# Does a discontinuous function for have an antiderivative?

1. Aug 7, 2013

### Andrax

Can F'(x) =f(x) even if f is not continuous
I tried making a function let
f(x) =5 if x<=5
f(x)=4 if x>5
f is not continuous at 5
Then F(x) =5x x<=5
F(x) =4x+5 x>5
Clearly F is continuous at 5 but F is not differentiable at 5..
So is there a discontinuous function that has F'(x) =f(x) for every x?

2. Aug 7, 2013

### micromass

Staff Emeritus
Sure, some discontinuous functions have anti-derivatives. For example, take the function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by

$$f(x) = \sin(1/x) + \frac{\cos(1/x)}{x}$$

with $x\neq 0$ and $f(0) = 0$. This is discontinuous, but has anti-derivative $F(x) = x\sin(1/x)$.

However, not all discontinuous functions admit anti-derivatives. For example, the function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x) = 0$ if $x\leq 0$ and $f(x) = 1$ if $x>0$. This can be shown to admit no anti-derivative: http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)

The question to characterize the functions which do admit anti-derivatives seems very hard, and is (as far as I know) unsolved.

3. Aug 8, 2013

### lugita15

What's an example of a Darboux function without an antiderivative?

4. Aug 8, 2013

### micromass

Staff Emeritus
It's not difficult to show that one must exist as follows: Any function can be written as the sum of two Darboux functions. Thus if all Darboux functions had an antiderivative, then all functions have one. This is a contradiction.

A specific example would probably be given by http://en.wikipedia.org/wiki/Conway_base_13_function This function is a Darboux function, but it is everywhere discontinuous. But it can be shown using the Baire category theorem, that any derivative is in fact continuous somewhere (and in fact, in a dense and $G_\delta$ set of points).

This might be of interest: http://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be