# Prove that $x^\alpha \cdot\sin(1/x)$ is absolutely continuous on $(0,1)$

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 $\alpha=3/2$.

Thanks!

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 $\delta\to 0$.

• Isn't this uniform continuity that you proved, and not absolute continuity? Feb 22, 2013 at 3:20
• I misread the question. My apologies to the questioner! I revised the answer. 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 $$f(x)=f(0) + \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]}dt \Longrightarrow_{L.D.C} f(x)=f(0) + \int^{1}_{0} f'(t) \chi_{[0 , x]}dt \\ \Rightarrow f(x)=f(0) + \int^{x}_{0} f'(t)dt \Longrightarrow f(x) \ is \ AC \ if a > 1 \ \ \ \square$$

• I think showing $f$ is AC on $[\delta,1]$ for any $\delta>0$ is enough. See: math.stackexchange.com/a/4456/497335
– Bach
Aug 7, 2019 at 1:49
• @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. Aug 10, 2022 at 7:02