The Fourier transform of a "comb function" is a comb function?

Assume an arbitrary periodic function $f(t)$ with period $T$. Consider the Fourier series representation of $f(t)$, in which $\omega_0=\frac{2\pi}{T}$: $$f(t)=\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}$$ Take the Fourier transform of the sides: \begin{align} \mathcal{F}\{f(t)\}=&\mathcal{F}\{\sum_{n=-\infty}^{+\infty}c_n e^{i n \omega_0 t}\}\\ =&\sum_{n=-\infty}^{+\infty}c_n\mathcal{F}\{ e^{i n \omega_0 t}\}\\ F(\omega)=&2\pi\sum_{n=-\infty}^{+\infty}c_n\delta(\omega-n\omega_0) \end{align} This means that the Fourier transform of a periodic signal is an impulse train where the impulse amplitudes are $2\pi$ times the Fourier coefficients of that signal.

With $f(t)=\delta(t)$, the Fourier series coefficients are $c_n=\frac{1}{T}$ for all $n$.

Hence, $$\mathcal{F}\{\sum_{n=-\infty}^{+\infty}\delta(t-nT)\}=\frac{2\pi}{T} \sum_{n=-\infty}^{+\infty}\delta(\omega-n\omega_0)$$ or in comb notation: $$\boxed{\mathcal{F}\{\text{comb}_T(t)\}=\omega_0\ \text{comb}_{\omega_0}(\omega)}$$

where $$\text{comb}_A(x)\triangleq\sum_{n=-\infty}^{+\infty}\delta(x-nA)$$

Intuitive Explanation

The Comb is a sum of Time Shifted Dirac Delta.

The Fourier Transform of a Dirac Delta is known to be a constant.

The Fourier Transform of a Time Shifted Function is known to be Fourier Transform of the function multiplied by a complex exponential factor which is $ \exp(-i 2 \pi f T) $

Just apply this points to the Comb Function considered as a sum of Time Shifted Dirac Delta with distance $ kT $ and you get a sum of Frequency Shifted exponential functions, each of which multiplied by a constant.

Finally use Euler’s Formula to consider complex exponentials as a periodic sinusoidal function and observe that you have constructive interference only in frequencies which are integer multiple of $ \frac{1}{T} $