antiderivative always exists? [closed]

Question

It is well known that many real valued real functions are not Riemann integrable on subsets of $\mathbb{R}$, but formally an antiderivative may still exist. May I see an example of a function having no antiderivative? Or does any function (without additional hypothesis) always have an antiderivative?

Edit: Does a continuous function always satisfy Choquet's characterization? If yes, take $f$ differentiable and such that $f'$ is bijective. In order to explicitly compute the antiderivative of $f'^{-1}$ I find myself needing that $f'$ is also differentiable. Could such an additional hypothesis be ruled out somehow?

Answer 1

There are elementary necessary conditions on real functions to be the derivative of a real function. For instance, Darboux's theorem states that a derivative must satisfy the intermediate value theorem. The Baire category theorem implies that a derivative is continuous on a dense set. It allows to find functions with no antiderivative easily.

There is also a characterization due to Choquet :

A real function $f$ admits an antiderivative if and only if there exists an homeomorphism $\phi$ such that $f \circ \phi$ is of Baire class 1 and satisfies the intermediate value theorem.

Reference : French Wikipedia page on Gustave Choquet

Answer 2

Any distribution $T$ on the real line has an anti-derivative, i.e. there exists a distribution $S$ such that $$S'=T\tag{$\ast$}.$$ Here is a constructive proof: with a given $T$, define the distribution $S$ by $$ \langle S, \phi\rangle_{\mathscr D',\mathscr D }=-\langle T, \psi_\phi\rangle_{\mathscr D',\mathscr D }, \quad \text{with}\quad (\psi_\phi)(x)=\int_{-\infty}^x \phi(t) dt-\chi_0(x)\int_{\mathbb R} \phi(t) dt, $$ where the function $\chi_0$ smooth, equal to 1 on $[\max\text{supp} \phi,+\infty)$ and equal to $0$ on $(-\infty, \min \text{supp}\phi]$. This makes sense since the function $\psi_\phi$ is smooth (obvious) and also compactly supported: if $x\ge \max\text{supp} \phi$, then $$ (\psi_\phi)(x)=(1-\chi_0(x))\int_{\mathbb R} \phi(t) dt=0. $$ If $x\le \min\text{supp} \phi$, then $ (\psi_\phi)(x)=-\chi_0(x)\int_{\mathbb R} \phi(t) dt=0. $ We check now $S'=T$. Indeed we have $$ \langle S', \phi\rangle_{\mathscr D',\mathscr D }= -\langle S, \phi'\rangle_{\mathscr D',\mathscr D } =\langle T, \psi_{\phi'}\rangle_{\mathscr D',\mathscr D }. $$ We note that $\int_\mathbb R\phi'(t) dt =0$, so that $ (\psi_{\phi'})(x)=\phi(x) $ and thus, we find $$ \langle S', \phi\rangle_{\mathscr D',\mathscr D }=\langle T, \phi\rangle_{\mathscr D',\mathscr D },\quad \text{i.e.}\quad S'=T. $$ A simple addendum: if $S_1, S_2$ are two antiderivatives, the $S_2-S_1=\text{constant}.$