IMC 1998 · Day 2 · P12

a) We first construct a sequence $c_n$ of positive numbers such that $c_n \to \infty$ and $\sum\limits_{n=1}^{\infty} c_n b_n < \frac{1}{2}$. Let $B = \sum\limits_{n=1}^{\infty} b_n$, and for each $k = 0, 1, \dots$ denote by $N_k$ the first positive integer for which $$\sum_{n=N_k}^{\infty} b_n \le \frac{B}{4^k}.$$ Now set $c_n = \dfrac{2^k}{5B}$ for each $n$, $N_k \le n < N_{k+1}$. Then we have $c_n \to \infty$ and $$\sum_{n=1}^{\infty} c_n b_n = \sum_{k=0}^{\infty} \sum_{N_k \le n < N_{k+1}} c_n b_n \le \sum_{k=0}^{\infty} \frac{2^k}{5B} \sum_{n=N_k}^{\infty} b_n \le \sum_{k=0}^{\infty} \frac{2^k}{5B} \cdot \frac{B}{4^k} = \frac{2}{5}.$$ Consider the intervals $I_n = (a_n - c_n b_n,\ a_n + c_n b_n)$. The sum of their lengths is $2 \sum c_n b_n < 1$, thus there exists a point $x_0 \in (0,1)$ which is not contained in any $I_n$. We show that $f$ is differentiable at $x_0$, and $f'(x_0) = 0$. Since $x_0$ is outside of the intervals $I_n$, $x_0 \ne a_n$ for any $n$ and $f(x_0) = 0$. For arbitrary $x \in (0,1) \setminus \{x_0\}$, if $x = a_n$ for some $n$, then $$\left| \frac{f(x) - f(x_0)}{x - x_0} \right| = \frac{f(a_n) - 0}{|a_n - x_0|} \le \frac{b_n}{c_n b_n} = \frac{1}{c_n},$$ otherwise $\left| \frac{f(x) - f(x_0)}{x - x_0} \right| = 0$. Since $c_n \to \infty$, this implies that for arbitrary $\varepsilon > 0$ there are only finitely many $x \in (0,1) \setminus \{x_0\}$ for which $$\left| \frac{f(x) - f(x_0)}{x - x_0} \right| < \varepsilon$$ does not hold, and we are done.

Remark. The variation of $f$ is finite, which implies that $f$ is differentiable almost everywhere.

b) We remove the zero elements from sequence $b_n$. Since $f(x) = 0$ except for a countable subset of $(0,1)$, if $f$ is differentiable at some point $x_0$, then $f(x_0)$ and $f'(x_0)$ must be 0.

It is easy to construct a sequence $\beta_n$ satisfying $0 < \beta_n \le b_n$, $\beta_n \to 0$ and $\sum\limits_{n=1}^{\infty} \beta_n = \infty$.

Choose the numbers $a_1, a_2, \dots$ such that the intervals $I_n = (a_n - \beta_n,\ a_n + \beta_n)$ ($n = 1, 2, \dots$) cover each point of $(0,1)$ infinitely many times (it is possible since the sum of lengths is $2 \sum \beta_n = \infty$). Then for arbitrary $x_0 \in (0,1)$, $f(x_0) = 0$ and $\varepsilon > 0$ there is an $n$ for which $\beta_n < \varepsilon$ and $x_0 \in I_n$ which implies $$\frac{|f(a_n) - f(x_0)|}{|a_n - x_0|} > \frac{b_n}{\beta_n} \ge 1.$$