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I know that the derivative of a differentiable function doesn't have to be continuous. How discontinuous can a derivative be?.

Inspired by Limits and continuity of a derivative, I was thinking of defining the notion of pseudo-continuous: $f:(a,b) \to \mathbb R$ is pseudo-continuous at $x \in (a,b)$ if $$ f(x) = \lim_{y\to x} \frac1{y-x} \int_x^y f(t) \, dt .$$ And then I wanted to show that a function is the derivative of a differentiable function if and only if it is pseudo-continuous.

But then I realized that the derivative doesn't have to be Lebesgue integrable, for example $$ f(x) = \frac x{\log|x|} \sin\left(\frac1x\right) , \quad x \in (-\tfrac12,\tfrac12) ,$$ or $$ f(x) = x^2 \sin\left(\frac1{x^2}\right) ,$$ Does there exist a differentiable function $f:(0,1) \to \mathbb R$ such that its derivative restricted to any subinterval of $(0,1)$ fails to be in $L^1$?

Stephen Montgomery-Smith
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  • Searching through similar questions, I came across the Henstock–Kurzweil integral https://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral. I could use this in my definition of pseudo-continuous. – Stephen Montgomery-Smith Aug 26 '20 at 04:44

2 Answers2

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The derivative is the (pointwise) limit of a sequence of continuous functions, e.g. $$g_n(x) = \frac{f(x + h_n(x)) - f(x)}{h_n(x)}$$ where we can take $h_n(x) = \frac{1}{n+1}$ if $b = +\infty$, and if $b < +\infty$ we can take $h_n(x) = \frac{b-x}{n+1}$. It follows that the family $\{ \lvert g_n\rvert : \in \mathbb{N}\setminus \{0\}\}$ is pointwise bounded.

Take an arbitrary nonempty interval $(u,v) \subset (a,b)$. For each $k \in \mathbb{N}$ the set $$A_k = \bigl\{ x \in (u,v) : \lvert g_n(x)\rvert\leqslant k \text{ for all } n\bigr\}$$ is relatively closed, and since the family is poinwise bounded we have $$(u,v) = \bigcup_{k \in \mathbb{N}} A_k\,.$$ Furthermore, $(u,v)$ is a Baire space (it's completely metrisable), hence there is a $k \in \mathbb{N}$ such that $$V = \operatorname{int} A_k \neq \varnothing\,.$$ Then $\lvert f'(x)\rvert \leqslant k$ for all $x \in V$

Thus every nonempty open interval in $(a,b)$ contains a nonempty open interval on which $f'$ is bounded. This means the set of points $x$ such that $f'$ is Lebesgue integrable on some neighbourhood of $x$ is a dense open subset of $(a,b)$, hence topologically very large.

However, the measure of this set would be the more important type of size. I don't know whether it can be arbitrarily small (of course it's nonzero), but I suspect it can.

Daniel Fischer
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    I can answer your last question. If $G$ is any closed nowhere dense subset of $\mathbb R$, its complement can be written as a countable disjoint union of open intervals $(a_n,b_n)$. Let $f(x) = (x-a_n)^2 (x-b_n)^2 \sin(\frac{1}{(x-a_n)^2 (x-b_n)^2}$) if $x\in (a_n,b_n)$, and let $f(x) = 0$ on $G$. Then $f$ is differentiable, and $f'$ is not in $L^1$ in any neighborhood of any point in $G$. And we can make the measure of the complement of an closed nowhere dense subset arbitrarily small. – Stephen Montgomery-Smith Aug 29 '20 at 18:45
  • I expected the result, but not that the construction would be that nice. Thanks. – Daniel Fischer Aug 29 '20 at 18:47
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I assert the answer is "no."

Theorem: Let $f:(a,b) \to \mathbb R$ be differentiable. Then there exists a subinterval $[u,v] \subset (a,b)$ such that $f'$ is uniformly bounded on $[u,v]$.

Proof: Suppose the converse is true. Suppose $\epsilon = \frac1{100}$.

Recursively pick sequences $y_n$, $\delta_n$ as follows.

Pick $y_1$ such that $|f'(y_1)| \ne 0$. Given $y_n$, choose $\delta_n > 0$ so that $[y_n - \delta_n, y_n+\delta_n] \subset (a,b)$, and so that if $|h| < \delta_{n}$, we have that $$ (1-\epsilon) |f'(y_n)| \le \left| \frac{f(y_n) - f(y_n+h)} h \right| \le (1+\epsilon)|f'(y_n)| ,$$ and so that if $n > 1$, then $$ \delta_n \le \frac{|f'(y_{n-1})|}{8|f'(y_{n})|} \delta_{n-1} .$$ Next, by hypothesis, there exists a point $$ y_{n+1} \in (y_n-\delta_n,y_n-\tfrac12\delta_n) $$ such that $|f'(y_{n+1})| \ge 2 |f'(y_n)|$.

