We want to find an elementary function $f(x)$ that is smooth and strictly increasing on some interval $x\in(0,\epsilon)$, satisfying $\lim\limits_{\,x\to0^+}f(x)=0$, whose asymptotic for $x\to0^+$ is $f'(x)\sim2f(2x)$, that is, $\lim\limits_{\,x\to0^+}\frac{f'(x)}{2f(2x)} = 1$.

By a lengthy series of trial and error I found a solution. You can click “Reveal spoiler” below to see it, unless you want to ponder on this problem on your own for a while before looking at my version.

$\displaystyle\quad f(x)=\exp\left(-\left(\frac12+\frac1{\ln2}\right)\cdot\ln x-\frac1{\ln4}\cdot\ln^2\left(-\frac{\ln x}{x\cdot\ln2}\right)\right)$

Its derivative is rather cumbersome, but Mathematica says the limit $\lim\limits_{\,x\to0^+}\frac{f'(x)}{2f(2x)}$ is indeed $1$, which is also confirmed by manual calculations. The solution is not unique, there are possible variations — I picked one that looked more readable.

If we only impose a weaker condition $f'(x)=\mathcal O\left(f(2x)\right)$, then the solution becomes simpler:

$\displaystyle\quad f(x)=2^{\large-\frac{1}{2} \log _2^2\left(-\frac{\log_2\!x}{x}\right)}$

Is there a systematic approach to finding a solution to this problem?
Is there a simpler elementary function with required properties?

  • 2
    $\begingroup$ BTW, a solution of the exact functional equation $f'(x)=2f(2x)$ is given by non-elementary, non-analytic Fabius function, see math.stackexchange.com/a/1067358/19661, math.stackexchange.com/a/1764273/19661, math.stackexchange.com/q/3283519/19661, math.stackexchange.com/q/2497242/19661, en.wikipedia.org/wiki/Fabius_function $\endgroup$ Apr 26, 2021 at 19:25
  • $\begingroup$ I must be missing something but doesn't f(x) = 1+2x work? $\endgroup$ Apr 27, 2021 at 23:44
  • $\begingroup$ @MaximeWeill It would break both $\lim\limits_{\,x\to0^+}f(x)=0$ and $\lim\limits_{\,x\to0^+}\frac{f'(x)}{2f(2x)} = 1$ conditions. $\endgroup$ Apr 28, 2021 at 0:22
  • $\begingroup$ Well, it only breaks the zero condition. In general, you can show any such function that satisfies these conditions is not analytic around zero but that still leaves a whole lot of funcrions $\endgroup$
    – QC_QAOA
    Apr 28, 2021 at 4:56
  • $\begingroup$ @QC_QAOA Yes, you are right — it only breaks the first one, my mistake. Yes, if we look for a solution in terms of a power series, then all its coefficients must be zero, and the function will be constant and, hence, not strictly increasing as required. So, a solution cannot be analytic at zero. Surely, there must be many different solutions, and I was able to find a few, but only because I experimented a lot and finally had some luck. I wonder if there is a more systematic approach to find an explicit solution, and whether there are simpler ones than mine. $\endgroup$ Apr 28, 2021 at 19:20

1 Answer 1


This doesn't give a simpler function, but it suggests a systematic derivation of your results.

Let $f(x)=4^{g(\log_2{(x)})}$, where $g$ must be smooth, increasing, and $$\lim_{u\to-\infty}{g(u)}=-\infty\tag{0}$$ Then $$f'(x)=\ln{(4)}4^{g(\log_2{(x)})}\cdot\frac{g'(\log_2{(x)})}{\ln{\!(2)}x}=2x^{-1}g'(\log_2{(x)})4^{g(\log_2{(x)})}$$ Your condition is \begin{align*} 1&=\lim_{x\to0^+}{\frac{2f(2x)}{f'(x)}} \\ &=\lim_{x\to0^+}{\frac{2\cdot4^{g(\log_2{(x)}+1)}\cdot x}{2\cdot 4^{g(\log_2{(x)})}\cdot g'(\log_2{(x)})}} \\ &=\lim_{u\to-\infty}{\frac{2^u4^{g(u+1)-g(u)}}{g'(u)}} \end{align*} Taking logarithms base $2$, $$0=\lim_{u\to-\infty}{(2(g(u+1)-g(u))+u-\log_2{(g'(u))})}\tag{1}$$

(Note that $g$ is well-defined only up to constants: if we add one to $g$, conditions (0-1) are preserved. This allows us to freely drop constants of integration, after we find $g'$.)

