On the existence of $ \lim_{x \to 0^{+}} \frac{\log(f(x))}{\log(x)} $ under some constraints

Question

I am considering a smooth-enough real-valued function $ f: (0,1) \to (0,\infty) $ such that

1. $ f $ is decreasing,
2. $\lim_{x\rightarrow0^{+}}f(x)=\infty $,
3. $ x \mapsto x^{2} f'(x) $ is decreasing,
4. $\lim_{x\rightarrow0^{+}}x^{2} f'(x)=0 $.

QUESTION. Under these constraints, does the limit $$ \lim_{x\rightarrow0^{+}}\dfrac{\log(f(x))}{\log(x)}$$ exist?

Thanks a lot to anybody who has any thoughts or counterexamples or spent time reading this question!

Answer 1

Here's one way to get a counterexample.

Let $f(x) = g(1/x)$, so $g'(t) = -(1/t^2) f'(1/t)$. Your conditions say that as $t \to +\infty$, we have $g$ positive and increasing to $\infty$, with $g'$ decreasing to $0$. We can choose sequences $t_n$ and $y_n$ such that

1. $y_n = t_n^{1/4}$ if $n$ is even, $t_n^{1/2}$ if $n$ is odd.
2. The slopes $s_n = (y_{n+1}-y_n)/(t_{n+1}-t_n)$ are positive and decreasing to $0$ with $n$
3. $t_n \to \infty$ as $n \to \infty$.

Take the piecewise linear function whose graph has vertices $(t_n, y_n)$; after smoothing off the corners we get a suitable function $g$ where $\log(g(t))/\log(t) = -\log(f(1/t))/\log(1/t)$ oscillates infinitely often between $1/4$ and $1/2$.

Answer 2

Based on Robert's answer, how about this: Let $$ v(x) = -\frac{3}{8} + \frac{1}{8}\cos(-\log(-\log(x/2))) $$ so that $v(x)<0$, $\limsup_{x \to 0^+}v(x) = -1/4$, $\liminf_{x \to 0^+}v(x) = -1/2$. Then
$$ f(x) = x^{v(x)} $$

is decreasing and goes to $\infty$ at $0$. And $-x^2f'(x)$

is increasing and has limit $0$ at $0$. And of course $$ \frac{\log(f(x))}{\log x} = v(x) $$ does not converge as $x \to 0^+$.