Differentiability properties of $\psi(x)\cos(\phi(x)),\,\, \psi(x)\sin(\phi(x))$ at $x=0$

Question

Let $\psi:[0,\infty) \to [0,\infty)$ be a smooth strictly increasing function satisfying $\psi(0)=0$, $\psi'(0)>0$ and let $\phi:(0,\infty) \to \mathbb R$ be smooth.

Suppose that $\lim_{x \to 0^+}\phi'(x)\psi(x)=0$. Define $$f_1(x)=\psi(x)\cos(\phi(x)), \, \,\,\,f_2(x)=\psi(x)\sin(\phi(x))$$ on $(0,\infty)$, and extend them continuously to zero by setting $f_i(x)=0$.

Can the following properties hold simultaneously?

1. $\,f_i$ are infinitely (right) differentiable at $x=0$.

2. $\,$All the (right) derivatives of $f_i$ of even order vanish at zero.

3. At least one of the $\,f_i'(0)$ is non-zero.

Comment:

The assumptions imply that $\alpha:=\lim_{x \to 0^+} \phi(x)$ exists. Indeed,

$$ \frac{f_1(x)-f_1(0)}{x}=\frac{\psi(x)-\psi(0)}{x}\cos(\phi(x))\Rightarrow \\ \cos(\phi(x))=\frac{f_1(x)-f_1(0)}{x} \frac{1}{\frac{\psi(x)-\psi(0)}{x}} \Rightarrow \\ \lim_{x \to 0^+} \cos(\phi(x))=\frac{f_1'(0)}{\psi'(0)}, $$

and similarly $\lim_{x \to 0^+} \sin(\phi(x))=\frac{f_2'(0)}{\psi'(0)}$. So, both $\lim_{x \to 0^+} \cos(\phi(x)), \lim_{x \to 0^+} \sin(\phi(x))$ exist, and hence so does $\lim_{x \to 0^+} \phi(x)$.

Now, a direct calculation shows that $$ f_1'(x)=\begin{cases} \psi'(x)\cos(\phi(x))-\psi(x)\phi'(x)\sin(\phi(x)) & \text{if $x > 0$} \\ \psi'(0)\cdot \cos(\alpha) & \text{if $x=0$}\end{cases}$$

Now, I am not sure how to proceed from here. for $x>0$, we have $$ f_1''(x)=\psi''(x)\cos(\phi(x))-2\psi'(x)\phi'(x)\sin(\phi(x))-\psi(x)\phi''(x)\sin(\phi(x))-\psi(x)(\phi'(x))^2\cos(\phi(x)),$$

but since we don't know whether $\phi'(x),\phi''(x)$ have limits when $x \to 0$, it is not clear to me what to do next.

Answer 1

Yes, these properties can hold simultaneously.

For example, let $$\psi(x) = x, \quad \phi(x) = 0$$ These are both smooth on $\mathbb{R}$, so it suffices to drop any right-only or range-restriction considerations. For the conditions on $\psi$, it is clearly strictly increasing, and satisfies $\psi(0) = 0$, $\psi'(0) = 1 > 0$. Moreover, $\phi$ is smooth, and $\phi' \psi = 0$, so certainly $\lim_{x \to 0^+} \phi'(x) \psi(x) = 0$. Hence these satisfy your requirements on $\psi$ and $\phi$.

It follows that $$f_1(x) = x, \quad f_2(x) = 0$$ These are smooth, and all derivatives of order $n \geq 2$ (and hence all even ones) are identically zero. Lastly, we have that $f_1'(x) = 1$, so $f_1'(1) = 1 \neq 0$.