2

We know that if f is a differentiable fuction, then f' is a Darboux fuction due to the Darboux theorem. However, the Darboux theorem isn't know as the "characterization of fuctions with primitive theorem" so I guess there have to be at least one fuction that is Darboux that doesn't have a primitive.

I'm trying to find an example but I don't get anything. Is this an still open problem? Is there any property in between "darboux" and continuous that can caracterizate if a fuction has a primitive?

Orange
  • 21
  • Relevant: https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be?rq=1 – Sassatelli Giulio Jun 11 '22 at 14:32
  • 1
    An easy example is this: for any $a\in [-1,1]$, take $g_a(x)=\begin{cases}\sin\frac1x&\text{if }x\ne 0\ a&\text{if }x=0\end{cases}$. Every $g_a$ is Darboux, so call $G_a$ their purported primitive such that $G_a(1)=0$. Necessarily $G_a(x)=\int_1^x g_a(t),dt=\int_0^x \sin\frac 1t,dt$ for all $t>0$ and $G_a(0)=\lim_{t\to0^+}\int_1^t \sin\frac1x,dx=\int^0_1\sin\frac1x,dx$. Therefore $a=g_a(0)=\lim_{t\to 0^+}\frac{\int_1^t \sin\frac1x,dx-\int_1^0\sin\frac1x,dx}{t}$. So there is at most one $a$ such that $g_a$ has a primitive (it turns out to be $0$, and in fact $g_0$ has a a primitive). – Sassatelli Giulio Jun 11 '22 at 14:57
  • 1
    For others who might be interested, Darboux functions can be very discontinuous -- discontinuous at every point, and even graphs that are dense subsets of the plane (which is much stronger pathology than discontinuous at every point). Also this latter property along with not being Lebesgue measurable, and even not Lebesgue measurable on any set of positive Lebesgue measure (which is much stronger pathology than not being Lebesgue measurable). See the freely available paper Discontinuous functions with the Darboux property by Israel Halperin (1959). – Dave L. Renfro Jun 11 '22 at 16:02

0 Answers0