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Let $\Omega\subset\mathbb{R}^n$ be open and $f$ be a locally integrable function. The distribution associated with $f$, $\Lambda_{f}\in D'(\Omega)$, is defined via \begin{equation} \Lambda_{f}(\phi)=\int f\phi, \end{equation} where $\phi\in D(\Omega)$ is a test function.

For $\mu$, a regular measure, $\Lambda_{\mu}$ is defined to be $\Lambda_{\mu}(\phi)=\int\phi d\mu$.

The distribution derivative of $f$ is the distribution defined by \begin{equation} (D^{\alpha}\Lambda_{f})(\phi)=(-1)^{|\alpha|}\Lambda_{f}(D^{\alpha}\phi), \end{equation} where $D^{\alpha}$ is the derivative operator with multi-index $\alpha$.

One natural question is the commutability of $D^{\alpha}$ and $\Lambda$, that is, whether \begin{equation} D^{\alpha}\Lambda_{f}=\Lambda_{D^{\alpha}f}. \end{equation}

If $f\in C^{N}(\Omega)$ for some $N>|\alpha|$, then this is true. Rudin says in general this is false by giving a counterexample on $\mathbb{R}$. He takes $f$ to be a left-continuous function of bounded variation, then $Df$ exists a.e. Then he shows $D\Lambda_{f}(\phi)= \Lambda_{\mu}$, where $\mu([a,b])=f(b)-f(a)$.

This shows $D$ and $\Lambda$ commute only if $f$ is absolutely continuous.

However, this example is not quite satisfactory in the sense Rudin interprets $D$ as the a.e. derivative, rather than a regular derivative. In fact, if $f$ is differentiable then it is absolutely continuous.

A good counterexample would be one in which $f\in C^{N-1}(\Omega)$ with $N=|\alpha|$ and the commutativity fails.

Thanks very much!

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1 Answer 1

up vote 1 down vote accepted

The Cantor function belongs to $C^0$ and its distributional derivative of order $1$ is not described by the pointwise derivative (which exists a.e. This qualifies as "good counterexample" for $N=1$ under the terms of your question.

Counterexamples for higher values of $N$ can be obtained by taking the indefinite integral of the Cantor function $N-1$ times.

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