Proof of Aliasing Theorem

To show:

$$\zbox {\hbox{\sc Downsample}_N(x) \;\longleftrightarrow\;\frac{1}{N} \hbox{\sc Alias}_N(X)}$$

or

From the DFT case [235], we know this is true when $x$ and $X$ are each complex sequences of length $N_s$, in which case $y$ and $Y$ are length $N_s/N$. Thus,

$$x(nN) \;\longleftrightarrow\; Y(\omega_k N) = \frac{1}{N} \sum_{m=0}^{N-1} X\left(\omega_k + \frac{2\pi}{N} m \right), \; k\in [0,N_s/N)$$

where we have chosen to keep frequency samples $\omega_k$ in terms of the original frequency axis prior to downsampling, i.e., $\omega_k = 2\pi k/ N_s$ for both $X$ and $Y$. This choice allows us to easily take the limit as $N_s\to\infty$ by simply replacing $\omega_k$ by $\omega$:

$$x(nN) \;\longleftrightarrow\; Y(\omega N) = \frac{1}{N} \sum_{m=0}^{N-1} X\left(\omega + \frac{2\pi}{N} m \right), \; \omega\in[0,2\pi/N)$$

Replacing $\omega$ by $\omega^\prime =\omega N$ and converting to $z$-transform notation $X(z)$ instead of Fourier transform notation $X(\omega)$, with $z=e^{j\omega^\prime }$, yields the final result.