Is the square root of a monotonic function whose all derivatives vanish smooth? 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 $g(0)=g'(0)=0$).
Edit:
As pointed out by Igor Rivin, it seems that theorem 2.2 (on page 639) here (pdf) does the job. It states that
 any square root of $f$ "precised up to order $m$" is of class $C^m$. (The definition of a "square root precised up to order $m$" is Definition 1.1 on page 636).
This certainly settles the issue. 
However, I think it would still be nice to find a simpler approach, since here we assume much more-the strict monotonicity is a much stronger assumption than those assumed in the paper.

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 $h=g''$.
 A: Extend the domain of the function $g$ to $\mathbb R$ by letting $g(x):=0$ for real $x<0$. The resulting function, which we shall still denote by $g$, is $C^\infty$ on $\mathbb R$. 
Theorem 3.5 on page 144 implies that a nonnegative $C^4$ function $f$ on $\mathbb R$ has a $C^2$ square root if for any minimum $x_0$ of $f$ we have $f(x_0)=0$. 
This latter condition obviously holds for our function $g$ in place of $f$ -- because $g$ is strictly increasing on $[0,\infty)$ and thus has no minima in $(0,\infty)$, and $g=0$ on $(-\infty,0]$. 
Therefore, we can conclude that $\sqrt g$ is $C^2$ on $\mathbb R$, even without assuming that $g''>0$ in a neighborhood of zero. 
Yet, this conclusion falls short of your main goal, to show that $\sqrt g$ is $C^\infty$. Looking at the proof of the mentioned Theorem 3.5, this task may be too big for a usual MO answer and may require a full-blown paper. 
A: The answer is yes, by the results of 
Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico, On square roots of class (C^m) of nonnegative functions of one variable, Ann. Sc. Norm. Super. Pisa, Cl. Sci. (5) 9, No. 3, 635-644 (2010). ZBL1207.26004.
Here is the math review:

Clearly the condition is fulfilled in the OP (for any $m$).
