Let $f: \mathbb R \rightarrow \mathbb R$ be a smooth function. Is there a characterization of the space of functions $f$ whose ratios of higher order derivatives are bounded by at most an exponentially growing constant at infinity? That is, for some $K$ large enough, there is a constant $C > 0$ for which $$ \Bigg\lvert\frac{f^{(n)}{(x)}}{f(x)}\Bigg\rvert \leq C^n, \quad \forall |x| \geq K. $$ (It might be that the correct form to avoid issues with zeroes is $|f^{(n)}(x)| \leq C^n(L + |f(x)|)$ instead, where $L$ is large enough.)
I have an intuition the answer is functions which are bounded by an exponential but I do not know how to characterize the space to allow for bad behaviour like bounded oscillations with unbounded derivatives, e.g. $f(x) = e^x + \sin^2(e^x)$, say.
I also expect this space to contain functions of the form $f(x) = P(x)e^{cx}$ where $P$ is a polynomial and $c$ is a constant. For instance, if $f(x) = xe^x$ then $f^{(n)}(x) = (x + n)e^x$ and $|n/x| \leq n/K$ is growing linearly. More generally the $n$-th derivative of $P(x)e^x$ is $Q_n(x)e^x$ where $Q_n$ is a binomial sum of derivatives of $P$ up to order $n$, and so there at most $2^n$ terms in $Q_n$ and each of the terms can be bounded uniformly since $|P^{(n)}(x)/P(x)|$ is the ratio of a lower order polynomial to a higher order one for $n \geq 1$.
The motivation for this question is to have a class of functions which are "essentially exponential" in growth with respect to all derivatives. I suppose I could always claim this definition above is the characterization but I was wondering if there is a cleaner characterization given the examples seem relatively well behaved and almost commonplace.
Edit 1: after some reading, I have a feeling this might be related to Fourier transforms and that it might be possible to prevent bad behaviour by asking for a condition on the Fourier transform (e.g. that it has compact support, say). However I don't really know enough about the Fourier transform to make a tractable characterization.
Edit 2: I would be okay if someone can give a characterization of the case $n = 1$, although I expect this to lead to a characterization for all $n \geq 1$. There was a deleted answer that showed this is equivalent to boundedness of the log derivative which implies at most exponential growth. But this is not sufficient since one can have exponential growth with a bad derivative coming from a bounded oscillatory term, e.g. $f(x) = e^x + \sin^2(e^{e^x})$.
I want to give two characterisations of smooth functions $\newcommand{\N}{{\mathbb N}}\newcommand{\R}{{\mathbb R}}\newcommand{\C}{{\mathbb C}}\newcommand{\CC}{{\mathcal C}}\newcommand{\norm}[1]{\left|#1\right|}f:\R \to\R$ satisfying the condition $$\tag{C}\mbox{ There exist }C,K>0\mbox{ such that for }|x|\geq K:\ \ \ f(x)\neq0\mbox{ and }\Bigg\lvert\frac{f^{(n)}{(x)}}{f(x)}\Bigg\rvert \leq C^n. $$ The set of these functions is denoted by $\CC$.
The first characterisation is the following.
Claim 1: A smooth functions $f:\R \to\R$ satisfies condition (C) if and only if there are constants $C,K$ such that the restrictions of $f$ to $[K,\infty[$ and $]-\infty,-K]$ do not vanish and are restrictions of entire functions $f_+$ and $f_-$ of order at most 1, respectively, satisfying more precisely $$\tag{E} \norm{f_\pm(z)}\leq\norm{f(x)}\exp(C\norm{z-x})\mbox{ for }\pm x\geq K\mbox{ and }z\in\C.$$
It will be shown later, that it is not sufficient that $f_\pm$ are entire functions of order at most 1. It is important that (C) holds with a constant independent of $x$.
The second characterisation is based on the first one and concerns the zeros of the functions $f_\pm$.
Let $a_k(g)$, $k=1,2,\dots$, denote the sequence of zeros of the entire function $g$ such that each zero appears as many times as its multiplicity indicates and $\norm{a_k(g)}\leq\norm{a_{k+1}(g)}$ for all $k$. Let $n_g(R,x)$ be the number of zeros of $g$ in the open disk $D(R,x)=\{z\in\C\mid\norm{z-x}<R\}$.
