Fall 2013 - Problem 12

Solution.

1. Let $\func{f}{\br{0,1}}{\R}$ be given by $f\p{x} = x\sin\p{\frac{1}{x}}$ when $x \neq 0$ and $f\p{0} = 0$. Observe that if $\epsilon > 0$, then on $\br{\epsilon, 1}$,

   $$f'\p{x} = \sin\p{\frac{1}{x}} - \frac{1}{x} \cos\p{\frac{1}{x}} \implies \abs{f'\p{x}} \leq 1 + \frac{1}{\epsilon},$$

   so $f$ is Lipschitz on $\br{\epsilon, 1}$, hence absolutely continuous. To show that $f$ is not absolutely continuous on $\br{0, 1}$, let

   $$a_n = \frac{1}{\frac{\pi}{2} + 2\pi n} \quad\text{and}\quad b_n = \frac{1}{-\frac{\pi}{2} + 2\pi n},$$

   i.e., so $a_n < b_n$ and these are essentially points of local maximum and minimum, respectively. Observe that

   $$\begin{gathered} \abs{b_n - a_n} = \frac{\pi}{4\pi^2 n^2 - \frac{\pi^2}{4}} \leq \frac{1}{n^2} \\ \abs{f\p{b_n} - f\p{a_n}} = a_n + b_n = \frac{4\pi n}{4\pi n^2 - \frac{\pi^2}{4}}. \end{gathered}$$

   Thus, for any $\delta > 0$, we can pick $N \in \N$ large enough so that

   $$\sum_{n=N}^\infty \abs{b_n - a_n} < \delta.$$

   But

   $$\sum_{n=N}^\infty \abs{f\p{b_n} - f\p{a_n}} = \infty,$$

   so we can pick $M > N$ so large that

   $$\sum_{n=N}^M \abs{b_n - a_n} < \delta, \quad\text{but}\quad \sum_{n=N}^M \abs{f\p{b_n} - f\p{a_n}} \geq 1,$$

   so $f$ cannot be absolutely continuous.

2. We denote the variation of $f$ on $\br{a, b}$ via

   $$V_f\p{\br{a,b}} = \sup_{a = t_1 < \cdots < t_n = b; n \in \N} \sum_{i=1}^{n-1} \abs{f\p{t_{i+1} - f\p{t_i}}}.$$

   First, we will show that $x \mapsto V_f\p{\br{0,x}}$ is continuous. Let $\epsilon > 0$ and $\set{0 = t_0 < t_1 < \cdots < t_n = 1}$ be a partition such that

   $$\sum_{i=1}^{n-1} \abs{f\p{t_{i+1} - f\p{t_i}}} \geq V_f\p{\br{0,1}} - \epsilon.$$

   Since $f$ is continuous at $0$, we can pick $\delta \in \p{t_0, t_1}$ small enough so that $\abs{f\p{\delta} - f\p{0}} < \epsilon$. On the other hand, $\delta < t_1 < \cdots < t_n = 1$ is a partition of $\br{\delta, 1}$, and so

   $$V_f\p{\br{0,1}} - \epsilon \leq \abs{f\p{\delta} - f\p{0}} + \abs{f\p{t_1} - f\p{\delta}} + \sum_{i=2}^{n-1} \abs{f\p{t_{i+1} - f\p{t_i}}} \leq \epsilon + V_f\p{\br{\delta, 1}}.$$

   Rearranging, we see $V_f\p{\br{0, \delta}} = V_f\p{\br{0, 1}} - V_f\p{\br{\delta, 1}} < 2\epsilon$, which proves the claim.

   To show that $f$ is absolutely continuous on $\br{0, 1}$, let $\epsilon > 0$ and let $h > 0$ be such that $V_f\p{\br{0, h}} \leq \epsilon$. Since $f$ is absolutely continuous on $\br{0, 1}$, let $\delta > 0$ come from the definition of absolute continuity.

   Suppose $a_1 < b_1 \leq a_2 < b_2 \leq \cdots \leq a_n < b_n$ is such that $\sum_{i=1}^n \abs{b_i - a_i} < \delta$. Then shrinking $h$ if necessary, we may assume without loss of generality that $h \in \br{a_1, b_1}$ and so

   $$\begin{aligned} \sum_{i=1}^n \abs{f\p{b_i} - f\p{a_i}} &\leq \abs{f\p{a_1} - f\p{0}} + \abs{f\p{h} - f\p{a_1}} + \abs{f\p{b_1} - f\p{h}} + \sum_{i=2}^n \abs{f\p{b_i} - f\p{a_i}} \\ &\leq V_f\p{\br{0, h}} + \abs{f\p{b_1} - f\p{h}} + \sum_{i=2}^n \abs{f\p{b_i} - f\p{a_i}} \\ &\leq 2\epsilon. \end{aligned}$$

   Indeed, $\set{0, a_1, h}$ gives a partition of $\br{0, h}$, and $\set{h, b_1, a_2, b_2, \ldots, a_n, b_n}$ gives a disjoint collection of intervals in $\br{h, 1}$ with total length less than $\delta$, and this completes the proof.