6

Let $f$ be a real-valued function defined on the open unit interval.

What assumptions you have to make about $f$ to be sure that it posseses an antiderivative?

I'm interested in the weakest (most general) possible assumptions, so some nontrivial equivalent condition would be optimal.

Thank you in advance!

Dave L. Renfro
  • 38,688
user150316
  • 83
  • 4
    I believe there are (at least) two different questions. The weakest condition to have an antiderivative is to simply be a derivative, and there is a lot of literature on attempts to classify the property of being a derivative (in ways besides just restating "is a derivative"). See Andrew M. Bruckner's book Differentiation of Real Functions for an entry point into this area. The other question has to do with methods for obtaining an antiderivative, and for this various integration methods have been developed -- the Riemann integral, the Lebesgue integral, the Denjoy integral, etc. – Dave L. Renfro May 15 '14 at 19:53
  • My question is adressing precisely the first issue. The reference you mentioned is kind of thing I was looking for, but I'm afaid I don't have acces to it. Anyway, thank you! – user150316 May 15 '14 at 20:43
  • @DaveL.Renfro In addition and maybe I'm not understanding fully, is it safe to say that if $f$ is continuous on $(a,b)$, there exists another function $F$ which is differentiable on $(a,b)$ down to $f$? While continuity doesn't imply the function itself is differentiable, continuity does imply the existence of another function which is differentiable down to $f$? – DWade64 Oct 26 '18 at 19:54
  • 1
    @DWade64: Yes, given any continuous function $f$ defined on $(a,b),$ then there exists a function $F$ defined on $(a,b)$ such that $F'(x) = f(x)$ for each $x \in (a,b).$ And you don't need advanced mathematical ideas for this such as Lebesgue integration and measure theory --- for continuous functions $f,$ the corresponding functions $F$ can be defined by using the Riemann integral. This is one half of the usual 2-part version of the Fundamental Theorem of Calculus. See Is every continuous function a derivative? – Dave L. Renfro Oct 26 '18 at 20:30
  • @DaveL.Renfro Thank you for the help as well as the answer you gave! – DWade64 Oct 26 '18 at 21:14
  • @DWade64: If f is continuous on the Real line ( at least a.e.) , then $F:= \int_0^x f $ is an antiderivative, and, wherever F is continuous, F'=f. – MSIS Nov 07 '21 at 22:13

2 Answers2

6

This is in response to your reply to my comment.

Here are two useful expository papers that are freely available on the internet:

Andrew M. Bruckner, Derivatives: why they elude classification, Mathematics Magazine 49 #1 (January 1976), 5-11. (curtsey of Andres Caicedo and google)

Andrew M. Bruckner, The problem of characterizing derivatives revisited, Real Analysis Exchange 22 #1 (1995-96), 112-133.

Dave L. Renfro
  • 38,688
-2

For a simpler, more basic take, if f is a.e. differentiable ( thus a.e. continuous), on the Real line, define:

$$F(x):= \int_0^x f(t)dt +C $$, where $C$ is a Real constant. Then $F'=f$ a.e.

Edit: In order for a function f to be recovered as the integral of its derivative, i.e., to have $F(x):= \int_{0}^x f'+C $ everywhere, $F$must be Absolutely Continuous.

For a function f to be a derivative, i.e., the derivative of another function, it must satisfy the Darboux property.

MSIS
  • 815
  • And , do tell, downvote, just what is wrong with this? – MSIS Nov 08 '21 at 05:45
  • Yet another cowardly downvote. – MSIS Oct 03 '24 at 02:23
  • 1
    Very strange to see this downvoted! At least one explanation would be in order. – A rural reader Nov 25 '24 at 23:27
  • I took some time to edit it and no one offered anything constructive. Bet they can't offer a counterexample. – MSIS Nov 25 '24 at 23:37
  • They downvoted , anonimously another question of mine. I asked for an explanation politely . A few weeks after not receiving one, I called them up on it. Then they started downvoting other questions of mine. Anonymously, of course. – MSIS Nov 26 '24 at 19:20
  • At any rate, I've notified Admin here and they told me they would look into it. – MSIS Nov 26 '24 at 19:20