C-nx} \ge x^C-nx+1 $$ and therefore $$ \lim_{y \to \infty} \frac{f(y)}{y^n} = \lim_{x \to \infty} \frac{f(e^x)}{e^{nx}} = +\infty \, . $$

Proof of the claim: $f(0) > 0$ and $f$ has no fixed points (see for example thoughts about $f(f(x))=e^x$), therefore $f(x) > x$ for all $x \in \Bbb R$. Then $$ d = \min \{ f(x) - x \mid 0 \le x \le 1 \} $$ is strictly positive. Define the sequence $(E_k)$ recursively by $E_0 = 0$, $E_{k+1} = e^{E_k}$: $$ 0, 1, e, e^e, e^{e^e},\ldots $$ $f(x) \ge x+d$ holds for $E_0 \le x \le E_1$, and if it holds for $E_k \le x \le E_{k+1}$ then $$ f(x) = e^{f(\ln x)} \ge e^{\ln x + d} = xe^d \\ \ge x(1+d) = x + xd \ge x+d $$ for $E_{k+1} \le x \le E_{k+2}$. This concludes the proof of the claim, since $E_k \to \infty$.

Martin R
  • 128,226
4

remark
It is not enough to assume (for example) that $f(f(x)) = e^x$ for all $x>1$. You must assume (as the OP says) $f(f(x))=e^x$ for all $x$. This is used in Edit 2.

The following function $f : \mathbb R \to \mathbb R$ is continuous and $f(f(x)) = e^x$ for all $x > 1$ but fails the required conclusion.

graph

$$ f(x) = \begin{cases} -2x,\quad &x\ge T \\ e^{-x/2},\quad &x < T \end{cases} $$ Here $T \approx -0.7148$ is one of the points where the two graphs $-2x,e^{-x/2}$ cross.

GEdgar
  • 117,296
  • But according to your function $f$, we have: $f(0)=0$ hence $f(f(0)) = f(0) = 0 \neq 1 = e^0 $ ! – NN2 Mar 06 '21 at 19:46
  • 5
    @WillJagy This answer doesn't solve the question or provide anything that we can use to solve the question. So what is its purpose here? – NN2 Mar 06 '21 at 20:05