Composition by the Cube Root is Smooth

[caffeinemachine]

Let $f:\mathbf R\to \mathbf R$ be a smooth map and $g:\mathbf R\to \mathbf R$ be defined as $g(x)=f(x^{1/3})$ for all $x\in \mathbf R$.

Problem. Then $g$ is smooth if and only if $f^{(n)}(0)$ is $0$ whenever $n$ is not an integral multiple of $3$.

One direction is easy. Assume $g$ is smooth. Then we have $f(x)=g(x^3)$ for all $x$. Differentiating and using the chain rule gives that the required derivatives of $f$ vanish.

I am struggling with the converse. Assume $f^{(n)}(0)=0$ whenever $n$ is not an integral multiple of $3$. We need to show that $g$ is smooth. Since $x^{1/3}$ is smooth at all points except $x=0$, we see that $g$ too is so. So the only problem is at $0$. I am only able to show that the first derivative of $g$ at $0$ exists. Here is what I have done.

Let $x>0$ be arbitrary. By Taylor we know that
$$f(x)= f(0)+f'(0)x+ \frac{f''(0)}{2}x^2+ \frac{f''(\lambda_x x)}{6}x^3$$
for some $0<\lambda_x<1$.

This gives by hypothesis that
$$f(x) - f(0) = \frac{f''(\lambda_x x)}{6}x^3$$
Thus
$$g(x)-g(0)=\frac{f'''(\lambda_{x^{1/3}} x^{1/3})}{6} x$$
and therefore
$$\frac{g(x)-g(0)}{x}=\frac{f'''(\lambda_{x^{1/3}} x^{1/3})}{6}$$
Since $\lim_{x\to 0}f'''(\lambda_{x^{1/3}} x^{1/3}))$ exists, we see that $g'(0)$ exists.

But I am not able to extend this argument to show that $g''(0)$ etc. also exist.

 

[jvicens]

To show that $g''(0)$ exists, we can use the same argument as before. We know that $f''(x)$ exists for all $x\in \mathbf{R}$, so we can write
$$f(x)=f(0)+f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \frac{f^{(4)}(\lambda_x x)}{24}x^4$$
for some $0<\lambda_x<1$. Plugging this into the definition of $g(x)$, we get
$$g(x) = g(0) + \frac{f''(0)}{2}x + \frac{f'''(0)}{6}x^2 + \frac{f^{(4)}(\lambda_x x)}{24}x^3$$
Using the fact that $f^{(n)}(0)=0$ for all $n$ not divisible by $3$, we can see that the first three terms are equal to $0$ at $x=0$. Then, using the fact that $\lim_{x\to 0}f^{(4)}(\lambda_x x)$ exists, we can conclude that $g''(0)$ exists.

We can continue this process for higher derivatives as well. For example, to show that $g^{(3)}(0)$ exists, we can use the same argument as before, but with $n=6$ in the Taylor expansion of $f(x)$. This will give us
$$g(x) = g(0) + \frac{f^{(6)}(\lambda_x x)}{120}x^5$$
Using the fact that $f^{(n)}(0)=0$ for all $n$ not divisible by $3$, we can see that the first term is equal to $0$ at $x=0$. Then, using the fact that $\lim_{x\to 0}f^{(6)}(\lambda_x x)$ exists, we can conclude that $g^{(3)}(0)$ exists.

We can continue this process for all higher derivatives, using the fact that $f^{(n)}(0)=0$ for all $n$ not divisible by $3$ and the existence of