Claim 2: A smooth functions $f:\R \to\R$ satisfies condition (C) if and only if there are constants $B,C,K$ such that its restrictions to $[K,\infty[$ and $]-\infty,-K]$ do not vanish and are restrictions of entire functions $f_+$ and $f_-$ of order at most 1, respectively, such that their (finite or infinite) sequences of zeros satisfiy $$\tag{N} n_{f_\pm}(R,x)\leq C\,R\ \ \mbox{ and } \delta_{f_\pm}(R,x):=\sum_{\norm{a_n(f_\pm)-x}<R}\frac1{a_n(f_\pm)-x}\leq B\mbox{ for }\pm x\geq K \mbox{ and }R>0.$$ Again it is important that the constants $B,C$ are independent of $x$. Observe that the first inequality implies that $f_{\pm}$ do not have zeros in the disks $D(1/C,x)$ and therefore not in the strips $\renewcommand{\Re}{\mbox{Re}\,}\renewcommand{\Im}{\mbox{Im}\,} \{z\in\C\mid\pm\Re z\geq K,\norm{\Im z}<1/C\}.$
Before giving the proofs, we present two examples. First, let $(\varepsilon_k)_{k=-\infty}^\infty$ be an arbitrary bi-infinite sequence of complex numbers of absolute value smaller than $1/2$ and $\newcommand{\Z}{\mathbb Z}b_k,$ $k\in\Z$, the bi-infinite sequence of the numbers $k+i+\varepsilon_k$, $k\in\Z$. Consider an entire function $f$ of order at most 1 having the $b_k$ and $\bar b_k$ as its zeros, for example the canonical product $$f_1(z)=\prod_{k\in\Z}\left[\left(1-\frac z{b_k}\right)e^{\frac z{b_k}} \left(1-\frac z{\bar b_k}\right)e^{\frac z{\bar b_k}}\right].$$ Clearly, $f_1$ has real values for real $z$ by construction and it satisfies the conditions of the second claim. Therefore it satisfies the wanted condition (C) even on the entire real axis.
As a second example, consider the product $$f_2(z)=\prod_n\left[\left(1-\frac{z}{2^{n}+i}\right)^n\left(1-\frac{z}{2^{n}-i}\right)^n \right].$$ Clearly, $f_2$ is an entire function of order 0, real valued, nonvanishing on the real axis and has no zeros in the strip $\{z\in\C\mid\norm{\Im z}<1\}$, but since its zeros $2^n\pm i$ have multiplicity $n$, it does not satisfy the condition of Claim 2. Not surprisingly, we calculate $$\frac{f_2'(2^n+1)}{f_2(2^n+1)}\approx\frac n{1-i}+\frac n{1+i}=n$$ and condition (C) is evidently not satisfied. This illustrates that the additional conditions in (E) or (N) cannot be omitted.
Proof of Claim 1: We begin with deriving necessary conditions for $f\in\CC$. First, for a function $f\in\CC$ $$\tag{1}\mbox{ There exist }C,K>0\mbox{ such that for }y,z\geq K\mbox{ or }y,z\leq-K:\ \ \norm{f(z)}\leq \norm{f(y)}e^{C\norm{y-z}}.$$ Proof: By (C), there exist $C,K$ such that $\norm{f'(x)}\leq C\norm{f(x)}$ for $\norm x\geq K$. We consider first only $x\geq K$. By the intermediate value theorem, we can assume that $f(x)>0$ for $x\geq K$. Thus we have $f'(x)\leq C\, f(x)$ and hence $\dfrac d{dx}\left(f(x)e^{-Cx}\right)=e^{-Cx}(f'(x)-C\,f(x))\leq0$ for $x\geq K$. Therefore $f(x)e^{-Cx}$ is decreasing and we have $$f(z)e^{-Cz}\leq f(y)e^{-Cy}\mbox{ for }z\geq y\geq K.$$ Using $f'(x)\geq -C\, f(x)$, we find similarly that $f(y)e^{Cy}\geq f(z)e^{Cz}$ for $y\geq z\geq K$. This proves (1) for $y,z\geq K$. The proof for $y,z\leq -K$ is analogous.
