Does there exist a smooth function $g:[-1,1]\to[0,1]$ such that it is flat at $x=0$ and $g^{(i)}(x)\leq M g(x)$ for some $M?$

Question

Question: Fix $p\in\mathbb{N}_{\geq 1}.$ Does there exist a smooth function $g:[-1,1]\to[0,1]$ such that $g$ is flat at $x=0$ (meaning $g^{(i)}(0) = 0$ for all natural numbers $i$), $g(x) = 0$ if and only if $x = 0$ and $g^{(i)}(x)\leq M g(x)$ for some $M>0,$ all $x\in [-1,1]$ and all $i\leq p?$

I tried the classic smooth flat function $g(x) = e^{-\frac{1}{x^2}}$ if $x\neq 0$ and $g(0) = 0.$ It satisfies all properties above but fails $\lim_{x\to 0} \frac{g'(x)}{g(x)} = \lim_{x\to 0} \frac{1}{x^3} = \infty.$ So the $M>0$ does not exist.

This post contains some examples of flat functions at $x=0.$ But they do not answer my question.

Maybe some function in Schwartz space and flat at $x=0$ satisfies all properties above.