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Prove that the function $$g(x)=\begin{cases} \sin (\frac 1x) & x \neq 0 \\ 1 & x=0 \end{cases}$$ is not a derivative of any function that is differentiable along the whole line.

We want to prove that $g$ has no antiderivative. Assume the opposite - let $g$ have an antiderivative $G$. Every antiderivative has the Darboux property, so for any $a,b$ in the domain of $G$, if $G(a)G(b)<0$, then there is $c\in (a,b)$ that $G(c)=0$.

I further tried to use mean value theorem to contradict the Darboux property assumption, but I didn't get anything meaningful. Could you please help me?

Sine of the Time
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qerty149
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1 Answers1

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This $g$ obeys the intermediate value (Darboux) property. To see this, note that it is continuous everywhere except at $x = 0$, but in any interval around $0$, no matter how small, it attains all values in $[-1, 1]$ and no other values (exercise: explain why this implies $g$ has the Darboux property).

Your idea to start with an antiderivative $G$ is a good one. Note that we can explicitly define what the antiderivative is at all points $x \neq 0$: $G(x) = \int_{0}^{x}\sin(1/t)dt$. This follows by the fundamental theorem of calculus, since $g$ is continuous for $x \neq 0$. This means that, for all $x \neq 0$, we have that $G'(x) = g(x)$, and further, we know that any antiderivative will equal (some translation of) $G$ for all $x \neq 0$. This means that, if $G$ fails to be an antiderivative for $g$, the failure needs to happen at $x = 0$.

What does it mean for $G$ not to be an antiderivative of $g$ at $x = 0$? It means precisely that $G'(0) \neq g(0)$. You have an integral expression for $G(x)$, which means you can compute $G'(0)$, and if you can show that it's distinct from $g(0)$, you're done. To that end, The antiderivative of $\sin(1/x)$ shows that $G'(0) = 0$, and so $g$ admits no antiderivative.

Nicholas Priebe
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