Consider now $f$ satisfying (C) with constants $C,K$ and fix $z>x\geq K$. We apply Taylor's formula with large $n$ $$\begin{array}{l} f(z)=f(x)+{f'(x)}(z-x)+\dfrac{f''(x)}{2!}(z-x)^2+\cdots+ \dfrac{f^{(n-1)}(x)}{(n-1)!}(z-x)^{n-1}+R_n(f,x,z),\\\mbox{ where } {R_n(f,x,z)}=\dfrac{f^{(n)}(\xi)}{n!}(z-x)^n \end{array}$$ and $\xi$ is a certain number between $x$ and $z$ depending upon $x,z,f,n$. Using (C) and (1), we estimate $$\norm{R_n(f,z)}\leq \frac{C^n\norm{f(\xi)}}{n!}(z-x)^{n}\leq \frac{C^n(z-x)^n}{n!}e^{z-x}f(x).$$ Now it is well known that $L^n/n!\to 0$ as $n\to\infty$ for any constant $L$. Therefore the restriction $f\mid_{]K,\infty[}$ of $f$ to $]K,\infty[$ is real analytic and the sum of its Taylor series. The above estimate also shows that the series converges for any complex $z$. Hence $f\mid_{]K,\infty[}$ is the restriction of an entire function we call $f_+$. Finally for $x\geq K$ and $z\in\C$ we can estimate the series for $f(z)$ by $\sum_n \frac1{n!}C^n|z-x|^n\,|f(x)|=\norm{f(x)}e^{C\norm{z-x}}$. This proves the necessity of the conditions in Claim 1 for $f_+$ and $x\geq K$. The proof for $f_-$ and $x\leq -K$ is analogous.
Conversely, suppose that a smooth function $f:\R\to\R$ satisfies the conditions of Claim 1 and let $f_+$ denote the entire extension of $f\mid_{[K,\infty[}$ which satisfies (E) with certain constants $C,K$. Fixing $x\geq K$, we apply Cauchy's integral formula $$f^{(n)}(x)=f_+^{(n)}(x)=\frac{n!}{2\pi i}\oint_{\norm{z-x}=R}\frac{f_+(z)}{(z-x)^{n+1}}\,dz$$ for positive integers $n$ and positive $R$. Using (E), this yields $\norm{f^{(n)}(x)}\leq n!\,e^{CR}R^{-n}\norm{f(x)}$. With the optimal choice $R=n/C$, we obtain that $$\norm{f^{(n)}(x)}\leq n!\,e^nC^n{n^{-n}}\norm{f(x)}\leq (eC)^n\norm{f(x)}.$$ This proves (C) for $x\geq K$ with the constant $eC$ replacing $C$. The proof for $x\leq -K$ is the same.
Proof of Claim 2: We first introduce some notation and collect four statements of the theory of entire functions of exponential growth.
For an entire function $g$, let $M_g(R,x)$ be the maximum of $\norm{g(z)}$ over the circle $\norm{z-x}=R$.
Lemma: 1. If $g$ is an entire function with $g(0)=1$ then $n_g(R,0)\leq \log M_g(eR,0)$ for $R>0$.
2. There exists a constant $D_1$ such that for every entire function $g$ of order at most 1 with $g(0)=1$ and Hadamard representation $g(z)=e^{az}\prod_n\left[(1-\frac z{a_n})e^{\frac z{a_n}}\right]$ we have $$\left|a+\sum_{\norm{a_n}<R}\frac1{a_n}\right|\leq {D_1}\frac{ M_g(eR,0)}R\mbox{ for }R>0. $$ 3. There exists a constant $D_2$ such that for every sequence $a_n$, $n=1,2,\dots$ tending to infinity with $0<\norm{a_1}\leq\norm{a_2}\leq\dots$ and $\sum_n\norm{a_n}^{-\rho}<\infty$ for all real $\rho>1$, the canonical product $\Pi(z)=\prod_n\left[(1-\frac z{a_n})e^{\frac z{a_n}}\right]$ satisfies $$\log M_\Pi(R,0)\leq R\,\delta_\Pi(R,0)+D_2\left[\int_0^R\frac{n_\Pi(t,0)}t\,dt +R^2\int_R^\infty\frac{n_\Pi(t,0)}{t^3}\,dt \right]\,\mbox{ for }R>0.$$ 4. If $d_1,d_2$ are two entire functions with $d_1(0)=d_2(0)=1$ such that their quotient $\phi(z)=d_1(z)/d_2(z)$ can be analytically continued to an entire function also called $\phi$, then $$M_\phi(R,0)\leq 3M_{d_1}(eR,0)+3M_{d_2}(eR,0)\mbox{ for }R>0.$$
Statement 1 can be found in Levin's book on entire functions, section 2.5, formula (15). Statement 2 follows by modifiying the first part of the proof of Theorem 5.2 in Levin's book for functions with $g(0)=1$, statement 3 by modifying its second part for canonical products. Statement 4 is at the end of section 2.4.
