Let $f: [0, 1] \to \mathbb R$ be a measurable function. Define the (possibly infinite valued) upper and lower Dini derivative $D^+ f, D^- f: [0, 1] \to [-\infty, \infty]$ by

$$D^+ f (x) := \limsup_{y \to x} \frac{f(y) - f(x)}{y - x},$$

$$D^- f (x) := \liminf_{y \to x} \frac{f(y) - f(x)}{y - x}.$$

Question: Suppose $D^+ f, D^- f$ are everywhere finite and are in $L^1$. Does it follow that $f$ is absolutely continuous?

Some comments:

  1. If it is known that a function is everywhere differentiable, with derivative in $L^1$, then it is absolutely continuous, but this is not trivial to prove, or rather it does not follow directly from the usual statement of the Lebesgue FTC.

  2. At almost all points of non differentiability of a function, the upper and lower derivatives are infinite. It follows that $f$ is differentiable a.e., and of course $D^+ f = D^- f$ wherever $f$ is differentiable. So this is inherently a question about the null set on which they possibly differ.

  • 2
    $\begingroup$ How about $f(x)=x^2\cos(\pi/x^2)$ (with $f(0)=0$)? This is differentiable everywhere. But the function is not absolutely continuous. $\endgroup$ Commented Jan 20, 2023 at 5:10
  • $\begingroup$ Hm, is the derivative in $L^1$? $\endgroup$
    – Nate River
    Commented Jan 20, 2023 at 7:00
  • 1
    $\begingroup$ Ah... nope... missed that condition. $\endgroup$ Commented Jan 20, 2023 at 7:09
  • $\begingroup$ Do we know $|f(a)-f(b)|\le \int_a^b\max(|D^+f(x)|,|D^-f(x)|)\,dx$? (If so we would presumably be in business. $\endgroup$ Commented Jan 20, 2023 at 7:12
  • $\begingroup$ That sounds like it should be true… now how to prove it.. $\endgroup$
    – Nate River
    Commented Jan 20, 2023 at 7:32

1 Answer 1


Having finite Dini derivatives implies that the function is continuous, and we even get differentiable a.e. (eg."Mean Value Theorems And Functional Equations")

enter image description here

Once we have continuity, we have integral formulas (eg."Recovering a Function from a Dini Derivative") in terms of upper/lower Riemman (but with a restricted partition)

enter image description here

and so since we have L1, we also get absolute continuity.

  • 1
    $\begingroup$ This result is certainly close to what I want, but it seems the authors use a different notion of integration here. Can we say that it agrees with the Lebesgue integral when $D^+ f$ is in $L^1$? $\endgroup$
    – Nate River
    Commented Jan 29, 2023 at 10:55
  • 1
    $\begingroup$ yes, see their section (5d) page 42. $\endgroup$ Commented Jan 29, 2023 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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