International Mathematics Competition
for University Students
2016

IMC2016: Day 1, Problem 1

1. Let $f\colon [a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $f$ has infinitely many zeros, but there is no $x\in(a,b)$ with $f(x)=f'(x)=0$.

(a) Prove that $f(a)f(b)=0$.

(b) Give an example of such a function on $[0,1]$.

Proposed by Alexandr Bolbot, Novosibirsk State University

Solution. (a) Choose a convergent sequence $z_n$ of zeros and let $c=\lim z_n\in [a,b]$. By the continuity of $f$ we obtain $f(c)=0$. We want to show that either $c=a$ or $c=b$, so $f(a)=0$ or $f(b)=0$; then the statement follows.

If $c$ was an interior point then we would have $f(c)=0$ and $f'(c)=\lim \dfrac{f(z_n)-f(c)}{z_n-c}=\lim \dfrac{0-0}{z_n-c}=0$ simultaneously, contradicting the conditions. Hence, $c=a$ or $c=b$.

(b) Let $$ f(x)=\begin{cases}x\sin\dfrac1x & \text{if $0\lt x\leq 1$}\\ 0 & \text{if $x=0$.}\end{cases} $$ This function has zeros at the points $\dfrac1{k\pi}$ for $k=1,2,\ldots$, and it is continuous at $0$ as well.

In $(0,1)$ we have $$ f'(x) = \sin \dfrac1x - \frac1x\cos\frac1x. $$ Since $\sin\tfrac1x$ and $\cos\tfrac1x$ cannot vanish at the same point, we have either $f(x)\ne0$ or $f'(x)\ne0$ everywhere in $(0,1)$.