$f$ be a smooth function on real line , $f(0)=0$ , $f(x)>0, \forall x \ne 0$ and any $f^{(n)}(0)=0$ ; is $\sqrt f$ smooth? Let $f: \mathbb R \to \mathbb R$ be an infinitely differentiable function such that $f(0)=0$ , $f(x)>0 , \forall x \ne 0$ and 
$f^{(n)}(0)=0$ ( the $n$-th derivative ) $,  \forall n \in \mathbb N$ , then is the function $g(x):=\sqrt{f(x)}$ infinitely 
differentiable ? I am having difficulty checking whether any $n$-th derivative exist at $0$ . Please help . Thanks in advance 
 A: EDIT. It turns out that my proof contains an error: $f'(x)$ may take zero infinitely often as $x \to 0$. Then the L'Hospital argument breaks down.
Also, the reference in @levap's comment claims that $g$ need not be smooth, citing
$$ f(t) = \begin{cases}
\mathrm{e}^{-1/|t|} (\sin^2 (\pi/|t|) + \mathrm{e}^{-1/t^2}), & t \neq 0 \\
0, & t = 0
\end{cases} $$
as a counter-example.
Heuristics of the above example. The original reference is French, which I barely know, that I was unable to check how they came up with this counter-example. But I guess that this function is designed so that the following asymptotics is true:
$$ g(t) = \sqrt{\smash[b]{f(t)}} \approx \mathrm{e}^{-1/2|t|} |\sin(\pi/|t|)|.$$
Of course, the wedges appearing in the graph of $\mathrm{e}^{-1/2|t|} |\sin(\pi/|t|)|$ is mollified by the extra term $\mathrm{e}^{-1/t^2}$ so that $g$ is smooth away from $t = 0$. But such contribution is so small, even compared to the main envelope term $\mathrm{e}^{-1/2|t|}$, that the effect is almost negligible as $t \to 0$. So $g'(t)$ has 'almost-jump' at each point where $\sin(\pi/t)$ vanishes:

(Green line: derivative of $\mathrm{e}^{-1/2|t|} |\sin(\pi/|t|)|$, Red line: $g'(t)$)
This causes $g''(t)$ to have a train of high picks near $t = \frac{1}{n}$ ($n = 1, 2, \cdots$) and thus prevents $g''(t)$ from being differentiable. Actual computation shows that $g''(1/n)$ grows super-exponentially as $n \to \infty$, confirming the heuristics.

OLD ANSWER. I checked that $g$ is twice differentiable and $g''(0) = 0$. I think my computation will generalize to prove the smoothness of $g$, but I have no clean idea how to proceed.
Here is an observation that simplifies the computation:

Observation. Let $h$ be a real-valued function defined on a neighborhood of $0$. Then
  $$\lim_{x\to0} h(x) = 0 \qquad \text{ if and only if } \qquad \lim_{x\to0} h(x)^2 = 0.$$

