If $f,f',f''$ are bounded a.e., is $f'$ of bounded variation everywhere?

Question

Assume the function $f$ is such that everywhere except in $0$:

• $f$ is bounded on $\mathbb{R}$
• $f$ is twice differentiable everywhere except in $0$
• $f'$ and $f''$ are bounded everywhere except in $0$

Question I the derivative of $f$, in the weak sense, of bounded variation?

What I think $f'$ and $f''$ are not defined on $\mathbb{R}$ but by by extension, in the measure sense, $f'(t)=v(t)+\alpha H(t)$ where $v$ is differentiable on $\mathbb{R}$ and $H$ is the Heaviside function. Similarly, $f''(t)=a(t)+\alpha\delta(t)$ where $a=v'$ and $\delta$ is the Dirac function.

$\alpha H$ is of bounded variation, $a$ is bounded on $\mathbb R$, so $v$ is Lipschitz hence of bounded variation, and so is their sum $f'$.

Does it make sense?

Answer 1

I don't think what "by by extension, in the measure sense" means, so I'm ignoring everything after that.

1. Since $f''$ is bounded on $(0,\infty)$, it follows that $f'$ is Lipschitz on $(0,\infty)$. Hence, $f'$ has bounded variation on any interval of the form $(0,b]$.
2. Since $f''$ is bounded on $(-\infty,0)$, it follows that $f'$ is Lipschitz on $(-\infty,0)$. Hence, $f'$ has bounded variation on any interval of the form $[a,0)$.
3. Regardless of the value at $0$ (as long as it's somehow defined), $f'$ is BV on $[a,b]$; this follows by considering any partition of $[a,b]$, inserting $0$ into the partition, and adding things up.

But check whether your definition of BV requires the function to be defined everywhere (since $f'(0)$ appears to be undefined).