An elementary function with asymptotic $f'(x)\sim2f(2x)$ for $x\to0^+$

Solution 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.