# "If a function has an antiderivative, then it is continuous." What conditions can we add to make this true?

Definition. We say that a function $$f$$ is antidifferentiable or has an antiderivative if there exists a function $$g$$ such that $$g^{\prime}=f$$.

We know that this result is true:

Proposition. Suppose a function is defined on an interval with more than one number. If that function is continuous, then it is antidifferentiable.

Conjecture. Suppose a function is defined on an interval with more than one number. If that function is antidifferentiable, then it is continuous.

Update: Turns out that Wikipedia has something about this, including several counterexamples showing that the above Conjecture is false. Nonetheless, I still have this question:

Question. What conditions can we add to above Conjecture to make it true?

• consider step function Feb 19, 2020 at 2:35
• @J.W.Tanner: I thought the step function isn't antidifferentiable?
– user694131
Feb 19, 2020 at 2:44
• the step function is not differentiable, but it's the derivative of the ramp function Feb 19, 2020 at 3:01
• @J.W.Tanner: But it isn't because the ramp function isn't differentiable (in particular at $0$).
– user694131
Feb 19, 2020 at 3:03
• the step function is integrable meaning you can calculate the definite integral on any closed interval around the step, as the sum of areas of two rectangles. however is not continuous at the step Feb 19, 2020 at 3:04

No - though you are right in thinking that a counterexample is a bit hard to construct: in particular, something like a step function such as

$$f_1 : \mathbb{R} \rightarrow \mathbb{R},\ f_1(x) := \begin{cases} 0\ \mbox{if x < 0}, \\1\ \mbox{if x \ge 0}\end{cases}$$

is not antidifferentiable on its whole domain, because at all domain points except $$0$$, any antiderivative of such a function has to satisfy

$${F_1}_\mathrm{putative}(x) = \begin{cases} C_1\ \mbox{if x < 0}\\x + C_2\ \mbox{if x > 0}\end{cases}$$

for some constants $$C_1$$ and $$C_2$$ - but there is no way to define a value exactly at $$0$$ that will make this differentiate to $$f_1$$ there, because this function is simply not differentiable at $$0$$ (the closest you can get is if $$C_1 = C_2 = C$$, and then if you define it such that $${F_1}_\mathrm{putative}(0) = C$$, it will be continuous; but it is not differentiable there as it has a corner there).

Instead, you have to get a bit cleverer. A standard counterexample is (if I remember right)

$$f_2: \mathbb{R} \rightarrow \mathbb{R}, f_2(x) := \begin{cases} \sin\left(\frac{1}{x}\right)\ \mbox{if x \ne 0}\\ 0\ \mbox{if x = 0} \end{cases}$$

The antiderivatives of this function satisfy

$$F_2(x) = \begin{cases}x \sin\left(\frac{1}{x}\right) - \mathrm{Ci}\left(\frac{1}{x}\right) + C\ \mbox{if x \ne 0}\\ C\ \mbox{if x = 0}\end{cases}$$

including at $$x = 0$$, for some constant $$C$$. It can be checked that these functions are differentiable at $$x = 0$$, with derivative $$F'_2(0) = 0$$.

ADD: Looks like I remembered wrong - commenters above showed as more "standard" example that where we start with $$F_3(x) := \begin{cases}x^2 \sin\left(\frac{1}{x}\right)\ \mbox{if x \ne 0}\\ 0\ \mbox{if x = 0}\end{cases}$$ as antiderivative ... but meh! The above works and its exposition to me follows the direction of the implication in the question better: construct a weird "derivative" and then try and antidifferentiate it.