An Example Where $Df$ Is Not Integrable

Question

I have been trying to find an example where $f$ is continuous on $[0,1]$ and is differentiable on $(0,1)$, however, $Df$ (gradient) is not integrable on $[0,1].$

I thought of one example: $f(x,y)=arctan(xy).$ Am I right? Some more examples more help.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$Some hints, on the assumption that "integrability" means "Riemann integrability". (Throughout, $f$ denotes a continuous, real-valued function on $[0, 1]$ that is differentiable on $(0, 1)$, and $g$ denotes an arbitrary function on $[0, 1]$.)

1. What properties are necessary for $g$ to be integrable? (The types of condition to consider include: continuity, continuity except at finitely many points, boundedness, etc.)

Rationale: If you construct a function $f$ such that $g := f'$ does not satisfy a necessary condition for integrability, you're done.

2. What properties are sufficient for $g$ to be integrable? (Again, consider properties such as continuity, continuity except at finitely many points, boundedness, etc.)

Rationale: If you construct a function $f$ such that $g := f'$ satisfies a sufficient condition for integrability, you do not have a counterexample. In retrospect, this means that if you start writing down differentiable functions naively, you're sort of unlikely to stumble across an example of the type you seek, because $f'$ is likely to be integrable.

3. The formula $$ g(x) = \begin{cases} x^{2} \sin \frac{1}{x} & x \neq 0, \\ 0 & x = 0, \end{cases} $$ is well-known (and easily shown; calculate the derivative yourself) to define a function differentiable everywhere in $\Reals$, but to have discontinuous derivative at $0$.

Unfortunately, the derivative of this $g$ is integrable. Fortunately, a suitable modification (there are multiple possibilities) will produce a function of the type you seek.

(Limiting this answer to the preceding hints is my attempt to help you think systematically without depriving you the pleasure of finding an actual counterexample yourself, and thereby seeing why your function is a counterexample.)