I have been thinking about concepts related to differentiability. In particular, being continuous, having an antiderivative (i.e. being the derivative of something), and the Intermediate Value Property (satisfying the conclusion of the Intermediate Value Theorem, sometimes called being a Darboux function).
Continuous functions have the IVP, by the Int Value Thm, but the converse is false. Continuous functions have antiderivatives, by the Fund Thm of Calc, but derivatives need not be continuous. And functions which are derivatives have the IVP, by Darboux's Theorem.
The implication I haven't been able to find is if functions with the IVP must have antiderivatives. The classical counterexamples for these types of questions are functions $f(x)=x^n\sin\left(\frac1x\right)$. But if we look at, say $f(x)=\sin\left(\frac1x\right)$, this has the IVP, but what methods are there to show this has no antiderivative since we can't use Darboux's Theorem?
Best Answer
A function with IVP does not necessarily have an antiderivative. The key reason is that an IVP function can be extremely discontinuous, while a function with an antiderivative cannot.
In a bit more detail, the Conway Base 13 function is a function that has the IVP, but is nowhere continuous.
On the other hand, if $f$ has an antiderivative then $f$ must be a Baire class 1 function (that is, a pointwise limit of continuous functions), which we can see just by looking at the definition of being a derivative. The set of discontinuities of a Baire class 1 function is always a meagre set.
Put together, these two things show that the Conway function is an IVP function without an antiderivative.