Let \(y\in \mathbb{C}^N\), \(S=\supp y\), \(\lvert S\rvert\le s\) and \(\hat{y}_k=0\) for \(s\) “konsekutive” \(k\in \mathbb{Z}_N\), where \(\hat{y}\) is the discrete Fourier transform of \(\hat{y}\).Then \(y\equiv 0\).
Proof
For \(k=t+1, t+2, \ldots , t+s\in \mathbb{Z}_N\) we have \(\hat{y}(k)=0\), i.e.
\[ 0=\sum_{l\in \mathbb{Z}_N} e_l(-k)y(l). \]In matrix notation that is
\[ 0=\bigl([\e^{-2\pi i\frac{l}{N}}]^k\bigr)_{l\in S,\, k=t+1,\ldots , t+s}\cdot y_S. \]Multiply this equation with the scalar \(\e^{2\pi i\frac{t+1}{N}}\). This gives us
\[ 0=\bigl([\e^{-2\pi i\frac{l}{N}}]^k\bigr)_{l\in S,\, k=0,\ldots , s-1}\cdot y_S. \]The matrix is invertible and therefore \(y_S\equiv 0\). Since \(\supp y=S\) the whole vector vanishes.