Tightness for Nyquist Frequency

Question

If $f$ is a function whose Fourier transform is supported on $[-R,R]$, then $f$ is determined by it's values $\{ f(2nR) : n \in \mathbf{Z} \}$. In particular, if the Fourier transform is supported on $[-1/2,1/2]$, then $f$ is determined by it's values $\{ f(n) : n \in \mathbf{Z} \}$. Is there a proof that $f$ is not necessarily determined by it's values $\{ f(n/\lambda) : n \in \mathbf{Z} \}$ for any $\lambda < 1$? If the proof is too technical to give here, where might I find a proof?

Answer 1

I found an answer. Fix $f_0 \in \mathcal{S}(\mathbf{R})$. Then the Poisson summation formula, appropriately rescaled, tells us that for each $\xi \in \mathbf{R}$,

$$ \sum_{n = -\infty}^\infty f(n/\lambda) e^{-2 \pi n i \xi} = \lambda^d \sum_{n = -\infty}^\infty \widehat{f}(\xi - \lambda n). $$

One can determine all the coefficients $\{ f(n/\lambda) \}$ if one knows the right hand side for all values $\xi \in \mathbf{R}$. Thus if $f_1,f_2 \in \mathcal{S}(\mathbf{R})$ are distinct functions such that $\widehat{f_1}$ and $\widehat{f_2}$ are supported on $[-1/2,1/2]$, but are equal to one another at a periodization of scale $\lambda$, then $f_1(n/\lambda) = f_2(n/\lambda)$ for any $n \in \mathbf{Z}$. It is certainly possible to choose such functions if $\lambda < 1$.