Note that for $n \ge m > 1$ $$ |y_n - y_{n+1}| \le \delta_n \le 16^{m-n} \delta_m \le 2 \times 16^{m-n} |y_{m} - y_{m+1}| ,$$ In particular, the points $y_n$ form a Cauchy sequence. Let $y = \lim_{n\to \infty} y_n$.

Then \begin{align} |y - y_m| &= \left|\sum_{n=m}^\infty y_{n+1} - y_n \right| \\ & \le \sum_{n=m}^\infty |y_{n+1} - y_n| \\ & \le |y_{m+1} - y_m| \sum_{n=m}^\infty 2\times 16^{m-n} \\ & \le \frac{32}{15} |y_{m+1} - y_m| . \end{align} Also \begin{align} |f(y) - f(y_m)| &= \left|\sum_{n=m}^\infty (f(y_{n+1}) - f(y_n)) \right| \\ & \ge |f(y_{m+1}) - f(y_m)| - \sum_{n=m+1}^\infty |f(y_{n+1}) - f(y_n)| \\ & \ge (1-\epsilon) |f'(y_m)| |y_{m+1} - y_{m}| - (1+\epsilon) \sum_{n=m+1}^\infty |f'(y_n)||y_{n+1} - y_{n}| \end{align} Now $$ |f'(y_{n+1})||y_{n+2} - y_{n+1}| \le |f'(y_{n+1})|\delta_{n+1} \le \frac18 |f'(y_{n})|\delta_{n} \le \frac14 |f'(y_n)||y_{n+1} - y_{n}| . $$ So \begin{align} |f(y) - f(y_m)| & \ge (1-\epsilon) |f'(y_m)| |y_{m+1} - y_{m}| - (1+\epsilon) |f'(y_m)| |y_{m+1} - y_{m}| \sum_{n=m+1}^\infty 4^{m-n} \\ & \ge \frac18 |f'(y_m)| |y_{m+1} - y_{m}| . \end{align} Thus $$ \left| \frac{f(y) - f(y_m)}{y-y_m} \right| \ge \frac{15}{256} |f'(y_m)| .$$ But $|f'(y_m)| \ge 2^m |f'(y_1)| \to \infty$, and this contradicts that $f$ is differentiable at $y$.

Stephen Montgomery-Smith
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  • I haven't checked your computations (I'd say they're probably correct), but I think it's easier to use Baire's theorem. Since $f'$ is the pointwise limits of a sequence of continuous functions on a Baire space, it follows that each nonempty open subset of the domain contains a nonempty subset on which the sequence is uniformly bounded, and hence $f'$ is. – Daniel Fischer Aug 29 '20 at 12:29
  • @DanielFischer I was unaware of that theorem. Thank you. – Stephen Montgomery-Smith Aug 29 '20 at 15:00
  • @DanielFischer Where do I find this theorem? When I do a google search, I get Baire's category theorem. – Stephen Montgomery-Smith Aug 29 '20 at 15:14
  • It's a consequence (corollary if you wish) of the category theorem (or rather, the definition of a Baire space). If you look for a reference and can read (mathematical) German, look at Theorem 13.31 (first Edition, it's 13.32 in the third) in Boto v. Querenburg Mengentheoretische Topologie. Let ${f_i : i \in I}$ be a family of continuous real-valued functions on a Baire space $X$ such that $f(x) = \sup_i f_i(x) < +\infty$ for all $x$. Then for every nonempty open $U \subset X$ there is a nonempty open $V\subset U$ and $M \in \mathbb{R}$ such that $f(x) < M$ for all $x\in V$. – Daniel Fischer Aug 29 '20 at 15:27
  • Proof: $U$ is itself a Baire space. $A_n = { x \in U : f(x) \leqslant n}$ is closed in $U$ for all $n \in \mathbb{N}$. Since $f$ is finite, $U = \bigcup_n A_n$. Since $U$ is Baire there is an $n$ with $V := \operatorname{int} A_n \neq \varnothing$. Then $f(x) < n+1$ on $V$. – Daniel Fischer Aug 29 '20 at 15:27
  • If you post that as an answer, I will accept it. – Stephen Montgomery-Smith Aug 29 '20 at 15:32