Now cross-posted at MO. Let $g:[0,\infty] \to [0,\infty]$ be a smooth strictly increasing function satisfying $g(0)=0$ and $g^{(k)}(0)=0$ for every natural $k$.
Is $\sqrt g$ is infinitely (right) differentiable at $x=0$?
I know that $\sqrt g \in C^1$ at zero*, and that in complete generality, one cannot expect for $\sqrt g$ to be even $C^2$. However, in the counter-example given in the linked question, $g$ was not monotonic.
Does this additional assumption of (strict) monotonicity save us? I tried to look at the literature, but did not find a treatment of this particular case.
*The proof that $\sqrt g \in C^1$ goes by rewriting $g(x)=x^2h(x)$ where $h \ge 0$ is smooth (this is possible since $g(0)=g'(0)=0$).
Comment:
If we assume that $g''>0$ in a neighbourhood of zero (which implies that $g'>0$), then $\sqrt g \in C^2$. (details below).
I think that there is a chance for smoothness under the additional assumption that $g^{(k)}>0$ in a neighbourhood of zero for every $k$, but I am not sure. The calculations become quite messy even when trying to establish $\sqrt g \in C^3$.
A proof $\sqrt g \in C^2$ when $g',g''>0$ near zero: (We use these assumptions when applying L'Hôpital's rule).
$$\sqrt{g}'' = \frac{g''}{2\sqrt{g}} - \frac{(g')^2}{4g^{3/2}}.$$
Thus it is enough to prove that $(g'')^2/g\to 0$ and $(g')^4/g^3\to 0$.
$$ \lim_{x\to 0^+} \frac{(g'')^2}{g} = \lim_{x\to 0^+} 2\frac{g''g^{(3)}}{g'} = \lim_{x\to 0^+} 2\frac{g''g^{(4)}+(g^{(3)})^2}{g''} = 0, $$ where in the last equality we applied $\frac{(h')^2}{h}\to 0$ above for $h=g''$.
$$ \lim_{x\to 0^+} \frac{(g')^4}{g^3} = \lim_{x\to 0^+} \frac{4(g')^2g''}{3g^2} = \lim_{x\to 0^+} \frac{8(g'')^2 + 4g' g^{(3)}}{6g} = \lim_{x\to 0^+} \frac{2g' g^{(3)}}{3g} = \lim_{x\to 0^+} \left(\frac{2g^{(4)}}{3} + \frac{2g''g^{(3)}}{3g'}\right)=\lim_{x\to 0^+} \frac{2g''g^{(3)}}{3g'} = \lim_{x\to 0^+} \frac{2g^{(4)}}{3}+\frac{2(g^{(3)})^2}{3g''} = 0,$$
where in the first row we used the first calculation, and in the second we again applied $\frac{(h')^2}{h}\to 0$ to $h=g''$.
