I am attempting to calculate the total variation of the function $f$ on [0,1] defined as $$f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{2x}\right) \text{ if } x\neq 0 \\ 0 \qquad \qquad \text{ if } x=0\end{cases}$$

Total variation is defined as usual (almost everywhere in the internet you can get to know that this is in fact a function whose variation is bounded). However solving explicitly its value seems to be an impossible task.

My attempt was to show it via the integral of the absolute value of its derivative, as we know (from a Theorem) that:

$$ V_{[0,1]}(f) = \int_0^1 \vert f'(x)\vert dx$$

But this seems almost as impossible as calculating explicitly the supremum of the sum. Is there a workaround I am missing, for example a theorem or an identity on the bounded variation from $x^2$ and $sin(\pi/2x)$ or something of that style?

Edit: So I browsed the complete bounded-variation tag and it seems that no one has been brave enough to calculate bounded variations... on any function, not even common sines and such. So with that new information, is this even something worth even trying?

Edit 2: A code on MATLAB yields that this integral is around 1.49, I can even say that it is 1.5; but I don't see how I would be able to solve this analytically.


1 Answer 1


No, there is no nice identity to give you the total variation. And an attempt to do this here would likely be a waste of paper.

is this even something worth even trying?

No. If you had that number, what would you do with it?

To say something about this variation, we need to know the extrema of $f$. Those are the points $x_n\in (0,1)$ with $x\tan(1/x)=1/x$; we may assume this sequence is decreasing. Note that $x_n=1/y_n$ where $\tan y_n = y_n/2$. This transcendental equation comes up in other places too (Robin boundary problem for ODE and PDE); the qualitative behavior of its solutions is not too hard to understand, but there is no closed form.

The total variation is $2\sum x_n^2|\sin (1/x_n)|$, plus a term to account for the interval $[x_1,1]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.