# Function dominated by convex function is eventually convex

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

• I'm trying hard to even picture an $f(x)$ that even matches your original definition in terms of elementary functions. Nothing is coming to me. It seems that if the second derivative is positive, then the first derivative will continuously increase - so it will be tough for it to remain negative. Same with the fourth condition. If you have an example, I think that would help.
– Mark
Jun 23, 2015 at 19:07
• What about $\frac{1}{1+x}$ or $e^{-x}$
– Boby
Jun 23, 2015 at 19:08
• About your condition (3), note that $g'(x) < f'(x) < 0$ for all $x\in [x_1,\infty)$ together with $\lim_{x\to\infty} f(x) = 0$ and $g(x) > 0$ implies that $g(x) > f(x)$ on $[x_1,\infty)$, which contradicts your other assumptions. Do you mean that $0 < |g'(x)| < |f'(x)|$? In that case see my comment on my answer below. Jun 23, 2015 at 19:53
• @user225318 but since $f(x)>g(x)$ then $g(x) \to 0$
– Boby
Jun 23, 2015 at 19:57
• $$g(x) = g(x) - g(\infty) = - \int_x^\infty g'(x) \mathrm{d}x > - \int_x^\infty f'(x) \mathrm{d}x = f(x) - f(\infty) = f(x)$$ by the fundamental theorem of calculus. You cannot have $f(x) > g(x)$ and $g(x)$ decaying faster than $f(x)$ always. Jun 23, 2015 at 20:00

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 $g''((2k+1)\pi) > 0$.

• thanks. really cool example. What if I put one more assumption. That the derivative of $g(x)$ is also dominated by the derivative of $f(x)$? That is for some $x_1$, $g'(x)<f'(x)$ for all $x\in [x_1,\infty)$.
– Boby
Jun 23, 2015 at 19:40
• In my example above, change $f(x)$ to $f(x) = 10 e^{-x}$. Now $g'(x)$ is dominated by $f'(x)$. Jun 23, 2015 at 19:48
• oh wait, you have $f'(x) < 0$. Do you mean dominated as in $|g'(x)| < |f'(x)|$ or exactly as you wrote $g'(x) < f'(x)$? Jun 23, 2015 at 19:49
• I want $g(x)$ do deacrease faster then $f(x)$. So, $g'(x)<f'(x)$, right?
– Boby
Jun 23, 2015 at 19:51
• That's impossible; it contradicts $0 < g(x) < f(x)$ (by the fundamental theorem of calculus). See my comment on your question. Jun 23, 2015 at 19:54