We are now in a position to prove Claim 2. It is sufficient to consider an entire function $g$ of order at most 1 and to show that (E) and (N) are equivalent for $x\geq K$. For $x\leq -K$, the proof is the same.
First suppose that there exists $C>0$ such that $|g(z)|\leq\norm{g(x)}e^{C\norm{z-x}}$ for all $x\geq K$ and all complex $z$. For $g_x$ defined by $g_x(z)=g(z+x)/g(x)$ this means that $M_{g_x}(R,0)\leq e^{CR}$ for all $R>0$. Using Lemma 1, statement 1, this implies that $n_{g_x}(R,0)\leq eC\, R$ for all positive $R$ and therefore $n_g(R,x)\leq eC\, R$ for all $x\geq K$ and $R>0$.
Applying statement 2 to $g_x$, there exists $D_1>0$ such that for all positive $R$ we have $\norm{a+\delta_{g_x}(R,0)}\leq D_1\,eC$. Using $R>0$ so small that $n_{g_x}(R,0)=0$ by the above inequality for $n_{g_x}$, we find in particular that $\norm a\leq D_1\,eC$. Hence we have $\norm{\delta_{g_x}(R,0)}\leq 2D_1\,eC$, that is $\norm{\delta_g(R,x)}\leq 2 D_1\,eC$ for all positive $R$. So $g$ satisfies (N).
Now suppose that $g$ satisfies (N) and consider again $g_x(z)=g(z+x)/g(x)$ for arbitrary $x\geq K$. It satisfies $n_{g_x}(R,0)\leq CR$ and $\delta_{g_x}(R,0)\leq B$ for all positive $R$. We write down the Hadamard representation of $g_x$: $$g_x(z)=e^{a_x z}\Pi_x(z)\mbox{ with some constant }a_x\mbox{ and } \Pi_x(z)=\prod_{n=1}^\infty\left[\left(1-\frac z{a_n(g)-x}\right) e^{\frac z{a_n(g)-x}}\right].$$ We now apply statement 3 of the Lemma to $\Pi_x$ and use the inequalities (N). We obtain with some constant $D_2$ that $$\log M_{\Pi_x}(R,0)\leq BR+2D_2CR\mbox{ for }R>0. $$ This shows that $\log M_{g_x}(R,x)\leq (\norm{a_x}+B+2D_2C)R$ for $x\geq K$ and positive $R$. This implies (E) provided we can show that the $a_x$, $x\geq K$ form a bounded familiy of numbers.
Since $g(z)=g(K)g_K(z-K)$ we obtain $\norm{g(z)}\leq \norm{g(K)}e^{LK}\,e^{L\norm z}$ for all $z$ with $L=\norm{a_K}+B+2D_2C$ and hence $$g_x(z)\leq N_x\,e^{L\norm z},\mbox{ where }N_x=\norm{\tfrac{g(K)}{g(x)}}e^{LK+L\norm x}$$ for $x\geq K$.
Fixing temporarily some $x\geq K$, we now apply statement 4 of the Lemma to $d_1=g_x$ and $d_2=\Pi_x$. Then $\phi$ is given by $\phi(z)=e^{a_xz}$. We obtain $$\norm{a_x}R\leq 3LeR+3\log N_x+3e(B+2D_2C)R\mbox{ for }R>0.$$ This implies $\norm{a_x}\leq 3eL+3eB+6eD_2C$ for $x\geq K$ and the proof is complete.
Remarks: 1. Observe that the product of functions in $\CC$ also belongs to $\CC$. This can be shown directly using (C) or using Claims1 or 2. In cs89's answer, it is shown that this does not hold for sums.
2. In the same way, the nonvanishing smooth functions $f:\R\to\R$ such that there exists a constant $c>0$ with $|f^{(n)}(x)|\leq c^n|f(x)|$ for all real $x$ and all $n\in\N$ can be characterised. They are those nonvanishing smooth functions $f:\R\to\R$ restrictions of entire functions also called $f$ such that $\norm{f(z)}\leq\norm{f(x)}\exp(C\norm{z-x})\mbox{ for } x\in\R \mbox{ and }z\in\C$. They are also those nonvanishing smooth functions $f:\R\to\R$ restrictions of entire functions of order at most 1 also called $f$ such that there exist $B,C>0$ with $$n_{f}(R,x)\leq C\,R\ \ \mbox{ and } \delta_{f}(R,x)\leq B\mbox{ for }x\in\R \mbox{ and }R>0.$$