For smooth $f$ on $[0,1]$ such that $f^{(k)}(0)=0$ for all $k\in\mathbb{N}$, is it true that $\lim_{x\to0}\frac{xf'(x)}{f(x)}=\infty$?

Question

Got asked this question and I got a bit surprised at how messy it became.
Suppose $f$ is a smooth function on $[0,1]$ such that the $k^{\text{th}}$ derivative $f^{(k)}(0)=0,\forall k\in \mathbb{N}$.
Consider the limit $ L = \lim_{x\to0}\frac{xf'(x)}{f(x)}.$ Is it true that $L = \infty$ ?

I have tried a wide variety of functions that all failed, I tried to consider functions that decay fast enough or slow enough but they all failed that I started to think it might true, but I cant quite get how to show it.

COMMENT:

1. $\mathbb{N}=\{0,1,2,3,4,5 \cdots \}$
   That means $f^{(0)}(0)=f(0)=0$
2. f is defined to be non-zero near 0 so that the limit makes sense.

Answer 1

I don't think that $L= \infty$ is always true: Let $f:[0,1] \to \mathbb{R}$ be defined as $$ f(0)=0, \quad f(x)= (\frac{11}{10} + \sin(\frac{1}{x}))\exp(-\frac{1}{x})~~ (x \in (0,1]). $$ Then $f \in C^\infty([0,1])$ with $f^{(n)}(0)=0$ $(n \in \mathbb{N}_0)$ and $f(x)>0$ $(x \in (0,1])$. Now $$ f'(x)=\cos(\frac{1}{x})(-\frac{1}{x^2})\exp(-\frac{1}{x})+ (\frac{11}{10} + \sin(\frac{1}{x}))\exp(-\frac{1}{x})(\frac{1}{x^2}) $$ $$ =\frac{1}{x^2}(\frac{11}{10} + \sin(\frac{1}{x}) - \cos(\frac{1}{x})) \exp(-\frac{1}{x}), $$ so $$ \frac{x f'(x)}{f(x)}= \frac{1}{x}\frac{\frac{11}{10} + \sin(\frac{1}{x}) - \cos(\frac{1}{x})}{\frac{11}{10} + \sin(\frac{1}{x})} = \frac{1}{x}(1-\frac{ \cos(\frac{1}{x})}{\frac{11}{10} + \sin(\frac{1}{x})}). $$ The function $t \mapsto \frac{ \cos(t)}{\frac{11}{10} + \sin(t)}$ is $2\pi$-periodic with image $[-10/\sqrt{21},10/\sqrt{21}]$ (checked with WolframAlpha) and $10/\sqrt{21} >2$. Hence there are sequences $(x_n),(y_n)$ in $(0,1]$, both with limit $0$, such that for all $n$ $$ 1-\frac{ \cos(\frac{1}{x_n})}{\frac{11}{10} + \sin(\frac{1}{x_n})} = 1-\frac{10}{\sqrt{21}} < 0, \quad 1-\frac{ \cos(\frac{1}{y_n})}{\frac{11}{10} + \sin(\frac{1}{y_n})} = 1+\frac{10}{\sqrt{21}} > 0, $$ so $$ \frac{x_n f'(x_n)}{f(x_n)} \to -\infty, \quad \frac{y_n f'(y_n)}{f(y_n)} \to \infty \quad (n \to \infty). $$

Answer 2

Following the strategy in the following paper https://doi.org/10.1080/00029890.2019.1656025 with a small modification i claim the following.

$\textbf{Theorem}:$ Let f be a smooth function on [0,1] such that $f^{n}(0)=0$ for every $n\in \mathbb{N}\cup \{0\}$. If $\delta>0$ and \begin{equation}|xf'(x)|\leq C|f(x)| \text{ for some $C>0$ and for every $x\in [0,\delta] \qquad$ (1)} \end{equation} then $f(x)=0$ for every $x\in [0,\delta]$

$\textit{proof:}$ Once again following the proof adapted above, no matter what sign f(x) has by (1) : $$(x^{-2C}f^{2}(x))'=2x^{-2c-1}(xf(x)f'(x)-Cf^{2}(x))\leq 0$$ that is $x^{-2C}f^{2}(x)$ is decreasing on $(0,\delta]$ and so $$x^{-2C}f^{2}(x)\leq \lim_{t\longrightarrow 0}t^{-2C}f^{2}(t)=0$$ So that $f(x)=0$ on $[0,\delta]$ This ends the proof.

