Let E be an arbitrary ellipse in $\mathbb{R}^2$ with boundary $\partial G$. We denote by $E_T$ the dilated ellipse $$ \{ Tx:x \in E \}. $$ Furthermore, let $\tau(x)$ be the unit outward normal vector of $E$ at $x\in \partial G$.

We assume that the sum $$ \sum_{k\in \mathbb{Z}^2 : k\cdot \tau(x)<0}f(k) $$ is equal to A. [$\cdot$ is the standard scalar product]

I would like to proof that $$ \lim_{T \rightarrow \infty} \sum_{ k\in \mathbb{Z}^2 : k\in E_T-Tx} f(k)=A. $$

Can anyone help me find an approach or give a reference to literature for these kinds of problems.


1 Answer 1


By definition, requiring that the sum : $$\sum_{k\in \mathbb{Z}^2 : k\cdot \tau(x)<0}f(k)$$ converges to $A$ means that for any non-decreasing sequence of finite subsets $X_n \subset \{k\in \mathbb{Z}^2 : k\cdot \tau(x)<0\}$, we have : $$\lim_{n\to +\infty} \sum_{k\in X_n} f(k) = A$$

Therefore, we just have to check that $\{k\in\mathbb Z^2 |k\in E_T - Tx\} $ is a finite subset of $ \{ k\in\mathbb Z^2 |k\cdot \tau(x) < 0\}$ and that : $$\bigcup_{T\to \infty} \{k\in\mathbb Z^2 |k\in E_T - Tx\} = \{ k\in\mathbb Z^2 |k\cdot \tau(x) < 0\}$$

It is easy to see that this is true.

  • $\begingroup$ Thank you very much! Do you know of a book/paper that discusses your definition of convergence? $\endgroup$
    – HyyFly
    Sep 22 at 17:39
  • $\begingroup$ This is basically Lebesgue integration using the counting measure. $\endgroup$ Sep 24 at 17:58

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