Harmonic Analysis Paper Question This question is not homework, I am simply trying to decipher a proof in a paper.
Let $\varepsilon>0$ be fixed but arbitrary.  Define a function $\phi$ via its Fourier transform, where $\widehat{\phi}$ will be a mesa function with the following properties:
(1) $\widehat{\phi}\in C^\infty(\mathbb{R})$
(2) $\widehat{\phi}(\xi)=1$ for $\xi\in[-\pi+\varepsilon,\pi-\varepsilon]$
(3) supp$(\widehat{\phi})\subset[-\pi-\varepsilon,\pi+\varepsilon]$
(4) $\widehat{\phi}$ goes from 1 to 0 monotonically on the intervals $[-\pi-\varepsilon,-\pi+\varepsilon]$ and $[\pi-\varepsilon,\pi+\varepsilon]$.
(5) $\underset{n\in\mathbb{Z}}\sum\widehat{\phi}(\xi-2\pi n)=1$ for every $\xi\in\mathbb{R}$.
A lot of typing to say, this is a very commonly constructed function.  Now, assume $f$ is as nice a function as you would like to make calculations go smoothly, and define $g$ as follows:
$$g(x):=\underset{j\in\mathbb{Z}}\sum f(jh)\phi\left(\frac{x}{h}-j\right)$$
$h>0$ is fixed but again arbitrary.
Two questions: First they claim that $g$ interpolates $f$ on $h\mathbb{Z}$.  That is, $g(kh)=f(kh)$ for $k\in\mathbb{Z}$.
Plugging in gives $g(kh)=\underset{j\in\mathbb{Z}}\sum f(jh)\phi(k-j)$.  I am not sure why all terms but the $j=k$ term should go away?
Second, they split $f$ into the sum of two functions: $f=f_0+f_1$, where they define
$$\widehat{f_0}(\xi)=\widehat{f}(\xi)\widehat{\phi}(2h\xi)$$
and define
$$g_0(x)=\underset{j\in\mathbb{Z}}\sum f_0(jh)\phi\left(\frac{x}{h}-j\right)$$
And their claim is that
$$\widehat{g_0}(\xi)=\underset{j\in\mathbb{Z}}\sum\widehat{f_0}\left(\xi-\frac{2\pi j}{h}\right)\widehat{\phi}(h\xi)$$
I assume this is some clever use of the Poisson Summation Formula though it escapes me.  Again, a direct calculation yields
$$\widehat{g_0}(\xi)=\underset{j\in\mathbb{Z}}\sum hf_0(jh)e^{-ij\xi h}\widehat{\phi}(h\xi)$$
So why are these two sums the same?
(The Fourier Transform convention used is $$\widehat{f}(\xi)=\int_\mathbb{R}f(x)e^{-ix\xi}dx$$
Thanks for any help (and for reading this monstrosity of a question).
The paper is 
T. Hangelbroek, W. Madych, F. J. Narcowich and J. D. Ward, Cardinal Interpolation with Gaussian Kernels, J. Fourier Anal. Appl., 18 (2012), no. 1, 67--86
http://www.springerlink.com/openurl.asp?genre=article&id=doi:10.1007/s00041-011-9185-2
The proof starts on page 72.
 A: Answer 1: $\phi(k)=0$ when $k\in \mathbb Z\setminus\{0\}$. In other words, $\widehat \phi$ is orthogonal to $e^{i\xi k}$. 
Indeed, since $e^{i\xi k}$ is $2\pi$-periodic, we can chop $\widehat \phi$ into pieces of size $2\pi$, translate them all to $[-\pi,\pi]$, and find that they add up to $1$. 
$$
\int_{\mathbb R} e^{i\xi k} \widehat \phi(\xi)\,d\xi = \sum_{j\in\mathbb Z} \int_{2\pi j-\pi}^{2\pi j+\pi}  e^{i\xi k} \widehat \phi(\xi)\,d\xi = \sum_{j\in\mathbb Z} \int_{-\pi}^{\pi}  e^{i\xi k} \widehat \phi(\xi+2\pi j)\,d\xi = \int_{-\pi}^{\pi}  e^{i\xi k} \,d\xi =1
$$
Answer 2: first, your definition of $g_0$ has a typo: $f$ should be $f_0$. Second, 
  $$\sum_{j\in\mathbb Z} f_0(jh) e^{-i\xi jh}=\sum_{j\in\mathbb Z} \widehat {f_0}(\xi-2\pi j/h)$$ is indeed the Poisson summation formula in the form with dual lattices (at which Wikipedia hints, but does not state). The point is that the dual lattice of $h\mathbb Z$ is $h^{-1}\mathbb Z$. The formula is stated with lattices here. Or you can derive it by applying the standard form of Poisson summation  to $F_0(x)=f_0(hx)e^{-i\xi h x}$.