Now By negation as $f(x)$ is not identically $0$ on $[0,\delta] $ for arbitary delta we have that $$\forall C>0 , \frac{|xf'(x)|}{|f(x)|}>C \text{ for some $x\in (0,\delta]$}$$ Now as $\frac{xf'(x)}{f(x)}$ is smooth on some neighborhood of delta away from zero the following set$\{x:x\in (0,\delta] \text{ and }\frac{xf'(x)}{f(x)}>C\}$ is open for some $\delta$ small enough which means that $$\forall C>0 , \exists \delta>0 \text{ such that } 0<x<\delta \implies |\frac{xf'(x)}{f(x)}|>C $$ and we are done by applying the same proof to $[-\delta,0)$

Answer 3

As @Ace shows it, the quotient $\frac{xf'(x)}{f(x)}$ can't be bounded near $0$, thus if $L \in [-\infty,+\infty]$ exists, then $L = -\infty$ or $+\infty$. However, there are some cases in which it doesn't exist. For example, take, $$ f : x \mapsto e^{-\frac{1}{x^2}}, \qquad g : x \mapsto e^{-\frac{1}{x^2}}\cos\left(\frac{1}{x^2}\right), \qquad h = cf + g, $$ where $c$ is any real constant such that $1 < c < \sqrt{2}$ ($c = 1.2$ for example). $f$ and $g$ is defined on the whole $\mathbb{R}$ (with $f(0) = g(0) = 0$) and using the fact that $g(x) = \Re\left(e^{(i - 1)\frac{1}{x^2}}\right)$, we can prove that for all $n$, $f^{(n)}(x) = F_n(x)e^{-\frac{1}{x^2}}$ and $g^{(n)}(x) = \Re\left(G_n(x)e^{(i - 1)\frac{1}{x^2}}\right)$ if $x \neq 0$ where $F_n$ and $G_n$ are some rational fraction ($F_n$ is real, $G_n$ is complex) that only have poles in $0$, thus $f^{(n)}(x) \rightarrow 0$ and $g^{(n)}(x) \rightarrow 0$ when $x \rightarrow 0$. By the theorem of the limit of the derivative, $f$ and $g$ are smooth and for all $n$, $f^{n}(0) = g^{(n)} = 0$. We deduce that the same holds for $h$. Moreover, $f$ is positive outside $0$ and $|g| \leqslant f$ because $|\cos| \leqslant 1$ hence $h = cf + g \geqslant cf - f = (c - 1)f > 0$ outside $0$. Therefore, $\frac{xh'(x)}{h(x)}$ is well defined for all non zero $x$.

Now, we can compute that for all $x \neq 0$, \begin{align*} h'(x) & = \frac{2}{x^3}e^{-\frac{1}{x^2}}\left(c + \cos\left(\frac{1}{x^2}\right)\right) + \frac{2}{x^2}e^{-\frac{1}{x^2}}\sin\left(\frac{1}{x^2}\right)\\ & = \frac{2}{x^3}h(x) + \frac{2}{x^3}h(x)\frac{\sin\left(\frac{1}{x^2}\right)}{c + \cos\left(\frac{1}{x^2}\right)}\\ & = \frac{2}{x^3}h(x)\frac{c + \cos\left(\frac{1}{x^2}\right) + \sin\left(\frac{1}{x^2}\right)}{c + \cos\left(\frac{1}{x^2}\right)}. \end{align*} Therefore, for all $x \neq 0$, $$ \frac{xh'(x)}{h(x)} = \frac{2}{x^2} \cdot \frac{c + \cos\left(\frac{1}{x^2}\right) + \sin\left(\frac{1}{x^2}\right)}{c + \cos\left(\frac{1}{x^2}\right)}. $$ this ratio is, as expected, unbounded, and we can show that $L$ doesn't exist. Indeed, for $x_n = \frac{1}{\sqrt{2\pi n}} \rightarrow 0$, we have $\frac{1}{x_n^2} = 2\pi n$ so $\frac{x_nh'(x_n)}{x_n} = 2 \cdot 2\pi n \cdot \frac{c + 1 + 0}{c + 1} = 4\pi n \rightarrow +\infty$, but for $y_n = \frac{1}{\sqrt{\pi n + \frac{5\pi}{4}}}$, we have $\frac{1}{y_n^2} = 2\pi n + \frac{5\pi}{4}$ so $\frac{y_nh'(y_n)}{y_n} = 2 \cdot \left(2\pi n + \frac{5\pi}{4}\right) \cdot \frac{c - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{c - \frac{\sqrt{2}}{2}} = \left(8\pi n + 5\pi\right)\frac{c - \sqrt{2}}{2c - \sqrt{2}} \rightarrow -\infty$. $L$ doesn't exist and by the theorem of intermediate values, for all $l \in \mathbb{R}$ and all $n$ large enough, we can find some $z_n \in [x_n,y_n]$ (so $z_n \rightarrow 0$) such that $\frac{z_nh'(z_n)}{z_n} = l$.