# Distributional derivative of bounded functions

Let $f$ be a bounded measurable function on $\mathbb{R}$. We may consider its derivative $f'$ as a distribution on $\mathbb{R}$.

Is there a reasonable description of those distributions $\psi$ which arise in this way, i.e. are of the form $\psi = f'$ for some bounded measurable $f$?

For instance, $f'$ need not be a measure (it is a measure iff $f$ has bounded variation). On the other hand, the doublet $\psi = \delta_0'$ is not of this form.

Considering $f \in L^1_{\mathrm{loc}}$ would also be interesting.

More generally, on $\mathbb{R}^n$, I would also like to know which distributions are of the form $\psi = \operatorname{div} F$ for $F : \mathbb{R}^n \to \mathbb{R}^n$ bounded and measurable (or $L^1_{\mathrm{loc}}$).

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This does not answer the question, but seems worth mentioning; a differential form $\omega$ is called flat if $\omega\in L^\infty$ and $d\omega\in L^\infty$ (exterior derivative, taken in the distributional sense). These are studied in Whitney's book Geometric Integration Theory. –  user31373 Jun 12 '12 at 18:10
At least for $\mathbb{R}$, isn't this precisely the Sobolev space $W^{1,\infty}$? And the elements of this space are the bounded Lipschitz continuous functions. –  Vobo Jun 13 '12 at 12:40
@Vobo That happens if you integrate an $L^{\infty}$ function. Differentiation moves you in the opposite direction: from $L^{\infty}=W^{0,\infty}$ you get to $W^{-1,\infty}$. Which is the dual of $W^{1,1}$, but I'm not sure that Nate would accept this as a reasonable description. –  user31373 Jun 13 '12 at 14:57
Ah, I misunderstood the question. Sorry. –  Vobo Jun 13 '12 at 21:41