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If $f(x)=g(x)$, then $f'(x)=g'(x)$, is it right?

Assume there is a function $f(x)=m(x)\delta(x)$，where $m(x)$ is any function and $\delta(x)$ is Dirac-δ function. We know that $m(x)\delta(x)=m(0)\delta(x)$, so we have $f(x)=m(0)\delta(x)$.

But why $[m(0)\delta(x)]'\ne[m(x)\delta(x)]'$? $[m(0)\delta(x)]'=m(0)\delta'(x)$, and $[m(x)\delta(x)]'=m'(0)\delta(x)+m(x)\delta'(x)$, and I think it's obviously not equal.

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  • $\begingroup$ @RafaelWagner There is still a product rule for the derivative of the product of a distribution and a $C^\infty$ function. See for instance users.math.yale.edu/~auel/papers/docs/distalk1.pdf Theorem 13 $\endgroup$
    – Ian
    Mar 25, 2017 at 1:13
  • $\begingroup$ Why do you think that they are different? Although a care is needed as we have to work with distributions and their derivatives, your computation is essentially justified to give the identity $$ m(0)\delta'(x) = m'(0)\delta(x) + m(x)\delta'(x). $$ In some sense this is simply a complicated way of writing the product rule. $\endgroup$
    – Sangchul Lee
    Mar 25, 2017 at 1:22
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    $\begingroup$ @SangchulLee I think the point is to ask: consider $m(x)=kx+1$. Why, if $m \delta$ only "sees" that $m(0)=1$, can $(m \delta)'$ "see" $m'(0)=k$? $\endgroup$
    – Ian
    Mar 25, 2017 at 1:32
  • $\begingroup$ @Ian, I guess that this is the point where we have to depart from a naive picture of Dirac delta and move to the distribution theory, where the fact that $m\delta'$ can see both $m(0)$ and $m'(0)$ can be understood as the product rule written in the distribution side. $\endgroup$
    – Sangchul Lee
    Mar 25, 2017 at 1:52
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    $\begingroup$ @SangchulLee I think I have a cleaner explanation that I'm writing up now. It's more about $\delta'$ seeing more than it should be able to see, and the first term canceling that out. $\endgroup$
    – Ian
    Mar 25, 2017 at 1:53

1 Answer 1

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They are actually the same: $[m\delta]'=m(0)\delta'$ and $[m\delta]'=m'\delta+m\delta'$.

The first equality follows from the definition of the distributional derivative:

$$[m\delta]'(\phi)=-[m\delta](\phi')=-m(0)\phi'(0)=m(0)\delta'(\phi).$$

The second equality follows from the distributional product rule.

How can we reconcile these, when it seems like the first version does not involve $m'(0)$ while the second one does? We notice that the $m\delta'$ term in the second version has its own product rule:

$$[m\delta'](\phi)=\delta'[m\phi]=-m'(0)\phi(0)-m(0)\phi'(0).$$

The first term here serves to cancel out the first term arising in the distributional product rule (the one depending on $m'(0)$). In the end only the second term survives.

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