# Bounded variation and Integration

This is an exercise in section 6.3 of Royden & Fitzpatrick's Real Analysis.

Consider the function

$$f(x)= \begin{cases} x^a\sin\left(\frac{1}{x^b}\right) & \text{if }0 < x \leq 1 \\ 0 & \text{if }x=0. \end{cases}$$

show that if $a>b$, $f$ is of bounded variation on $[0,1]$.

MY ATTEMPT:

A theorem of the chapter says

$f$ of bounded variation $\Leftrightarrow f$ is differentiable a.e. on $(a,b)$ and $f'$ is integrable over $[a,b]$.

So I want to somehow show that

$$f'(x)= \begin{cases} -bx^{a-1-b}\cos(x^{-b})+ax^{a-1}\sin(x^{-b}) & \text{if }0 < x \leq 1 \\ 0 &\text{if }x=0.\end{cases}$$

is integrable. Now, by definition, $f$ is integrable if $|f|$ is integrable, and since $|f|$ is nonnegative, it is integrable if its integral is bounded. If this is the correct approach to this, then since the lebesgue integral is equal to the riemann integral on closed bounded intervals, then all I have to do is show the riemann integral of $f'$ is bounded. But the riemann integral of this thing is way too complicated (mathematica gives me a mess). Is this supposed to be this cumbersome?

Any help is much appreciated.

• I think you assumed $a>0$?
– user99914
Commented Nov 16, 2015 at 10:35
• Would it not be easier to calculate the variation with respect to a partition directly and show that this is in fact bounded when $a>b$?
– SamM
Commented Nov 16, 2015 at 10:43
• John Ma I forgot to mention $a$ and $b$ are both positive, thanks for pointing that out. @SamM, Your Ad here provided a nice answer using a main theorem of the chapter I'm working on. With your approach I would have to find the Total Variation, which is the supremum of the variations over all partitions, and I'm not sure how feasible or practical that would be, at least for me.
– Mike
Commented Nov 16, 2015 at 10:45
• I was just thinking about how I world attempt to show this. However, I'm not sure how the condition $a>b$ vines in which suggests that my approach may not hold merit.
– SamM
Commented Nov 16, 2015 at 10:58
• Can you show why $f'(0) = 0$. Commented Aug 5, 2021 at 20:54

First, the correct assumption should be $a>b>0$ (check if you overlooked something in the statement of the exercise).

Your basic approach is correct. But to show that the integral of $|f'|$ is bounded you don't need to calculate it explicitly. It is enough to give an upper bound. Note that $|\cos(x^{-b})|\le 1$ and $|\sin(x^{-b})|\le 1$. Using that and the triangle inequality,

$$\int_0^1 |f'(x)| dx \le b \int_0^1 x^{a-b-1} dx + a \int_0^1 x^{a-1} dx = \frac{b}{b-a}+1<\infty$$

Note that in the last step we have used that $a-b-1>-1$ (i.e. that $a>b$) and $a-1>-1$ (that $a>0$), because that is the condition for a power $x^\alpha$ to be integrable at $0$.

To be a bit pedantic, since you mentioned the Riemann integral, $|f'|$ is in general not Riemann integrable on $[0,1]$, because it may be unbounded (if $a-b-1<0$ or $a-1<0$). If you want to keep the perspective of Riemann integrals, you need to look at them as improper Riemann integrals.

• Thank you very much for your thorough answer. You are right, I forgot to mention that $a$ and $b$ are positive. So, for part b) of the question, if $a\leq b$ then it could be the case that $a-b-1\leq-1$ and $x^{a-b-1}$ would not be integrable, and that's why we need $a>b$, correct?
– Mike
Commented Nov 16, 2015 at 10:43
• In principle correct, but this is not (yet) complete. You need to set $a=b$ and look at $f'$ again and see if it is abs. integrable (it really is not, because $x^{-1}$ is not integrable, but you need to look at the role of $\cos$ and the other term, why can they not conspire to make it integrable?).
– J.R.
Commented Nov 16, 2015 at 10:47