If $g$ is slowly-varying, then $g(u+1)-g(u)\approx g'(u)$. I'm going to allow slightly more freedom: we'll take $g(u+1)-g(u)\approx g'(u)+c$, where $c$ is a constant.

Suppose (1) with this approximation holds exactly: that is, $$2g'(u)-\log_2{(g'(u))}=-u-c\tag{2}$$ This has no elementary function solutions, but the solutions to (2) are very well-studied. The standard convention is to describe them in terms of the Lambert W function, which solves the analogous equation $W(a)e^{W(a)}=a$: $$g'(u)=-\frac{W\left(-\ln{(4)}2^{u+c}\right)}{\ln{(4)}}$$

The natural thing to do here is to trace this back to $f$, Taylor-expand $W$, and hope that keeping the first few terms is enough to produce the desired behavior. In principle, this ought to work, but each time I try, I can't get $f$ to satisfy both conditions (0-1) at once.

Luckily, an there's an alternate approximation scheme for (2) that does seem to work. Define the sequence of functions $\{h_n\}_n$ by $h_0(u)=1$ and, for $n>0$, $$2h_n(u)+u+c=\log_2{(h_{n-1}(u))}$$ Describing the convergence of this sequence is subtle, but we're only working heuristically here. (That is, I'll skip it.)

Computing directly the first few terms, we have \begin{align*} h_0(u)&=1 \\ h_1(u)&=-\frac{u+c}{2} \\ h_2(u)&=-\frac{u+c+\log_2{\left(-\frac{u+c}{2}\right)}}{2} \\ h_3(u)&=-\frac{u+c+\log_2{\left(-\frac{u+c+\log_2{\left(-\frac{u+c}{2}\right)}}{2}\right)}}{2} \\ &=-\frac{u+c+\log_2{\left(-\frac{u+c}{2}\right)}+\log_2{\left(1+\frac{\log_2{\left(-\frac{u+c}{2}\right)}}{u+c}\right)}}{2} \end{align*}

As $u\to-\infty$, The difference between $h_3$ and $h_2$ is already $o(1)$; further correction terms will be likewise.

Since we only care that (1) should hold in the limit, this suggests that it suffices to take $g'=h_2$. If we do so, then we find that $$g(u)=\frac{-\frac{\log(2)}{2}(c+u)^2+(c+u)\log\left(-\frac{u+c}{2}\right)-(u+c)}{\log(4)}$$

Correspondingly, $$c\approx g(u+1)-g(u)-g'(u)=\frac{2(u+c+1)\log(\frac{u+c}{u+c+1})+2+\log(2)}{\log(16)}\to-\frac{1}{2}$$ as $u\to-\infty$.

Finally substituting back into $f$, we thus have $$f(x)=e^{-\frac{\log(x)^2}{\log(4)}} x^{-\frac{3}{2}-\frac{1}{\log(2)}}\left(1-\frac{2\log(x)}{\log(2)}\right)^{\frac{\log (x)}{\log(2)}-\frac{1}{2}}$$ Mathematica can then verify that $f$ does satisfy the desired conditions.

  • 2
    $\begingroup$ Thanks! Here is a solution in terms of the non-principal real branch of the Lambert W-function: $\exp\left(-\frac{\ln x}2-\left(1+\frac1{\ln2}\right) W_{\small\!-1}\!\left(-\frac{x\ln2}2\right)-\frac1{\ln4} W_{\small\!-1}\left(-\frac{x\ln2}2\right)^2\right),$ I had found it before I found a solution in elementary functions. $\endgroup$ Apr 30, 2021 at 1:12

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