Let $\epsilon > 0$ and $L >1$ such that $\frac{L}{2\epsilon} \in \mathbb{N}$. Take $\Lambda_{\epsilon, L} :=\epsilon\mathbb{Z^{d}}/L\mathbb{Z}^{d}$. Suppose $f \in C^{1}(\mathbb{R}^{d}/L\mathbb{Z}^{d},\mathbb{C})$ and define a sum: $$\epsilon^{d}\sum_{x \in \Lambda_{\epsilon, L}}f(x)$$ Is is possible to prove that: $$\epsilon^{d}\sum_{x \in \Lambda_{\epsilon, L}}f(x) \stackrel{\epsilon \to 0}\to \int_{\mathbb{R}^{d}/L\mathbb{Z}^{d}}f(x)dx$$ using, idk, Riemann sums or something like this? I'm really stuck here. Any help would be appreciated.

  • $\begingroup$ Shouldn't that be $\epsilon^d\sum_{x \in \Lambda_{\epsilon, L}}f(x)$? For positive $f$, as $\epsilon \to 0$ both $\frac 1{\epsilon^d}$ and $\sum_{x \in \Lambda_{\epsilon, L}}f(x)$ go to $\infty$. With that correction, the sum is indeed a Riemann sum for the integral. $\endgroup$ – Paul Sinclair May 24 at 22:19
  • $\begingroup$ @PaulSinclair you are completely right! I will edit it. Thanks! $\endgroup$ – IamWill May 24 at 23:19

With the correction, it is a Riemann sum. For purposes of integration, we can equate the hyper-torus with $[0,L)^d$. Since $f$ is continuous and is being integrated on a compact set (integral is the same for $[0,L]^d$), the integral converges. As such all Riemann sums must converge to the same limit.

A Reimann sum consists of dividing up the domain in non-overlapping regions summing over all the regions the product of the function at a sample point in the region by the region's area. The lattice points of $\Lambda_{\epsilon,L}$ are equally spaced. By the identification with $[0,L)^d$, they are the points $(n_1, n_2, ..., n_d)\epsilon$ for integers $0 \le n_i <\frac L\epsilon$. We can consider that the sample point for a cell $$[n_1\epsilon, (n_1+1)\epsilon]\times \ldots \times[n_d\epsilon,(n_d+1)\epsilon]$$ whose $d$-volume is $\epsilon^d$. The Riemann sum is therefore $$\sum_{x \in \Lambda} f(x)\epsilon^d$$

Since the norm of this partition $\sim \epsilon$, it goes to $0$ as $\epsilon \to 0$. Therefore the sum must converge to the integral as $\epsilon \to 0$ as well.

FYI - I have no idea why the $2$ is there in the requirement that $\frac L{2\epsilon}$ is integer. What is useful is that $\frac L\epsilon$ is integer. But even that could be dropped and the result would still hold. It would just take some additional argument.

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  • $\begingroup$ Thank you so much! I think the $2$ is here to demand $\frac{L}{\epsilon}$ to be an even number, so the lattice $\Lambda_{\epsilon, L}$ is symmetric about the origin. $\endgroup$ – IamWill May 25 at 16:25

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