Infinitely differentiable function $f$ from $\mathbb{R}_{\geq 0}$ onto $\mathbb{R}_{\geq 0}$ where $f^{(k)}(0)=0$ for all $k \geq 0$

Question

I'm trying to a find an onto function $f:\mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ such that the $f^{(k)}(0)=0$ for all $k \geq 0$ I'm familiar with functions such as $e^{-1/x^2}$ but the main problem I'm having is finding one that is onto. I suspect it needs to be some kind of infinite series of functions (to ensure the derivative is always zero at zero for all derivatives), but I'm truly lost on how to proceed. Most attempts I make eventually end up with a $k$th derivative that's non-zero at zero (such as $e^x-1$) or isn't onto (such as $e^{-1/x^2}$) and I really have no idea how to solve this issue. Any help or hints would be sincerely appreciated. Thanks.

Answer 1

Consider

$$f(x) =\frac{e^{-1/x}}{1-e^{-1/x}}$$ for $x>0$, and $0$ for $x\le 0$.

The idea is that the function $e^{-1/x}$ maps $[0, \infty)$ bijectively onto $[0,1)$, and $\frac{t}{1-t}$ maps $[0, 1)$ onto $[0, \infty)$.

Answer 2

Here's an idea based on smooth transition functions. Take $g(x)$ to be the function defined in the accepted answer to this question:

How to build a smooth "transition function" explicitly?

Now define a piecewise function $t(x): \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ by:

$$t(x) = g(x - n) + n \text{ on the interval } [n, n + 1), n \in \mathbb{N}$$

Since $g(x)$ is smooth, $g^{(0)}(0) = 0$, and $g^{(k)}(0) = g^{(k)}(1) = 0$ for $k > 0$, $t(x)$ is smooth and $t^{(k)}(0) = 0$ for all $k \geq 0$. Clearly, $t(x)$ is surjective.