# Problem of the Week # 239 - Oct 28, 2016

• MHB
• Ackbach
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Ackbach
Gold Member
MHB
Here is this week's POTW:

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Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that $f(a)\cdot f'(a) \cdot f''(a) \cdot f'''(a)\geq 0 .$

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Re: Problem Of The Week # 239 - Oct 28, 2016

This was Problem A-3 in the 1998 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his (of course) correct answer, which follows:

For any function $\phi$ with three continuous derivatives, define the function $P\phi$ by $(P\phi)(x) = \phi(x)\phi'(x)\phi''(x)\phi'''(x).$

If $f$ or any of its first three derivatives vanishes at some point $a$ then $Pf(a) = 0$ and the problem is solved. So we may assume that none of these functions is ever zero. Therefore, by the intermediate value theorem, none of the functions ever changes sign. So we may assume that each of them is either always positive or always negative.

If $f$ is always negative then the function $-f$ is always positive. Also, $P(-f)(x) = (-f(x))(-f'(x))(-f''(x))(-f'''(x)) = Pf(x).$ So if $P(-f)$ is positive then so is $Pf$. Therefore, replacing $f$ by $-f$ if necessary, we may assume that $f$ is always positive.

Let $g$ be the function given by $g(x) = f(-x).$ Then $g'(x) = -f'(-x)$, $g''(x) = f''(-x)$ and $g'''(x) = - f'''(-x).$ So $Pg(x) = Pf(-x)$, and if $f'$ is always negative then $g'$ is always positive. Therefore, replacing $f$ by $g$ if necessary, we may assume that $f'$ is always positive and hence $f$ is an increasing function.

Assuming that both $f$ and $f'$ are always positive, $f$ is bounded below (by $0$) and therefore has a greatest lower bound, $\ell$ say. Since $f$ is an increasing function, $\ell = \lim_{x\to -\infty}f(x)$. Thus there exists some $x_0<0$ such that $f(x_0) - \ell < f'(0).$ Also, $0<f(x_0-1)-\ell <f(x_0) - \ell < f'(0).$ It follows that $0 < f(x_0) - f(x_0-1) < f'(0).$ But by the mean value theorem $f(x_0) - f(x_0-1) = f'(x_1)$ for some point $x_1$ between $x_0-1$ and $x_0$. So $f'(x_1) < f'(0).$ By the mean value theorem again, there is some point $x_2$ between $x_1$ and $0$ for which $f''(x_2) >0.$ Therefore the function $f''$ is (always) positive.

But now the same argument from the previous paragraph can be applied substituting $f'$ in place of $f$: the functions $f'$ and $f''$ are always positive, $f'$ is bounded below and therefore has a greatest lower bound (which in fact is $0$). Using the previous argument, we can find a point $x_3<0$ such that $f'''(x_3) >0,$ and it follows that $f'''$ is (always) positive.

Thus all the functions $f$, $f'$, $f''$, $f'''$ are positive and hence so is $P(f)$.

## 1. What is the "Problem of the Week # 239 - Oct 28, 2016"?

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