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In one paper there is a theorem:

For every real function $f(x)$ the set of points where $D^+<D_-$ is a countable set.

For example the function $$f(x)=\begin{cases} 1, x<0, \\ 0,x\geq 0.\end{cases}$$ in point $0$ has $D^+=0$ and $D_-=+\infty$. And we say that in any function there are at most contable points where this occures.

Question 1: Is my example is correct?

Question 2: I have started just now to read about this type of derivatives and I need some clarification. Why we can not say that the set of points where $D^+>D_-$ is also a countable set? Can you give some simple example which shows that we can not say it.

Edit : I will give some clarifications to why I have this question. The proof of the theorem is slightly messy but the intuition behind is very understandable. The idea is that we can construct around all such points open intervals such that one interval contains only one such point (where $D^+<D_-$). And ofcourse we can number all these intervals by some rational number from this interval. And we get countable numbers of this intetvals. So my question is why we can not say the same thing about the points where $D^+>D_-$? I think I need some example.

Mokhmad-Salekh Khekhaev
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  • Don't you want to define upper/lowe dini derivative for us? – Red shoes Oct 07 '21 at 22:42
  • @Red shoes: The four Dini derivates of a function is a standard notion in first year graduate real analysis courses. – Dave L. Renfro Oct 07 '21 at 22:53
  • @RedShoes I thought about this. But in six or seven different sources there is the same definition. – Mokhmad-Salekh Khekhaev Oct 07 '21 at 22:58
  • @RedShoes: But in either case I ask about unilateral derivatives. Upper/lower Dini derivatives are called bilateral derivatives. I am not interested in them in this question. – Mokhmad-Salekh Khekhaev Oct 07 '21 at 23:01
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    (Just passing through for a moment and don't have time now to say more.) William Henry Young proved in 1908 that for every $f:{\mathbb R} \rightarrow \mathbb R$ the set of points where $D^+ < D_-$ or $D^- < D_+$ is countable. Regarding the opposite inequality, this is probably overkill, but there exists a continuous strictly increasing function such that at all but a first (Baire) category set (hence, for a $c$-dense set) we have both lower derivates equal to $0$ and both upper derivates equal to $+\infty$ (continued) – Dave L. Renfro Oct 07 '21 at 23:17
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    (see (c) in this 4 November 2000 sci.math post). Also, "most" continuous functions (in the sense of Baire category, using the sup metric) are such that at all but a very small subset of the reals (a set that is simultaneously measure zero and first Baire category, and smaller still) both lower derivates are equal to $-\infty$ and both upper derivates are equal to $+\infty$ (see this answer). – Dave L. Renfro Oct 07 '21 at 23:17
  • @DaveL.Renfro: Thanks for sources! They are very usefull. – Mokhmad-Salekh Khekhaev Oct 07 '21 at 23:35
  • @DaveL.Renfro I asked that because I believe there are many ppl here who are not familiar with definitions but yet can solve problems much better than the ones who know the definitions. – Red shoes Oct 07 '21 at 23:46
  • @Red shoes: OK. In looking this over, I see that the words Dini derivative/derivate (upper right, lower left) are only mentioned in the title. At the very least these words/terms should be mentioned in the question text, and probably also give a link such as this. – Dave L. Renfro Oct 08 '21 at 14:30

1 Answers1

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Question 1: Yes it is correct. Your example is so trivial because the set of non-differentiable points is countable which includes the set of those points with $D^+ < D^-$. In your example, the set of non-differentiable points is $\{0\}$.

Question 2: Take

$$ f(x) = \left\{\begin{matrix} 1 & x \in Q\\ 0 & x \notin Q \end{matrix}\right. $$

Red shoes
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