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In one dimension, we can define the lattice $ f(x+a)=f(x) $ so the lattice is translation invariant.

But, can we give a lattice so $ f(p^{k}x)=f(x) $ for ALL the primes $ P= 2,3,5,7,11,....$ and prime powers $ k \in Z $ so if we increase the 'size' of the lattice, the system remains invariant? Is this possible?

In this case, should we look for the Schrödinger equation, $ -\hbar ^{2} \Theta f(x)+ V(x) $ with $ \Theta = x \frac{d}{dx} $ and $ V(p^{k}x)=V(x)$?

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up vote 2 down vote accepted

The only solution to $$f(p^k x) = f(x) \tag{1}$$ for all $x \in \mathbb{R}$, primes $p$ and $k \in \mathbb{N}$ is $f = \text{constant}$ (if we assume that $f$ is continuous). To see why: take $x' = p^k x$ in (1) to obtain $$f(x') = f(x'/p^k)$$ for all $x'$. Now fix some prime $p$ and take the limit $k \rightarrow \infty$ in this equation to obtain $$f(x') = f(0)$$ for all $x'$.

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I was also thinking that it might be a constant function. Anyway, you make wrong assumption in your proof: $\lim_{y\to 0}f(y) \ne f(0)$ unless the $f$ is continue, and the limit might not even exists. – hwlau Dec 26 '12 at 16:35
@hwlau: indeed I stated that assumption in my post. Smoothness, or at least continuity, of Wightman/Green's functions seems to be a natural requirement, but indeed it's possible to circumvent this issue by constructing specific counterexamples. – Vibert Dec 26 '12 at 20:29
Sorry, I havent noticed about that – hwlau Dec 26 '12 at 20:39

I'm not sure what OP wants, but let me here make the following observation.

Since any natural number $n=\prod_p p^{m_p}$ has a prime factorization, this implies that $\forall n\in\mathbb{N}:f(nx)=f(x)$.

(I implicitly assume that the lattice $\Gamma$ is additively written $\Gamma+\Gamma\subseteq\Gamma$, and that the point $x\in\Gamma$, so that $nx\in\Gamma$ as well.)

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