# Are all functions that have a primitive differentiable?

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.

• 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
• 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
• 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 but that is weaker than being continuous. – hmakholm left over Monica Dec 29 '18 at 15:55

## 3 Answers

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

• Almost. What you wrote is $f'(x)$ for $x\ne0$; you need an additional argument to ensure that $f'(0)$ does not exist. – Andrés E. Caicedo Dec 29 '18 at 13:55
• @AndrésE.Caicedo I edited to be more clarifying – Martín Vacas Vignolo Dec 29 '18 at 14:02

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

• 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

Any bounded function that has discontinuity at a single point is integrable but of course that function will be non differentiable.

• @Martin has justified my statement. – user630002 Dec 29 '18 at 13:54