# How to intuitively understand what it means for a function to be of bounded variation on a closed set and determine whether f(x) is BV or not.

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!

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

• 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. Apr 5, 2021 at 17:41
• 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 Apr 6, 2021 at 17:27

$$\newcommand{\BV}{\mathrm{BV}[a,b]} \newcommand{\BVV}{\mathrm{BV}[0,1]} \newcommand{\V}{\mathrm{V}[f;a,b]} \newcommand{\VV}{\mathrm{V}[f;0,1]} \newcommand{\PP}{\mathcal{P}} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\b}[1]{ \color{blue}{\frac{2}{#1\pi}}} \newcommand{\r}[1]{\color{red}{\frac{2}{#1\pi}}}$$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:

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 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?