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The distributional derivative of a function $f$ is the function $g$ that verifies $\int f\phi'=-\int g\phi$. In particular, if f is differentiable, the distributional derivatives is the same as the standard derivative. Now similarly, if $\mu$ is a measure, I think its distributional derivative is defined as the measure $\nu$ such that $\int \phi'd\mu = -\int \phi d\nu$. I was wondering if the distributional derivative of the measure $\mu$ had something to do with the Radon-Nykodym derivative, or if they absolutely have nothing to do with each other (exept for the word "derivative").

edamondo
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1 Answers1

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For the Radon-Nikodym derivative, see e.g. this question.

Regarding your definition of the distributional derivative: Keep in mind that the equation \begin{equation} \int f \phi' = - \int g \phi \end{equation} has to hold for all $\phi$ in some test function space e.g. $C^\infty_c$ or the space of Schwartz functions leading to related but different notions of weak derivatives.

Now, every sufficiently regular measure $\mu$ can be regarded as a distribution. Note that, e.g. measures of exponential growth are not tempered distributions such that, again, the space of test functions matters!

Regarding the derivative: Consider the Dirac measure $\delta_0$ at the origin. Its distributional derivative is the linear functional $\phi \mapsto -\phi'(0)$. Suppose that this linear functional could be represented by a measure. Then (by working out the technical details) you will find that, by the Riesz-Markov-Kakutani representation theorem, the functional should extend continuously to $C_c$ (the space of continuous functions with compact support. But it is now easy to construct a sequence of $C_c^\infty$ functions $\phi_n$ converging - in $\mathbf{C_c}$ - to a function $\phi$ that is not differentiable at the origin in such a way that $\sup_n \vert \phi_n'(0) \vert$ is unbounded. This is a contradiction!

Consequently, the distributional derivative of $\delta_0$ cannot be represented by a measure.

iolo
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  • Thank you for your explanation. I am new to the subject so maybe my questions makes no sense, however: if instead we take the distributional derivative of a function, will it always be a measure? – rod Sep 17 '23 at 11:46
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    It depends on the regularity of the function. If you look at those with bounded variation on $\mathbb{R}$, the first distributional derivative will be a signed Radon measure. On $\mathbb{R}^n$ the gradient will be a vector-valued Radon measure. Considering e.g. $x \to sin(1/x)$ (unbounded variation), its distributional derivative is not a signed measure. – iolo Sep 18 '23 at 12:16