6

Suppose that $f:(a,b)\to\mathbb{R}$ is a differentiable function. Does it follow that $f$ has bounded variation on some subinterval $[c,d]\subset (a,b)$?

Details and ideas

  • Being differentiable means only that $f'(x)$ exists for all $x\in (a,b)$. Continuity of $f'$ is not assumed.
  • $f$ need not be of bounded variation on every subinterval $[c,d]\subset (a,b)$. For example, $f(x)=x^2 \sin x^{-2}$ is differentiable on $\mathbb{R}$ but has infinite variation on any interval containing $0$.
  • One can add several copies of $f$ as above to create several points where variation blows up. But trying to add infinitely many of them, e.g., $\sum c_n f(x-q_n)$ with $q_n$ running over a dense set, appears likely to destroy differentiability somewhere.
  • A post by Dave L. Renfro gives a list of bad properties that the derivative of a differentiable function may have, but I didn't find anything inconsistent with being BV on some subinterval.
  • The question is motivated by this problem.
Community
  • 1

1 Answers1

7

As shown here A Fundamental Theorem of Calculus, any function which is everywhere differentiable with Lebesgue-integrable derivative is automatically absolutely continuous (and hence of bounded variation, see e.g. https://en.wikipedia.org/wiki/Absolute_continuity).

Now, measurability of $f'$ is clear. Thus, if $f'$ is bounded (and thus integrable) on some subinterval $[c,d] \subset (a,b)$, we are done. In this case, $f$ is even Lipschitz on some subinterval, so that we don't actually need the above theorem.

But since $f'$ is a pointwise limit of continuous functions on any compact subinterval of $(a,b)$ (by definition of the derivative), using a Baire category argument as here Pointwise limit of continuous functions is 1) measurable and 2) pointwise discontinuous, we see that $f'$ is continuous at some $x_0 \in (a,b)$. In particular, it is bounded on some neighborhood of $x_0$.

As noted above, this implies the claim.

Community
  • 1
PhoemueX
  • 35,087
  • 2
    In other words, a differentiable function cannot be nowhere Lipschitz. Nice. I knew the ingredients, but never put them together. –  Feb 04 '16 at 17:41
  • does the claim still follow when we only have $f'$ exists a.e.? – Sam Blattner Dec 19 '16 at 05:54
  • 1
    @SamBlattner: Good question. Let me think about it :) – PhoemueX Dec 20 '16 at 16:18
  • @Sam: Do you want to know whether the function is Lipschitz on some interval or of bounded variation on some interval? In the first case, I think I can produse a counterexample. In the second case, you probably need to start a new question. I still think the answer is no, but the arguments in my above post are not applicable anymore... – PhoemueX Dec 25 '16 at 08:50