I am trying to figure out whether the function below is of bounded variation or not. I know there are techniques to prove that a function is of bounded variation by using Jordan's Decomposition Theorem or Lebesgue's Differentiation Theorem for Integrals.

So my questions are:

  1. Intuitively (by looking at a graph of the function) what does it mean for f(x) to be of bounded variation? Is it to do with how much it oscillates in an interval?

  2. How can I prove whether this sin function is of bounded variation or not?

The function below is considered from closed interval f:[0,1] → R.

Thank you for the help!

$$ f(x)= \begin{cases} 42 & \text{for } x=0; \\ \sin\left(\frac1{\sqrt{x}}\right) & \text{for }x\ne 0. \end{cases} $$

  • 1
    $\begingroup$ Essentially, yes - a function not of bounded variation oscillates too much and too often in the domain of concern. For your function, try to quantify the points for which $\sin(x^{-1/2}) = 1$ and $\sin(x^{-1/2}) = -1$, and show that there are infinitely many such oscillations in $[0,1]$, and you can show it is not BV from the definition. $\endgroup$ Apr 5, 2021 at 17:41
  • $\begingroup$ Thank you. I get the intuition better but I am not sure how to show the infinitely many oscillations. I know that if I had xsin(1/sqrt(x)) I would be able to find xi's (in the partition) such that sin(1/sqrt(x))=1 and sin(1/sqrt(x))=-1. I don't know how to show that there are infinitely many oscillations. I am struggling to find helpful explanations online. @EeveeTrainer $\endgroup$ Apr 6, 2021 at 17:27

1 Answer 1


$ \newcommand{\BV}{\mathrm{BV}[a,b]} \newcommand{\BVV}{\mathrm{BV}[0,1]} \newcommand{\V}{\mathrm{V}[f;a,b]} \newcommand{\VV}{\mathrm{V}[f;0,1]} \newcommand{\PP}{\mathcal{P}} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\b}[1]{ \color{blue}{\frac{2}{#1\pi}}} \newcommand{\r}[1]{\color{red}{\frac{2}{#1\pi}}} $I'll go ahead and motivate this with a similar problem/function that might be easier to parse through.

Motivating Example:

So, from definition, $f \in \BV$ iff it has it has finite variation $\V$, defined symbolically as so... Let $\PP$ be the class of finite partitions $P := \{x_0,\cdots,x_{n_P}\}$ of $[a,b]$ (where $x_i \le x_{i+1}$); then $$ \V := \sup_{P \in \PP} \sum_{i=0}^{n_P - 1} \Big| f(x_{i+1}) - f(x_i) \Big| $$ Intuitively, $\V$ quantifies how much $f$ oscillates on the domain $[a,b]$.

Often, then, if you wish to show $f \not \in \BV$, you want to choose partitions $P$ such that $f(x_i)$ is some sort of maximum, then $f(x_{i+1})$ is some sort of minimum, then $f(x_{i+2})$ is some sort of maximum again, and so on.

A simpler example for $\BVV$ is $$ f(x) = \begin{cases} \sin(1/x) & x \in (0,1] \\ 0 & x = 0 \end{cases} $$ Looking at the graph, the graph oscillates between $1$ and $-1$ infinitely often:

enter image description here

We ask ourselves, then: for which $x$ does $\sin(1/x) = 1$? What about $\sin(1/x) = -1$?

Solving for $x$ is trivial enough and we get

\begin{alignat*}{99} f(x) &=+ &&1 &&\implies x &&= \frac{2}{\pi(4n+1)} &&\text{ for } n \in \mathbb{Z} \\ f(x) &=- &&1 &&\implies x &&= \frac{2}{\pi(4n+3)} &&\text{ for } n \in \mathbb{Z} \end{alignat*}

(Since we're dealing with $[0,1]$ though, we can restrict our concern to positive integers $n$.) Of course, these also happen to alternate, happily. So, how do we form a partition with this? Choose a sequence of partitions that, essentially, contains the first ever-so-many such $x$'s from the right. So we can have

$$P_n := \set{ \color{blue}{\frac{2}{\pi}} , \r 3 , \b 5 , \r 7, \b 9, \r {11}, \b{13}, \cdots }$$

and so on and so forth until we have $n$ items in $P_n$. The blue terms give $f(x) = 1$ and the red give $f(x) = -1$. Then you can show any such partition $P_n$ has variation

$$\sum_{i=0}^n \Big| \underbrace{f(x_{i+1}) - f(x_i)}_{\text{always } \pm 2} \Big| = 2(n-1)$$

But you can construct infinitely many such $P_n$, with increasing $n$. Want one with $10$ points? Sure. $100$? Sure. $10^{100}$? Go right ahead. Notice that each of the applicable $x$ will always alternate, and always lie further within $[0,1]$ (for $n > 0$) - successive $x$ are always smaller, but still in that interval. Then we can argue: the limit of such a sequence must be less than the supremum over all the partitions, by definition, but since that limit is infinity, the supremum and thus variation must also be infinity. Symbolically:

$$\infty = \lim_{n \to \infty} 2(n-1) = \lim_{n \to \infty}\underbrace{ \sum_{i=0}^n \Big| f(x_{i+1}) - f(x_i) \Big|}_{\text{variation for a } P_n} \le \sup_{P \in \PP} \sum_{i=0}^{n_P - 1} \Big| f(x_{i+1}) - f(x_i) \Big| = \VV$$

and hence

$$\VV = \infty \implies f \not \in \BVV$$

Your Problem:

Your problem is mostly the same, just with a slight modification. So, you need, for your function, to find $x$ such that $\sin(x^{-1/2}) = \pm 1$. Similarly to before, you should easily get

\begin{alignat*}{99} f(x) &=+ &&1 &&\implies x &&= \frac{4}{\pi^2} \cdot \frac{1}{(4n+1)^2} &&\text{ for } n \in \mathbb{Z} \\ f(x) &=- &&1 &&\implies x &&= \frac{4}{\pi^2} \cdot \frac{1}{(4n+3)^2} &&\text{ for } n \in \mathbb{Z} \end{alignat*}

(This only really differs from the motivating example in that you square both sides to find $x$ as your very last step.)

Can you see that $x \in (0,1]$ for every such $x$? That they alternate as well, and that if you take successively bigger $n$'s, you get successively smaller $x$'s?

With that in mind, you can construct a similar sequence of partitions as before, and show the variation to be infinite.

  • $\begingroup$ Thank you! That was really helpful! And this idea of finding the min and max of the function to show the variation is infinite works for all functions not of bounded variation? $\endgroup$ Apr 7, 2021 at 15:31
  • $\begingroup$ I wouldn't be so brazen as to say it works for them all, but it's certainly worked for all of the ones I've encountered myself $\endgroup$ Apr 7, 2021 at 15:40

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