Conjecture: $\lim_{x\to 0^+}f(2x)\log f(x)=0$.

Question

I encountered the following problem:

Suppose $f$ is a strictly increasing $C^1$ function defined on $[0,a]$ for some $a>0$, such that $f(0)=f'(0)=0$. Is it true that $\lim_{x\to 0^+}f(2x)\log f(x)=0$?

I tried to use the fact that $z^{\alpha}\log z\to 0$ as $z\to 0^+$ for any $\alpha>0$, and write $f(2x)\log f(x) = \frac{f(2x)}{f(x)^{\alpha}}f(x)^{\alpha}\log f(x)$. It would then be sufficient to show that there always exists an $\alpha>0$ such that $\frac{f(2x)}{f(x)^{\alpha}}$ is bounded at zero. However, it turns out this is not the case. Here's a counterexample: $f(x) = e^{-e^{1/x}}$ (augmented with $f(0)=0$). But even for this counterexample the initial conjecture works.

Any help would be appreciated.

Answer 1

We want $f$ to decrease really fast, so that decreasing it's argument two times increases even $\log$ by some large quantity. Also, note that it's enough to have such behavior only on points $x = 2^{-n}$, as long as function will remain in $C^1$.

Let $f\left(\frac{1}{2^n}\right) = \frac{1}{^n2}$, where $^n2$ is tetration, $^02 = 1$, $^{n+1}2 = 2^{(^{n}2)}$, and interpolate $f$ smoothly between with something like

$$f(x) = \frac{1}{^n2} + g\left(2^n \cdot \left(x - \frac{1}{2^n}\right)\right) \cdot \left(\frac{1}{^{n-1}2} - \frac{1}{^n2}\right)$$ when $x \in \left[\frac{1}{2^n}, \frac{2}{2^n}\right)$ and $g$ is some increasing smooth function on $[0, 1]$ with $g(0) = 0$, $g(1) = 1$ and having all derivatives in $0$ and $1$ equal to $0$, and having first derivative bounded by some constant $M$.

We have $f(x) < \frac{1}{^n2} < \frac{1}{2^{2n}}$ (for large enough $n$) if $x < \frac{1}{2^n}$, so $f(x) \to 0$ and $f'(0) = 0$.

For derivatives, if $x = \frac{1}{2^n}$, then $f'(x) = 0$ both from left and right by property of $g$.

Otherwise, if $x \in \left(\frac{1}{2^n}, \frac{2}{2^n}\right)$, we have $$f'(x) = g'\left(2^n \cdot \left(x - \frac{1}{2^n}\right)\right) \cdot 2^n \cdot \left(\frac{1}{^{n-1}2} - \frac{1}{^n2}\right) < \frac{M\cdot 2^n}{^{n-1}2} $$ which goes to $0$ as tetration growths much faster than exponent, so $f'(x) \to 0$.

Important property of tetration: $\log \left({^{n+1}2}\right) = {^{n}2}$.

For $x = \frac{1}{2^{n+1}}$ we have $$f(2x)\log_2 f(x) = \frac{1}{^n2} \cdot \log \frac{1}{^{n+1}2} = \frac{1}{^n2} \cdot \left(-{^n2}\right) = -1$$

Answer 2

Hint :

As $f$ is increasing so $f$ is positive now using inverse function of $f(x)$ try to show in the neightborhood of zero :

$$x\ln(x)\leq f\left(2x\right)\ln\left(f\left(x\right)\right)\leq 0$$

And then apply the squeeze theorem .