I have the following claim that I think have been proved by someone, but I can not find the reference, hence I would like to ask for help. Here is the claim:


Let $f\_1, \ldots, f\_n$ be continuous functions from $\mathbb R^d$ to $\mathbb R$ and $L\_1, \ldots, L\_n$ are $d$-dimensional lattices in $\mathbb R^d$ with the property that $$f\_i(x)=f\_i(x+l),$$ for all $x$ in $\mathbb R^d$ and all $l$ in $L\_i$. 


Suppose that $L\_i$ and $L\_j$ are not in a bigger lattice $L$ for all distinct $i$ and $j$, $f\_1, \ldots, f\_n$ are linearly independent, or one of them is a constant function.


I am thinking of something along the line of Fourier Transform to solve this problem. Any ideas? Thanks in advance.



Periodic functions over different lattices in $\mathbb R^d$ are linearly independent
Although it surely accomplishes nothing more of substance than more elementary-sounding arguments, taking Fourier transforms in the sense of tempered distributions gives a quick outcome: the Fourier transform of $e^{i\xi x}$ is a constant multiple of Dirac $\delta$ at $\xi$, so the Fourier transform of an $L$-periodic functions is supported on (edit: thx @Yoav Kallus) the dual lattice $\check{L}$, as tempered distribution. Thus, for $f\_i$ $L\_i$-periodic, $0=\sum\_i f\_i$ gives $0=\sum\_i \widehat{f}\_i$ as tempered distribution, and we look at supports. Depending what hypotheses you really want/need on the lattices, we easily reach conclusions: for example, if $L\_i\cap L\_j=\{0\}$ for any two $i\not=j$, (edit) then the same is true of the duals and the support of every $\widehat{f}\_i$ appearing must be (contained in) $\{0\}$. 


The pairwise intersections can be intermediate discrete subgroups of $\mathbb R^n$, of course, without being lattices or $\{0\}$.


