I'm trying to show that $F(x)=x^a\sin\left(x^{-b}\right)$ for $0<x \leq 1$ and $F(0)=0$ has bounded variation only if $a>b$.

I know I have to show there exist an $M< \infty$ such that for any partition $0=t_0<t_1<...<t_n=1$ we have

$$\sum_{j=1}^N |F(t_j)-F(t_{j-1})|<M \iff |F(t_1)| + \sum_{j=2}^N |F(t_j)-F(t_{j-1})|<M .$$

I'm stuck here.

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    $\begingroup$ Do you know the relation between the total variation of $F$ and $\lvert F'\rvert$? $\endgroup$ Nov 2, 2016 at 21:04
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    $\begingroup$ When a function $g$ is differentiable on the interval $[a,b]$ (we don't require continuous differentiability or so), then the total variation of $g$ on $[a,b]$ is $\int_a^b \lvert g'(t)\rvert\,dt$. The integral has to be interpreted as a Lebesgue integral in case of ugly $g'$, but here $F'$ is continuous on $(0,1]$, so we can do with the Riemann integral. Note that the integral is finite if and only if $g$ has bounded variation on $[a,b]$. So here, you'd look at $$\lim_{\varepsilon\to 0} \int_\varepsilon^1 \lvert F'(t)\rvert\,dt.$$ $F$ has bounded variation if and only if the limit is finite. $\endgroup$ Nov 2, 2016 at 21:28
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    $\begingroup$ Rudin, Real And Complex Analysis, Chapter 7. Particularly theorem 7.21 and exercise 13 are relevant. And I was talking more generally, not only about the well-behaved $F$ we have here, so I meant $g$ there, not $F$. (But let me stress that it's important that $g$ is differentiable at all points [well, we can drop finitely many points, that's no problem], it is not sufficient that $g$ be differentiable almost everywhere on $[a,b]$ to have the relation between the total variation and $\int \lvert g'\rvert$.) $\endgroup$ Nov 2, 2016 at 22:04
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    $\begingroup$ @DanielFischer: As you pointed out in the comment, $F'$ is only continuous on $(0,1]$. Do you have an argument for why $F$ is BV if and only if the limit $\lim_{\varepsilon\to0}\int_\varepsilon^1|F'(t)|\ dt$ is finite? How can one show that the total variation of $F$ on $[0,1]$ is the limit? $\endgroup$
    – user9464
    Jan 3, 2017 at 23:03
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    $\begingroup$ @Jack For $\varepsilon > 0$, the total variation of $F$ on $[\varepsilon,1]$ is $\int_{\varepsilon}^1 \lvert F'(t)\rvert\,dt$. For every function $f$, one has $TV(f, [0,1]) = \lim\limits_{\varepsilon \to 0} TV(f,[\varepsilon,1]) + \limsup_{\varepsilon \to 0} \lvert f(\varepsilon) - f(0)\rvert$. If the limit of the first term is finite, then $f$ is bounded, and the second term also is finite. $\endgroup$ Jan 4, 2017 at 10:40

2 Answers 2


[I'm assuming in this answer that $a,b>0$.]

Note that $f$ is continuous on $[0,1]$ and its derivative on $(0,1]$ reads as: $$ f'(x)=ax^{a-1}\sin(x^{-b})-bx^{a-b-1}\cos(x^{-b}),\quad x\in(0,1]. $$

We want to study the integrability of $f$ on $[0,1]$. On the one hand, since $a>0$, one has $$ I(x):=ax^{a-1}\sin(x^{-b})\in L^1([0,1]) $$ since $\int_0^1|I(x)|\,dx\le\int_0^1ax^{a-1}\,dx=1.$ Thus it suffices to study the integrability of $$ J(x):=x^{a-b-1}\cos(x^{-b}),\quad x\in(0,1]. $$

On the other hand, according to the accepted answer to a related question When is $\int_{0}^1|x^{a-b-1}\cos(x^{-b})|\ dx<\infty$?, one can conclude that

$f'\in L^1([0,1])$ if and only if $a>b$.

Now, we have the following two cases.

  • $\begingroup$ I have a few questions: 1. why is I(x) a L1-integrable function here based on the argument: a-1>-1? 2. If f' is not integrable, then why does that imply f is not of BV? $\endgroup$
    – ajl123
    Oct 25, 2020 at 18:27
  • $\begingroup$ @ajl123 I answered your two questions by editing this old answer. $\endgroup$ Mar 28, 2023 at 12:05

[I'm also assuming that $a,b >0$.]

Let $T_f(c,d)$ be the variation of $f$ on $[c,d]$. Suppose that $a\leq b$, we have $$T_f(0,1)\geq \sum_{k=1}^\infty T_f(\frac{1}{\sqrt[b]{k\pi+\frac\pi2}},\frac{1}{\sqrt[b]{k\pi-\frac\pi2}})\geq \sum_{k=1}^\infty \frac{1}{(k\pi+\frac\pi2)^{\frac ab}}+\frac{1}{(k\pi-\frac\pi2)^{\frac ab}}=\infty.$$ Therefore $f$ is not bounded variation on $[0,1]$. Now we assume that $a>b$. Recalling that $$ f'(x)=ax^{a-1}\sin(x^{-b})-bx^{a-b-1}\cos (x^{-b}), $$ we may have $|f'(x)|\leq 2\max(a,b)x^{a-b-1}:=Mx^{a-b-1}$. By Lagrange mean value theorem, $\forall x<y\in(0,1]$, $|f(x)-f(y)|\leq M\left(\max_{t\in[x,y]}t^{a-b-1}\right)(y-x)$. From this we estimate $$ \begin{align}T_f\left(\frac1{2^n},1\right)&=\sum_{k=0}^{n-1} T_f\left(\frac1{2^{k+1}},\frac1{2^k}\right)\\ &\leq \sum_{k=0}^{n-1}M\max\left(\frac1{2^{(k+1)(a-b-1)}},\frac1{2^{k(a-b-1)}}\right)\left(\frac1{2^{k}}-\frac1{2^{k+1}}\right)\\&<B<\infty,\end{align} $$ with $B$ a positive number independent of $n$. Hence $f$ is BV on $[0,1]$.

  • $\begingroup$ Now we assume that $𝑎<𝑏$ should be Now we assume that $𝑎>𝑏$. $\endgroup$ May 7, 2022 at 22:00
  • $\begingroup$ Thanks. You are right. $\endgroup$
    – jy zhang
    May 8, 2022 at 8:22
  • $\begingroup$ (+1) for an elementary (i.e. no integration knowledge needed) solution. I haven't checked the details, but I'm guessing it is essentially what I described in my comment to this question, at least for the $a\leq b$ case. I'm not sure about the $a > b$ case, in the sense of whether my approach is not as rigorous as what you've done. (Comment made while I'm busy with something else, so I can't really think about this now.) @Anne Bauval (in case you're interested). $\endgroup$ Mar 28, 2023 at 19:32

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