For $d \in 2\mathbb{N}$ and $0 \leq k,l \leq d-1$ the following double sum turned up in my computations: $$ f(k,l)=\sum_{x=0}^{d-1} \sum_{y=0}^{d-1} \varepsilon(x+y)\exp\left( \frac{2 \pi i\cdot(k x + l y)}{d}\right). $$ Here, $\varepsilon(z)=1$ for $0 \leq z \leq d-1$ and $\varepsilon(z) = -1$ for $d \leq z \leq 2d -2$.
Using Mathematica, I noticed that $f(k,l)\neq 0$ iff $k\cdot l=0$ or $k=l$ (at least for the cases I looked at). This would simplify further computations significantly for me, so my question is:
If this is true in general, why so? I was trying some finite Fourier transform yoga, but I didn't succeed.
Progress so far: Maybe it is helpful to turn $x+y$ into a summation variable. For this we need to expand the double sum as follows (I hope I made no mistake) so that instead of one "square matrix" of double indeces we get three "triangles" of double indeces:
\begin{align} \sum_{x=0}^{d-1} \sum_{y=0}^{d-1}&=\sum_{x=0}^{2d-1} \sum_{y=0}^{2d-1-x} -\sum_{x=d}^{2d-1} \sum_{y=0}^{2d-1-x} -\sum_{y=d}^{2d-1} \sum_{x=0}^{2d-1-y}\\ &=\sum_{z=0}^{2d-1} \sum_{y=0}^{z} -\sum_{z'=0}^{d-1} \sum_{x'=0}^{z'} -\sum_{z'=0}^{d-1} \sum_{y'=0}^{z'}, \end{align} where $z=x+y$, $z'=x+y-d$, $x'=x-d$ and $y'=y-d$. So in the last two summands we have $\varepsilon=1$ and in the first summand we can split the first sum in positive part and a negative part. Using the geometric series then it should follow.