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Are all functions that have a primitive differentiable?

For some background, I know that not all functions that are integrable are differentiable. For example:

$$ f = \begin{cases} 0 & x \neq 0 \\ 1 & x = 0 \end{cases} $$

is integrable over $\mathbb{R}$, and $\int_{a}^{b} f(x) dx = 0$. However, a function $F(x)$ such that $F'(x) = f(x) \space \forall x \in \mathbb{R}$ does not exist.

But I can't find a counterexample to the statement: "All functions that have a primitive are differentiable". These functions are guaranteed to be continuous due to the fundamental theorem of calculus, but not every continuous function is differentiable, hence the question.

Thanks !

Edit: here is a very important comment from @HenningMakholm that I thought would be useful for other students encountering this question

Also, (real) functions that have a primitive are not necessarily continuous. For example,

$$ f(x) = \begin{cases} 0 & \text{for } x = 0 \\ 2x \sin(\frac{1}{x}) - \cos(\frac{1}{x}) & \text{otherwise} \end{cases} $$

is discontinuous at $x=0$, but nevertheless the derivative of

$$ F(x) = \begin{cases} 0 & \text{ for } x = 0 \\ x^2 \sin(\frac{1}{x}) & \text{ otherwise} \end{cases} $$

Functions that have a primitive do have the intermediate-value property but that is weaker than being continuous.

amWhy
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ninivert
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  • Any continuous function on a closed bounded interval has a primitive, but not all continuous functions are differentiable. – Andrés E. Caicedo Dec 29 '18 at 13:50
  • Take a function that is $C^1$ but not $C^2$. – dmtri Dec 29 '18 at 13:51
  • @dmtri That is not enough. Your conditions allow for the existence of a discontinuous second derivative. – Andrés E. Caicedo Dec 29 '18 at 13:52
  • @Andres, you are right! I was focusing more on differentiability – dmtri Dec 29 '18 at 13:55
  • See also https://math.stackexchange.com/questions/292275/discontinuous-derivative – GEdgar Dec 29 '18 at 15:04
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    Note that, famously, this **is** true for functions defined on an open subset of $\mathbb C$. – hmakholm left over Monica Dec 29 '18 at 15:50
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    Also, (real) functions that have a primitive are _not_ necessarily continuous. For example, $$ f(x) = \begin{cases} 0 & \text{for }x=0 \\ 2x\sin(1/x) - \cos(1/x) & \text{otherwise} \end{cases} $$ is discontinuous at $x=0$, but is nevertheless the derivative of $$ F(x) = \begin{cases} 0 & \text{for }x=0 \\ x^2\sin(1/x) & \text{otherwise} \end{cases} $$ Functions that have a primitive [do have the intermetdiate-value property](https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)) but that is weaker than being continuous. – hmakholm left over Monica Dec 29 '18 at 15:55
  • @HenningMakholm I'm integrating (no pun intended) your comment in the question for future reference. Thanks a lot ! – ninivert Dec 29 '18 at 17:22
  • +1 because you make the question useful for a larger class of people –  Dec 29 '18 at 21:55

3 Answers3

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A primitive for $f(x)=x^{1/3}$ is $F(x)=\frac{3}{4}x^{4/3}$, but $f'(0)$ doesn't exist, because $$\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0}\dfrac{x^{1/3}}{x}=\lim\limits_{x\to0}\dfrac{1}{x^{2/3}}=\infty$$

Martín Vacas Vignolo
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Even more simply: $f(x) := |x|$.

The_Sympathizer
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    Of course, how did I not think about that ? I already accepted the @Martin's answer, but thanks for the example ! For further reference, I'll just state here that $\int |x| dx = \frac{1}{2}x|x|$ – ninivert Dec 29 '18 at 15:26
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Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.