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

thinkthat 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 $h=g''$.