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.


1 Answer 1


Hint: Take $x_{2k}=\frac{1}{\pi k}$ and $x_{2k+1}=\frac{1}{2\pi k+\frac{\pi}{2}}$

  • 1
    $\begingroup$ @Matthieu Could you please edit your comment? The TeX is not rendering properly. If I'm reading it correctly, your last summation is essentially the harmonic series (use the comparison at the limit test). $\endgroup$
    – Reveillark
    Mar 14, 2019 at 20:02
  • 1
    $\begingroup$ 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 $\sum_{i=1}^{n}\frac{1}{i}$. $\endgroup$
    – Tomislav
    Mar 14, 2019 at 21:42

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