If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \tag{1} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \tag{2} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \tag{4} \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \tag{1'} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \tag{2'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \tag{4'} \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?
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1$\begingroup$ You need to consider your formulas in the context of integrals such as $$\int\limits_{-\infty}^\infty f(z)\, \delta(z−a) \, dz=f(a)$$ and $$\int\limits_{-\infty}^\infty f(a)\, \delta(z−a) \, dz=f(a) \int\limits_{-\infty}^\infty \delta(z−a) \, dz=f(a)\times 1=f(a).$$ $\endgroup$Steven Clark– Steven Clark2026-02-27 17:05:10 +00:00Commented Feb 27 at 17:05
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$\begingroup$ These integrals are valid when $f(z)\in C^\infty_c(\Bbb{R})$ for example which is not a condition met by your function $$f(z)=\sum\limits_{n=-\infty}^{\infty} e^{2 i \pi n z}=1+2 \sum\limits_{n=1}^{\infty} \cos(2 \pi n z)$$ which is the Fourier series for the Dirac comb of period $1$. The Fourier series for the Dirac comb doesn't converge in the usual sense, rather it only converges in a distributional sense. $\endgroup$Steven Clark– Steven Clark2026-02-27 17:56:52 +00:00Commented Feb 27 at 17:56
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$\begingroup$ You're over-complicating things by considering the invalid integral $$\int\limits_{-\infty}^\infty f(z)\, \delta\left(z−\frac{1}{2}\right) \, dz\tag{1}$$ where $$f(z)=\lim\limits_{N\to\infty} \left(\sum\limits_{n=-N}^{N} e^{2 i \pi n z}\right)=1+2 \lim\limits_{N\to\infty} \left(\sum\limits_{n=1}^{N} \cos(2 \pi n z)\right)\tag{2}$$ is the Fourier series for the Dirac comb of period $1$ since this series doesn't converge at $z=\frac{1}{2}$ independent of the invalid integral in formula (1) above. $\endgroup$Steven Clark– Steven Clark2026-02-27 18:22:35 +00:00Commented Feb 27 at 18:22
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$\begingroup$ Your formula $$\frac{(-1)^n}{2 \pi i}\, \left(\delta\left(z-\frac{1}{2}\right)-\delta\left(z+\frac{1}{2}\right)\right)$$ corresponds to zero since $$\int\limits_{-\infty}^{\infty} \frac{(-1)^n}{2 \pi i}\, \left(\delta\left(z-\frac{1}{2}\right)-\delta \left(z+\frac{1}{2}\right)\right)\, dz=\frac{(-1)^n}{2 \pi i} \int\limits_{-\infty}^{\infty} \left(\delta\left(z-\frac{1}{2}\right)-\delta \left(z+\frac{1}{2}\right)\right)\, dz=0$$ $\endgroup$Steven Clark– Steven Clark2026-02-28 20:57:26 +00:00Commented Feb 28 at 20:57
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1$\begingroup$ @StevenClark In fact, we need not require that $f\in C_C^\infty$. We only need $f$ to be continuous at $a$ since $\delta(z-a)$ is a distribution with compact support on $\{a\}$. Aside, the objects written as integrals are not integrals. For general distributions (i.e., not necessarily the Dirac Delta.), they represent continuous linear functionals on $D(\mathbb{R})$. $\endgroup$Mark Viola– Mark Viola2026-03-18 19:17:40 +00:00Commented Mar 18 at 19:17
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The identity $$ f(z)\delta_a=f(a)\delta_a $$ is true for any test function $f$ (depending on the situation, these could be from different sets).
In you first example, when you consider $$ \sum_{n\in\mathbb Z}e^{2\pi i n z}, $$ this is not a function at all, as you write yourself, it is a distribution on its own, and what you consider is a product of two distributions, which is quite difficult to define. In most basic settings we assume that the distributions are not allowed to be multiplied between themselves at all, and hence the expression $$ \delta_a \sum_{n\in\mathbb Z}e^{2\pi i n z} $$ does not make any sense.
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$\begingroup$ I understand that you can't multiply non regular distributions together, but multiplication between a distribution and a regular distribution is perfectly defined. Hence, the question is not about product, but about the infinite summation of distributions, and why it doesn't work in each examples, the first of which is just a mere introduction for the second example. $\endgroup$Nolord– Nolord2026-02-28 00:37:37 +00:00Commented Feb 28 at 0:37
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$\begingroup$ @Nolord You lost me very quickly with your arguments. If your first example is not relevant, you can ask another question. The rest of your examples is just complete mess of notation. $\endgroup$Artem– Artem2026-02-28 01:23:27 +00:00Commented Feb 28 at 1:23
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$\begingroup$ The question is very clearly written at the end of the post. It's completely up to you if you want to spend time on it and write a proper answer instead. $\endgroup$Nolord– Nolord2026-02-28 05:02:36 +00:00Commented Feb 28 at 5:02
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2$\begingroup$ @Nolord Please clarify your use of the terms "non regular distribution", "regular distribution", and "distribution" and how they apply to the Fourier series $\sum\limits_{n\in\mathbb Z} e^{2 \pi i n z}$ for the Dirac comb $\mathrm{III_1}(z)$. $\endgroup$Steven Clark– Steven Clark2026-02-28 19:28:09 +00:00Commented Feb 28 at 19:28
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$\begingroup$ The distributional relationship is true whenever $f$ is continuous at $a$ $\endgroup$Mark Viola– Mark Viola2026-03-09 16:41:12 +00:00Commented Mar 9 at 16:41