I would appreciate any help with my HW exercise:

Prove that $f(x) = x^\alpha \cdot \sin(1/x)$ is absolutely continuous on $(0,1)$, when $\alpha>1$.

It's easy to find the derivative of $f$:

$$f'(x) = \alpha x^{\alpha-1} \sin(1/x) - x^{\alpha-2} \cos(1/x).$$

So, when $1<\alpha<2$, the function is not Lipschitz, and that is the main problem.

I searched for similar questions and found this: Examples of absolutely continuous functions that are not Lipschitz. and this: http://mathdl.maa.org/images/cms_upload/0002989049585.di021349.02p00072.pdf

But I couldn't understand the PDF file, which merely handles the case $\alpha=3/2$.



2 Answers 2


I assume that $1<\alpha<2$. If not, as remarked by the questioner, $f(x)$ is Lipschitz on $(0,1)$ and the problem is simpler.

You can use the following approach. Take a small $\delta>0$. You wish to bound $V=\sum_i |f(x_i)-f(y_i)|$ whenever $S=\{(x_1,y_1),\dots,(x_n,y_n)\}$ is a finite set of pairwise disjoint intervals contained in $(0,1)$ with $\sum_i |x_i-y_i|<\delta$.

Let $N$ be the smallest integer such that $N>(2\pi \delta)^{-1}$, and split the interval into two pieces, $(0,1/(2\pi N))$ and $(1/(2\pi N),1)$. By splitting an interval in $S$ if necessary, which does not decrease $V$, you can assume that each member of $S$ is in either $(0,1/(2\pi N))$ or $(1/(2\pi N),1)$.

Let $V_1$ be the portion of $V$ coming from the intervals in $(0, 1/(2\pi N))$. Here, the function $f(x)$ has alternating local maxima and minima. The maxima are close to the values $x=2/((4n+1)\pi)$; let $M_n$ be the value of $x$ close to $x=2/((4n+1)\pi)$ where $f(x)$ has a local maximum. Similarly, the minima are close to the values $x=2/((4n+3)\pi)$; let $m_n$ be the value of $x$ close to $x=2/((4n+3)\pi)$ where $f(x)$ has a local minimum.
Argue that $$V_1\le |f(M_N)|+|f(M_N)-f(m_N)|+|f(m_N)-f(M_{N+1})|+\cdots\ \ \ (1)$$ and that $$ f(M_n)=f(m_n)=O(n^{-\alpha}).\qquad (2) $$ Then, combining $(1)$ and $(2)$, conclude that $V_1=O(N^{1-\alpha})$.

Let $V_2$ be the portion of $V$ coming from the intervals in $(1/(2\pi N),1)$. In this portion of $(0,1)$, $|f'(x)|$ is bounded above, so the function $f(x)$ is Lipschitz. Argue that the Lipschitz constant is $O(\delta^{\alpha-2})$. Therefore, $V_2$ is $O(\delta^{\alpha-1})$.

Since $V=V_1+V_2$, adding the above estimates together should prove that $V\to 0$ as $\delta\to 0$.

  • $\begingroup$ Isn't this uniform continuity that you proved, and not absolute continuity? $\endgroup$
    – Julien
    Feb 22, 2013 at 3:20
  • $\begingroup$ I misread the question. My apologies to the questioner! I revised the answer. $\endgroup$ Feb 22, 2013 at 7:51

For any $ \delta > 0 $, the function $ f(x) $ is Lipschitz on $ [\delta,1] $, since $$ |f'(x)| = |a x^{a-1} \sin \frac{1}{x} - x^{a-2} \cos \frac{1}{x}| \leq |a x^{a-1} \sin \frac{1}{x}| + |x^{a-2} \cos \frac{1}{x}| \leq a |x|^{a-1} + |x|^{a-2} < a + \delta^{a-2} < \infty \ \ \ (**)$$ Therefore, for every $ x,y \in [\delta,1] $ I can find a number $ c $ such that $ \frac{|f(x)-f(y)|}{|x-y|} < c $ (since $ |f'(x)| < \infty $ on $ [\delta , 1] $) and therefore $ f(x) $ is AC on $ [\delta , 1] $.\

$$ f(x)=f(\delta)+\int^{x}_{\delta} f'(t)dt \Rightarrow f(x)=f(\delta)+\int^{1}_{0} f'(t) \chi_{[\delta , x]} dt $$ Choose $ \delta = \frac{1}{n} $ and take the limit as $ n $ approaches infinity $$ \lim_{n \rightarrow \infty} f(x)= \lim_{n \rightarrow \infty} f(\frac{1}{n})+ \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]} dt \ \ \ (*)$$ Moreover, $ f(x) $ is continuous at $x=0$. Because $$-1 \leq \sin \frac{1}{x} \leq 1 \Rightarrow -x^a \leq x^a \sin \frac{1}{x} \leq x^a \Rightarrow \lim_{x\rightarrow 0^+} -x^a = 0 \leq \lim_{x\rightarrow 0^+} x^a \sin \frac{1}{x} \leq \lim_{x\rightarrow 0^+} x^a = 0 \Rightarrow \lim_{x\rightarrow 0^+} x^a \sin \frac{1}{x}=0 $$ And because $ f(0)=0 $ we conclude that $ f(x) $ is continuous at $ x=0 $.

$$ (*) \Rightarrow f(x)=f(0) + \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]} dt $$ By $ (**) $ we know that $ |f'(t) \chi_{[\frac{1}{n} , x]}| \leq |f'(t)| \leq ax^{a-1} + x^{a-2} = g(x)$, if $ a>1 $ then $ g(x) $ is Riemann integrable over $ [0,1] $ (and therefore is Lebesgue-integrable) since $$ \int^{1}_{0} ax^{a-1} + x^{a-2} = a + \frac{1}{a-1} < \infty $$ And because $ f'(t) \chi_{[\frac{1}{n} , x]} $ converges pointwise a.e to $ f'(t) \chi_{[0 , x]} $, using the Lebesgue Dominated Convergence theorem, we would have $$ f(x)=f(0) + \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]}dt \Longrightarrow_{L.D.C} f(x)=f(0) + \int^{1}_{0} f'(t) \chi_{[0 , x]}dt \\ \Rightarrow f(x)=f(0) + \int^{x}_{0} f'(t)dt \Longrightarrow f(x) \ is \ AC \ if a > 1 \ \ \ \square$$

  • $\begingroup$ I think showing $f$ is AC on $[\delta,1]$ for any $\delta>0$ is enough. See: math.stackexchange.com/a/4456/497335 $\endgroup$
    – Bach
    Aug 7, 2019 at 1:49
  • $\begingroup$ @Bach I believe the extra steps are needed, the function $x\sin(1/x)$ is A.C. on $[\delta,1]$ but $[0,1]$. The differences come in where the calculation $a+\frac{1}{a-1}<\infty$ is used to show $L^1$ to apply LDC. $\endgroup$ Aug 10, 2022 at 7:02

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