# Determining if $f \in BV([0,1])$.

I am reading about Riemann-Stieltjes Integration in Carother's Real Analysis. We find that functions of Bounded Variation provide us with a rich class of integrators. Therefore, I am trying to learn how to quickly identify whether a function is of bounded variation. So far I have been using Jordan's Theorem is quite helpful.

Jordans's Theorem

If $$f = \alpha - \beta$$ where $$\alpha,\beta$$ are non-decreasing functions, then $$f \in BV([a,b])$$.

I used Jordan's Theorem to solve the following question below:

Question 1) Determine whether or not $$f(x)=x^{2}sin(\frac{1}{x})$$ if $$x \neq 0$$ and $$f(0)=0$$ is of bounded variation on $$[0,1]$$.

Solution 1): By Jordan's Theorem, we aim to find two non-decreasing functions, $$\alpha,\beta$$ such that $$f(x) = x^{2}sin(\frac{1}{x})$$ $$=$$ $$\alpha - \beta$$. Intuitively, I think to myself that $$f(x)$$ is bounded above by 1, therefore when trying to find a non-decreasing $$\alpha$$, I could add a non-decreasing function, say $$10x$$, that "dominates" $$f(x)$$ so that $$f(x) + 10x$$ is non-decreasing also. In other words, because $$10x$$ "dominates" $$f(x)$$ and $$10x$$ is non-decreasing, then their sums will be non-decreasing. We can then simply make $$\beta = 10x$$, which is non-decreasing itself to get $$f(x) = \alpha - \beta$$ which is the difference of two non-decreasing functions. We are done.

Now here is my problem.

Question 2) Determine whether or not $$f(x) = \sqrt{x}sin(\frac{1}{x})$$ if $$x \neq 0$$ and $$f(0)=0$$ is of bounded variation on $$[0,1]$$.

Solution 2): I would reuse the technique in Solution 1).

The problem is $$f(x)$$ in Question 2 is not of bounded variation. Therefore, my reasoning above must be incorrect somewhere but I cannot seem to find it. Also, let's say I knew that this function is not of bounded variation. Then I would aim to show the following: \begin{align*} V_{0}^{1}(f) = \underset{P}{Sup}\sum_{k=1}^{n}|f(x_{i})-f(x_{i-1})| \overset{n\to\infty}{\longrightarrow}\infty \end{align*} However, I can't seem to find a partition P that satisfies the statement above. I'm thinking that there is some special sequence of $$x's$$ that would make the summation above equal some sequence of values we know blows up to $$\infty$$. Any clarification would be greatly appreciated.

Hint: Take $$x_{2k}=\frac{1}{\pi k}$$ and $$x_{2k+1}=\frac{1}{2\pi k+\frac{\pi}{2}}$$
• Thank you for your reply. Using you hint I get the following: $V(f,p)$ $=$ $\sum_{i=1}^{n} | f(x_{i}) - f(x_{i-1})|$ $=$ $\sum_{i=1}^{n} | \sqrt{x_{i}} - \sqrt{x_{i-1}}|$ $=$ $x_{0} + x_{n} + \sum_{i=1}^{n-1} \sqrt{x_{i}} - \sqrt{x_{i-1}}$ $\geq$ $\sum_{i=1}^{n-1} \frac{2}{(2i+1)\pi}$. However, I don't see how the $\sum_{i=1}^{n-1} \frac{2}{(2i+1)\pi}$ $\overset{n\to\infty}{\longrightarrow}\infty$. Update I do now see this now as this is very close to the harmonic series $\sum_{i=1}^{n}\frac{1}{i}$. Mar 14, 2019 at 21:42