Does the distributional derivative of the Dirac Comb have the same properties as a single Dirac Delta?

Question

A single Dirac delta has a distributional derivative $\delta'$ defined in the sense that $$ \int_{-\infty}^{\infty} \textrm{d}x \, \delta'(x) \, \phi(x) = -\int_{-\infty}^{\infty} \textrm{d}x \, \delta(x) \, \phi'(x) \, . $$ I understand this comes from the integral of the total derivative of $\delta(x) \, \phi(x)$ vanishing, $$ \int_{-\infty}^{\infty} \textrm{d}x \, \frac{\textrm{d}}{\textrm{d} x} \big( \delta(x) \, \phi(x) \big) = \delta(x) \, \phi(x) \, \big|_{-\infty}^{\infty} = 0 \, , $$ which means the product rule for differentation gives the above expression defining $\delta'$. If we instead consider the Dirac comb $$ \mathrm{III}(t) = \sum_{k=-\infty}^{\infty} \, \delta(x - kT) \, , $$ which is periodic with period $T$, is the same total derivative trick valid - $$ \int_{-\infty}^{\infty} \textrm{d}x \, \frac{\textrm{d}}{\textrm{d} x} \big( \textrm{III}(x) \, \phi(x) \big) = 0 \, ? $$ Allowing us to define the distributional derivative of the Dirac comb analogously to the Dirac delta $$ \int_{-\infty}^{\infty} \textrm{d}x \, \textrm{III}'(x) \, \phi(x) = -\int_{-\infty}^{\infty} \textrm{d}x \, \textrm{III}(x) \, \phi'(x) \, . $$

Answer 1

In fact, if stop reasoning in terms of integrals ans start reasoning in terms of distributions and distributional derivatives, then this result is quite easy to show. Let us note $\phi $ - am arbitrary test function (i.e. compactly supported and $C^\infty$)

We start with a delta-distribution $\delta_p$ given by $$\langle \delta_p,\phi\rangle = \phi(p)$$ and its derivative $\delta_p'$, for which the definition of distributional derivative (yes, it has common roots with integration by parts) gives $$\langle \delta_p' ,\phi\rangle=-\langle \delta_p,\phi'\rangle = -\phi'(p).$$

Now the Dirac comb $B = \sum_{k\in\Bbb Z} \delta_{kT}$ is a well-defined distribution, because $$\langle B,\phi\rangle = \sum_{k\in\Bbb Z} \phi(kT),$$ and the latter series converges because $\phi$ is compactly supported. I will omit the continuity part of the definition, it is an easy exercise.

Since $B$ is a distribution, it has a distributional derivative - it is one of the major resons to use distributions. Moreover, we have an explicit definition: $$\langle B',\phi\rangle=-\langle B,\phi'\rangle = -\sum_{k\in\Bbb Z} \phi'(kT),$$ and the latter series converges for the exact same reasons as we mentioned earlier.

Finally, by testing against all possible test functions $\phi$ you can conclude that $B'$ has the form $ \sum_{k\in\Bbb Z} \delta'_{kT}$.

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Another approach would be to show that the distributions $B_n= \sum_{k\in\Bbb Z,\,|k|\le n} \delta_{kT}$ converge to $B$ in the distributional sense (easy to show by definition of this convergence). It implies - another great property of distributional derivatives - that $B'_n$ converges to $B'$, and the explicit formula for $B'_n$ is quite obvious.