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:

*

*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?


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