Function dominated by convex function is eventually convex

Question

Suppose that we have a twice-differentiable function $f$ on $x\in [0,\infty)$ such that

1. $f(x)>0$ on $x\in [0,\infty)$ (i.e. strictly positive)
2. $f'(x)<0$ on $x\in [0,\infty)$ (i.e. strictly decreasing)
3. $f^{''}(x) >0$ on $x\in [0,\infty)$ (i.e. strictly convex)
4. $\lim_{x \to \infty} f(x)=0$

and we have a twice-differentiable function $g(x)$ such that

1. $f(x)>g(x)>0$ on $x\in [0,\infty)$ (i.e. strictly less then $f(x)$ and strictly positive)
2. $g'(x)<0$ on $x\in [0,\infty)$ (i.e. strictly decreasing)

Is it true that there exist some $x_0$ such that $g^{''}(x)>0$ for all $x \in [x_0,\infty)$?

Or in other words does a function dominated by convex function eventually becomes convex.

As an example of $f(x)$ consider $e^{-x}$ or $\frac{1}{1+x}$.

Edit Thanks to the example given @user225318. The above is not true. What if we add more more assumption that is

3) There exists $x_1$ such that $g^{'}(x) < f^{'}(x)$ for all $x \in [x_1, \infty)$. (i.e. derivative of $g(x)$ is eventually dominted by the derivative of $f(x)$.

Edit assumption 3) makes no sense

Answer 1

Consider the function $g(x) = e^{-x} (1/2 + \sin(x) / 3 )$. Clearly on $[0,\infty)$ we have $0 < g(x) < e^{-x} = f(x)$.

$$g'(x) = - e^{-x}(1/2 + \sin(x) / 3) + e^{-x} \cos(x) / 3 = - e^{-x} (1/2 + [\sin(x) - \cos(x)]/3) < 0 $$

here we use that the maximum of $|\sin(x) - \cos(x)|$ is $\sqrt{2}$ and $\sqrt{2} / 3 < 1/2$. But

$$ g''(x) = e^{-x} \left[ \frac12 - \frac{2\cos(x)}{3}\right] $$

is not signed. In particular, $g''(2k\pi) < 0$ while $g''((2k+1)\pi) > 0$.