The proof is straightforward from the $\epsilon$-$\delta$ definition of the limit.
Step 1. $g'(0)$ exists and is equal to $0$.
Proof. Since $g(0) = 0$, it suffices to prove that $g(x)/x \to 0$ as $x \to 0$. By the lemma, we may instead prove that $g(x)^2/x^2 \to 0$ as $x \to 0$. But we know $g(x)^2 = f(x) = \mathcal{O}(x^3)$ from the Taylor's theorem, and hence the claim follows.
Step 2. $g''(0)$ exists and is equal to $0$.
Proof. As before, it suffices to prove that $g'(x)^2/x^2 \to 0$ as $x \to 0$. Using the fact that $f'(x)/x \to 0$ as $x \to 0$, we can apply the L'Hospital's theorem and
$$ \lim_{x\to0} \frac{g'(x)^2}{x^2}
= \lim_{x\to0} \frac{\left( \frac{f'\smash{(x)^2}}{4x^2} \right)}{f(x)}
\underset{\text{L'Hospital}}{=} \lim_{x\to0} \frac{\left( \frac{f'(x)( xf''(x) - f'(x))}{2x^3}\right)}{f'(x)}
= \lim_{x\to0} \frac{xf''(x) - f'(x)}{2x^3}
= 0. $$
A: I'm not sure about the general case yet, but the first derivative does exist. Put $g = \sqrt{f}$. By Taylor's theorem, for any $x > 0$ and positive $n$, there exists $\xi \in (0,x)$ such that
$$
f(x) = \sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}x^k + \frac{f^{(n)}(\xi)}{n!}x^n = \frac{f^{(n)}(\xi)}{n!}x^n.
$$
Since $f^{(n)}$ is bounded near $0$, $f(x) \leq C_nx^n$ for some constant $C_n> 0$. Take $n=4$, we have
$$
\lim_{x\to 0^+} \frac{g(x)}{x} \leq \sqrt{C_4} \lim_{x\to 0^+} \frac{x^2} {x} = 0.
$$
The case $x<0$ can be dealt with likewise. This proves that $g'(0) = 0$.
A: Yes, $g$ is smooth. We first prove the following proposition:
If $g$ is $k$ times differentiable, then the $k$ th derivative of $g^2$ is of the form $$ a_0 \ gg^{(k)} + a_1 \  g^{(1)}g^{(k-1)} + a_2 \  g^{(2)}g^{(k-2)} + \cdots   $$ where the $a_i$ are positive integers. 
Proof by induction: For $k=0$, the $k$ th derivative of $g^2$ is just 
$g^2 = 1 \ g g^{(0)}  $, so the result is true for $k=0$. Now suppose the result holds for $k$. Then the $k$ th derivative of $g^2$ is 
$$ a_0 \ gg^{(k)} + a_1 \  g^{(1)}g^{(k-1)} + a_2 \  g^{(2)}g^{(k-2)} + \cdots   $$ for some positive integers $a_i$ and so the $k+1$ th derivative of $g^2$ is $$ a_0 \ g g^{(k+1)} + a_0 \ g^{(1)} g^{(k)} + a_1 \ g^{(1)} g^{(k)} + a_1 \ g^{(2)} g^{(k-1)} + \cdots   $$ which completes the induction step after collecting terms, since the sum of positive integers are still positive integers. 
Now, $\sqrt{x}$ is a smooth function for $x>0$, and the composition of smooth functions is still smooth, and $f(x)>0$ for $x \neq 0$ so $g=\sqrt{f}$ is smooth for $ x \neq 0$. Hence the $k$ th derivative of $g$ exists for any $k$ at all points except $0$. We now prove that $g^{(k)}(x) \to 0$ as $ x \to 0$ $\forall$ $k$. The proof is by strong induction. $f(0)=0$ and $f$ is smooth, so continuous, so $f(x) \to 0$ as $x \to 0$. But $g^2 (x) = f(x) \to 0$ as $x \to 0$, so $g(x) \to 0$ as $x \to 0$, so the result is true for $k=0$. Now suppose it is true for $0$,$1$,$2$,$3$,$\cdots$, up to $k-1$, and look at the $2k$ th derivative of $g^2=f$, using the proposition:
$$ a_0 \ gg^{(2k)} (x) + a_1 \  g^{(1)}g^{(2k-1)} (x) + a_2 \  g^{(2)}g^{(2k-2)} (x) + \cdots + a_{k-1} \ g^{(k-1)} g^{(k+1)} (x) +  a_k \ g^{(k)} g^{(k)} (x) = f^{(k)}(x)   $$ If we let $ x \to 0$, all terms except the last one on the LHS vanish, and $a_k$ is a positive integer, so $g^{(k)} (x) \to 0 $ as $x \to 0$ as well, completing the induction step. 
Finally, we prove that $g$ is infinitely differentiable at $0$, and hence smooth. In fact, we prove that $g^{(k)} (0) = 0 $ $\forall$ $k$. The proof is once again by induction. $g(0) = \sqrt{f(0)} = \sqrt{0}= 0$, so the result is true for $k=0$. Suppose the result is true for $k$. Then
$$ \frac{g^{(k)}(x)-g^{(k)}(0)}{x-0} = \frac{g^{(k)}(x)}{x} $$ and since, from before, $g^{(k+1)} (x) \to 0 $ as $x \to 0$, by L'Hopital's rule, 
$$ \lim _{ x \to 0} \frac{g^{(k)}(x)}{x} = \lim _{ x \to 0} \frac{g^{(k+1)}(x)}{1} = 0  $$ and so $$ g^{(k+1)} (0) = \lim _{x \to 0} \frac{g^{(k)}(x)-g^{(k)}(0)}{x-0} = 0$$ completing the induction step. 
Edit: Upon re-reading my answer, I have noticed a flaw - one cannot conclude that all the terms vanish just because the smaller order derivatives tend to zero, because the higher order derivatives they are multiplied by may not be bounded and tend to infinity. The argument would still hold if one could show all the derivatives of $g$ are bounded near zero, but I cannot come up with a way to show that at the moment. 
