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As in the question. I'd really appreciate your help with this.

(1) $f:[0,1] \to [0,1]$ is increasing, that is, $x \leq y \implies f(x) \leq f(y)$

(2) $0 \leq f(x) \leq x$ for all $x \in [0,1]$

Does (1)+(2) imply that there exists $\epsilon>0$ such that $f$ is continuous on $[0,\epsilon)$? If no: Can f have infinitely many jumps on $[0,\epsilon)$ for any $\epsilon>0$?

NOTE: Clearly $f$ is continuous at $0$. The question is whether we can conclude that it is also continuous in some (small) neighborhood "to the right of $0$" when it is increasing/monotone.

My (probably wrong) intuition is that if we have infinitely many jumps then the "slope" cannot be below $1$ as it must be close enough to $0$.

Theo Bendit
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Miranda
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  • Maybe this https://math.stackexchange.com/q/172753/399263, but I don't know if $(2)$ is verified, these examples are wild. – zwim Oct 07 '21 at 19:47
  • You don't need increasing for that. Just $|f(x)| \le |x|$ is enough to make $f$ continuous at $0$. – principal-ideal-domain Oct 07 '21 at 19:58
  • @principal-ideal-domain The asker is aware of this. This is a question about continuity near $0$, not at $0$. – Theo Bendit Oct 07 '21 at 20:01
  • @TheoBendit I'm sorry. If you want, I can delete my comment. – principal-ideal-domain Oct 07 '21 at 20:01
  • @principal-ideal-domain No problem. I would advise against deleting, since you are not the first respondent to misread this question, and our comments together could serve as a caution to other would-be answerers. But, you may delete if you wish. :-) – Theo Bendit Oct 07 '21 at 20:04

1 Answers1

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No, you cannot infer continuity, nor indeed limit the number of jumps. You can say that the jumps must be countable (although this is true for any monotone function), and you can show that the total cumulative distance that the graph "jumps" is finite (i.e. $\sum_{x_0 \in [0, 1]} \left[\lim_{x \to x_0^+} f(x) - \lim_{x \to x_0^-} f(x)\right]$ converges), but there still could be infinitely many jumps.

As an example, let $$f(x) = \begin{cases}\left\lceil x^{-1} \right\rceil^{-1} & \text{if }x \in (0, 1] \\ 0 & \text{if }x = 0,\end{cases}$$ where $\lceil \cdot \rceil$ is the ceiling function, i.e. $\lceil x \rceil$ is the least integer greater than or equal to $x$. We have $$\left\lceil x^{-1} \right\rceil \ge x^{-1} > 0 \implies 0 < \left\lceil x^{-1} \right\rceil^{-1} \le x.$$ Since $\lceil \cdot \rceil$ is monotone increasing, and $x \mapsto x^{-1}$ is monotone decreasing, we see that $\left\lceil x^{-1} \right\rceil$ is decreasing, so $f$ is once again increasing (note $f(0) \le f(x)$ for $x > 0$).

The "jumps" occur whenever $x = \frac{1}{n}$ for some $n \in \Bbb{N}$, i.e. for countably many $n$. These points also accumulate around $0$, so $f$ is not continuous on any neighbourhood of $0$.

Theo Bendit
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    In case you wondered: $\int_0^1 f(x)dx = 2-\pi^2/6$. – principal-ideal-domain Oct 07 '21 at 20:16
  • Thank you very much - very clear answer. If you don't mind a follow-up, though I have a feeling that the answers may be no and no: Can you think of any natural conditions placed only on the function at 0 that would imply continuity in a neighborhood (I repeat, here I think the answer is no since we could even have it differentiable at 0 as far as I can see). How about if the function f is the pointwise limit of a sequence of continuous+increasing functions (again, I suspect this is of no help unless we actually assume uniform convergence). – Miranda Oct 09 '21 at 10:43
  • @Miranda Yes, I would say the answers are "no" and "no". We could, with a little bit of fiddling, make the $f$ above the pointwise limit of continuous increasing functions, but it would take more than a comment to establish this. There is a kind of "cheat" for the first question: I could say $f$ is analytic at $0$, i.e. equal to its Taylor series on a non-trivial interval about $0$. Analyticity is not really a property of a single point though, and that would be independent of any other bounds, like $0 \le x \le f(x)$. – Theo Bendit Oct 12 '21 at 20:42
  • @TheoBendit Thank very much for this. Just so I'm sure I understand, what you are saying is that we assume that f is analytic at 0 we get it continuous at 0. But doesn't analytic by itself imply local continuity, that is, any other assumptions imposed no longer matter? Even so, I find it very interesting because in specific application one might be able to show that analytic is implied. – Miranda Oct 13 '21 at 08:57
  • @Miranda Yes, that's what I was saying. We say something is analytic at a point, but it's not really a property of a single point, and it doesn't use any other imposed assumptions. Mind you, arguably continuity is not a property of a single point either... Still, it feels like a cheat! But, if it ends up helping, that's good! – Theo Bendit Oct 13 '21 at 09:05