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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.

usere5225321
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    Do you know the relation between the total variation of $F$ and $\lvert F'\rvert$? – Daniel Fischer Nov 02 '16 at 21:04
  • I know that if F is differentiable for all x in [a,b] and |F'|<M, then F has a bounded variation. The issue is that clearly F is not differentiable when x=0. – usere5225321 Nov 02 '16 at 21:05
  • $F$ is differentiable at $0$ when $a > 1$. If $a > 1+b$, then $F'$ is even continuous. But of course we need to consider less nice cases. I was aiming at the connection between the total variation and the integral of $\lvert F'\rvert$. – Daniel Fischer Nov 02 '16 at 21:15
  • @DanielFischer what is that connexion? – usere5225321 Nov 02 '16 at 21:22
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    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. – Daniel Fischer Nov 02 '16 at 21:28
  • @DanielFischer do you have any references for "integral is finite if and only if gg has bounded variation on [a,b]" – usere5225321 Nov 02 '16 at 21:50
  • @DanielFischer Also, did you mean " integral is finite if and only if F has bounded variation on [a,b]"? – usere5225321 Nov 02 '16 at 21:53
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    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 $\int \lvert g'\rvert$.) – Daniel Fischer Nov 02 '16 at 22:04
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    @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? –  Jan 03 '17 at 23:03
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    @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. – Daniel Fischer Jan 04 '17 at 10:40
  • @DanielFischer: Thanks for your comment! In particular, when $f$ is continuous, your argument (which is true for any function $f$) implies that $$ TV(f,[0,1])=\lim_{\varepsilon\to 0}TV(f,[\varepsilon,1]),$$ which gives a concise positive answer to this question, doesn't it? –  Jan 04 '17 at 12:59
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    @Jack It's not so concise when one adds a proof. Over there, user71352 gave the argument. – Daniel Fischer Jan 04 '17 at 13:15

2 Answers2

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[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.

Anne Bauval
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  • 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?
     – ajl123 Oct 25 '20 at 18:27
  • @ajl123 I answered your two questions by editing this old answer. – Anne Bauval Mar 28 '23 at 12:05
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[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 $[0,1]$.

Anne Bauval
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jy zhang
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  • Now we assume that $<$ should be Now we assume that $>$. – Petra Axolotl May 07 '22 at 22:00
  • Thanks. You are right. – jy zhang May 08 '22 at 08:22
  • (+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). – Dave L. Renfro Mar 28 '23 at 19:32