# 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 $$h=g''$$.

• Just a comment: if all derivatives of $g$ are nonegative in a common neighborhood of zero, then $g$ can be expanded in a power series, by Bernstein theorem, and would be equal to 0. However, the question changes if the neighborhhod is allowed to depend on the derivative. May 10, 2020 at 7:56

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 $$m$$).

• The DOI link doesn't work for me; the journal however released the back issues on numdam for free, so here's a direct link numdam.org/article/ASNSP_2010_5_9_3_635_0.pdf May 13, 2020 at 3:46
• @WillieWong Thanks! May 13, 2020 at 4:46
• Thank you. I see that theorem 2.2 there does the job. It says "that any square root of $f$ precised up to order $m$ is of class $C^m$". (The relevant definition of a "square root precised up to order $m$" is Definition 1.1 on page 636). May 13, 2020 at 5:22
• Link to Numdam abstract, rather than straight to PDF: Bony, Colombini, and Pernazza - On square roots of class $C^m$ of nonnegative functions of one variable. (The DOI link works for me; just to have it here.) Apr 21, 2021 at 19:03

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.

• "... may require a full-blown paper." Evidently Bony and collaborators thought the same :-) May 13, 2020 at 3:43
• Thank you. Actually I see now that theorem 3.1 on page 142 states exactly your corollary. May 13, 2020 at 5:08