# $f : \mathbb{R}^+ → \mathbb{R}$ with $f(0) = f'(0) = 0$ and $f(x) < x^2$ and $f',f'',f''' > 0$?

I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.

Here is the statement:

There is no function $$f(x)$$ on $$x \ge 0$$ such that $$f(0)=0$$, $$f'(0)=0$$, $$f(x) for $$x>0$$ and that the first, second and third derivative of $$f(x)$$ are strictly positive on $$x>0$$.

I appreciate any help!

• $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,\infty]$. Jan 17, 2019 at 18:14
• @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > \varepsilon$ for some constant $\varepsilon > 0$. Jan 17, 2019 at 18:36

I do not think it is true. Take for example

• $$f(x) = 1 - x + \frac12x^2 - e^{-x}$$
• $$f'(x) = - 1 + x + e^{-x}$$
• $$f''(x) = 1 - e^{-x}$$
• $$f'''(x) = e^{-x}$$

Here is a graph: note that the black line for $$x^2$$ is above the dark blue line for $$f(x)$$, while the green line for $$f''(x)$$ stays below $$1$$ while the second derivative of $$x^2$$ is $$2$$, and the red line for $$f'''(x)$$ tends towards $$0$$ from above as $$x$$ increases • Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < \frac{1}{2} x^2$ for $x > 0$. Jan 17, 2019 at 19:06
• @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$ Jan 17, 2019 at 19:15
• Many thanks for this counter example which together with other very nice hints makes the issue clear to me! Jan 17, 2019 at 20:44
• How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
– D777
Jan 19, 2019 at 14:02
• @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x \gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$. Jan 19, 2019 at 16:55

Your intuition is correct for a slightly different statement:

There is no function $$f(x)$$ on $$x \ge 0$$ such that $$f(0)=0$$, $$f'(0)=0$$, $$f(x) for $$x>0$$ and that the third derivative of $$f(x)$$ is bounded below by some $$\epsilon > 0$$ on $$x>0$$.

The argument is as follows: Suppose that such a function exists. Then we have $$f'''(x) \geq \epsilon$$, which means that (integrating both sides) $$f''(x) - f''(0) \geq \epsilon x.$$ Denote $$f''(0) = a$$, so that $$f''(x) \geq \epsilon x + a$$. Integrating two more times, we have $$f'(x) \geq \frac{1}{2} \epsilon x^2 + a x + f'(0) = \frac{1}{2} \epsilon x^2 + a x$$ $$f(x) \geq \frac{1}{6} \epsilon x^3 + a x^2 + f(0) = \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2$$ But if $$f(x) < x^2$$, we have $$x^2 > f(x) \geq \frac{1}{6} \epsilon x^3 + \frac{1}{2} a x^2$$ for all $$x > 0$$, which reduces to $$\frac{6(1 - a/2)}{\epsilon} > x.$$ for all $$x > 0$$. For any value of $$\epsilon > 0$$ and $$a \in \mathbb{R}$$, this will be violated by some $$x > 0$$, and we have a contradiction.

Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $$\frac{1}{6} x^3$$ times that value. But any function that grows as fast as $$x^3$$ (times any positive number) must have a greater value than $$x^2$$ at some point.

• You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1 Jan 17, 2019 at 20:46

As in the first answer and comment by Daniel, we can choose any $$f'''$$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $$0.$$

$$f'''(x) = \frac{1}{1+x^2}$$ $$f''(x) = \arctan x$$ $$f'(x) = x \arctan x - \frac{1}{2} \log \left(1+x^2 \right)$$ $$f(x) = \left( \frac{x^2 -1}{2} \right) \arctan x - \frac{x}{2} \log \left(1+x^2 \right) + \frac{x}{2}$$

As $$f'' < \frac{\pi}{2},$$ we get $$f'(x) = \int_0^x f''(t) dt < \int_0^x \frac{\pi }{2} dt = \frac{\pi x}{2}$$ $$f(x) = \int_0^x f'(t) dt < \int_0^x \frac{\pi t}{2} dt = \frac{\pi x^2}{4}$$

• I will check the functional forms soon, this looks quite interesting! +1 Jan 17, 2019 at 20:45

Just a comment: R+ is the set of STRICTLY POSITIVE reals, so I think R+ in opening statement needs to be replaced by the interval [0, infinity).

• Many thanks for the comment. To change the title or the text you may of course feel free to use the "edit" option to suggest a change in the text/title :-) Jan 25, 2019 at 8:21