I) Consider a function defined only on the vertices of a 1D regular lattice: $f_i \equiv f(x_i)$ for all $x_i$, $i \in \{ 1, 2, ..., N \}$ and $x_{i+1} - x_i = a$, where $a > 0$ is the “lattice spacing”.
The discrete Fourier transform (DFT) of the function is defined as $$ \tilde{f}(k) = \sum_{x_i} f(x_i) e^{- \mathrm{i} k x_i} \quad\quad (1) $$ where k is the wave-vector; the inverse DFT reads $$ f(x_i) = \frac{1}{N} \sum_{k} \tilde{f}(k) e^{\mathrm{i} k x_i} \quad\quad (2) $$ If $f$ is periodic with periodicity $Na$, that is $f(x_i + Na) = f(x_i)$, then the allowed k-values are $$ k_m = \frac{2\pi m}{Na}, m \in \mathbb{Z} \quad\quad (3) $$
Furthermore, since $f(k_m) = f(k_m + 2\pi/a)$, it is sufficient to consider $k_m$ with $m = 0, 1, ..., N - 1$; that is, wave-vectors are restricted to the “first Brillouin zone” (1BZ).
II) Consider the situation where the function $f$ is defined on $x_i$ and $x_i + b$ for all $x_i$, $i \in \{1, 2, \cdots, N\}$, where, as before, $x_{i+1} - x_i = a$, and $0 < b < a$.
$f$ is still periodic with periodicity $Na$, that is $f(x_i + Na) = f(x_i)$, therefore, the allowed k-values are $k_m = \frac{2\pi m}{Na}, m \in \mathbb{Z}$, as before. Yet, in contrast to case I), $f(k_m) \neq f(k_m + 2\pi/a)$ and hence, the wave-vectors are not restricted to the 1BZ.
Question: Is it still possible in case II) to find a countably-finite “restricted” set of k-values as in case I)?