Get ready Part 29

The solution

Proof by contradiction :

Suppose that there is $f\in C^3(\mathbb{R})$ such that $\forall x\in\mathbb{R}, \ f(x)f'(x)f''(x)f'''(x)<0$

Therefore $f'''$ is either $>0$ or $<0$ (if not by continuity, $f'''$ has a zero). Without loss of generality $(f\leftarrow -f)$ , we can suppose $f'''>0$ . Then because $f''$ is either $>0$ or $<0$ , if we let $a=\lim\limits_{-\infty}f''$ and $b=\lim\limits_{+\infty}f''$ (which exist because $f''$ is increasing) we have :

• if $f''>0$ $a\geq 0 \ \text{and} \ +\infty\geq b>a$
• if $f''<0$ $-\infty \leq a < b \ \text{and} \ b\leq 0$

The first case : $f''>0$

Because $b>0$ , necessarily $f'>0$ . Indeed, $f'(x)=\int_0^x f''(t)dt +C \underset{x\to+\infty}{\longrightarrow} +\infty$

The same argument shows that $f>0$ , therefore $ff'f''f'''>0$ absurd.

Second case : $f''<0$

Because $a<0$ , necessarily $f'>0$ . Indeed, $f'(x)=\int_0^x f''(t)dt +C' \underset{x\to-\infty}{\longrightarrow} +\infty$ Then, necessarily $f<0$ because $f(x)=\int_0^x f'(t)dt + C'' \underset{x\to-\infty}{\longrightarrow} -\infty$ And at the end $ff'f''f'''>0$ , absurd.

Let f(x) = x^3, f'(x) = 3x^2, f''(x) = 6x, f'''(x) = 6. Then f(x) f'(x) f''(x)*f'''(x) = 108x^6. Let x be any real number.

Not always. Take, for example $f(x)=\exp(-x), ~~ x \in \mathbb{R}.$ Then $f(a)f'(a)f''(a)f'''(a)= -\exp(-4a) < 0, \forall a \in \mathbb{R}.$

Let $f(x) = e^x$ , then for all real $a$ , $f(a).f'(a).f''(a).f'''(a) > 0$ .

$f(x) = 0$ satisfies this everywhere.

Moreover, if $f(x) = g(x)$ , such that $g'(a)g''(a)g'''(a) > or < 0$ , you can choose $f(x) = g(x) - g(a) \pm 1$ to make the equation positive.

e.g. $f(a)=e^a$ .

For $a \geq 0$ , $f(a).f'(a).f"(a).f'"(a)=(e^a)(e^a)(e^a)(e^a) \geq 0$ .

For $-a < 0$ , $f(a).f'(a).f"(a).f'"(a)=(e^{-a})(-e^{-a})(e^{-a})(-e^{-a}) \geq 0$ .