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Given is a function $f:I \rightarrow \mathbb{R}$ on a closed and bounded interval $I \subset \mathbb{R}$. If $f$ is of bounded variation on $I$, that is $\text{Var}_I(f) < \infty$, then we know that the number of discontinuities of $f$ is at most countable.

My questions:

  1. Does that also imply that the function is monotone on at most a countable number of subintervals in $I$?
  2. Does there exist an additional condition that guarantees a finite number of subintervals on which the function is monotone?

Restriction: One important restriction is that I cannot use the number of sign changes of the derivative.

Extra information: In case the above question is too general, then I wonder if such statements can be made about a function that is made up by adding up, composing, or multiplying integer root functions (such as square root and cube root), trigonometric functions and inverse trigonometric functions and where $I \subset \left[0,\pi\right]$.

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  • $\begingroup$ Why can't you do an analysis on $f'$? That seems (to me) like the most direct method. $\endgroup$
    – While I Am
    Commented Jun 14 at 15:22
  • $\begingroup$ @WhileIAm thanks a lot for your comment. I have considered this, one could analyse the number of roots of $f'$ to answer this question. If $f$ just consisted of products and sums of trigonometric functions, this would be easy since then $f'$ will be a trigonometric polynomial, for which we can bound the number of zeros depending on the degree. However, $f$ also involved roots, inverse trigonometric functions and compositions of those. Therefore the derivative of $f$ will not just be a trigonometric polynomial but something much more complex, for which I do not see an easy way to count roots. $\endgroup$
    – Nelus127
    Commented Jun 14 at 15:32
  • $\begingroup$ Certainly BV functions are usually not so simple as this. The derivative need not exist everywhere. A BV function could be monotone on no interval. $\endgroup$
    – GEdgar
    Commented Jun 14 at 15:42

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This is an answer to your question (2.). In general, a BV function is not guaranteed to be monotone on a finite number of subintervals. If $f(x) = x^2\sin(1/x)$, then $f' \in L^1[0,1]$, so $f$ is BV, but in every neighborhood of $0$ we have that $f'$ changes sign infinitely many times.

For your question (1.), if I am interpreting your question correctly I believe the answer is yes, as any infinite collection of subintervals contained inside a closed and bounded interval must be at most countable.

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  • $\begingroup$ If you have a more specific question about a class of functions (e.g., your functions in the extra information), I would recommend posting another question with the relevant details. $\endgroup$
    – While I Am
    Commented Jun 14 at 16:35
  • $\begingroup$ Thanks a lot for your answer, this is a useful example. However, I know that just having finite variance is not enough to ensure piecewise monotonicity on a finite number of subintervals. What I am asking is, on top of having finite variance, what is an additional condition to ensure that at function is piecewise monotone on a finite number of subintervals. $\endgroup$
    – Nelus127
    Commented Jun 17 at 8:30

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