Very recently, I discovered the field "irregularity of distributions". Given a infinite discrete subset X of $\mathbb{R}^2$, the aim is to get a lower bound : $$|\operatorname{Card}(X \cap B) - |B|| \geq C f(B)$$
Here $f$ is a positive function, $C > 0$ is a constant, $|.|$ denote the 2 dimensional lebesgue measure and $B$ is a ball.
In this paper : http://archive.ymsc.tsinghua.edu.cn/pacm_download/117/6400-11511_2006_Article_BF02392553.pdf, using Fourier techniques, it is proven (theorem 2A), that there exist a ball $B$ such that : $$|\operatorname{Card}(X \cap B) - |B|| \geq C |\partial B|^{1/2}$$
Here $|.|$ on the right-hand side denotes the 1 dimensional lebesgue measure and $\partial B$ the boundary of $B$.
I wonder is there is a way to obtain similar lower bound, but for smooth functions. Given $\phi : \mathbb{R}^2 \to \mathbb{R}^{+}$, $C^{\infty}(\mathbb{R}^2)$ with compact support (but other assumption can be done on $\phi$), is it possible to lower bound the following quantity ?
$$|\sum_{x \in X} \phi(x) - \int_{\mathbb{R^2}} \phi(x) dx | \geq C f(\phi)$$
Here $f$ is still a positive function.
Do you have reference for this problem ? Or any hints ?
Thanks,
I assume $\phi$ is nonnegative, $I=\int_{\mathbb{R}^n}\phi(x)dx$ exists, and $\phi(\mathbb{R}^n)$'s closure contains an interval $(0,a)$.
Theorem: $\,\inf\limits_X\,\Bigl|\sum_{x\in X}\phi(x)-\int_{\mathbb{R}^n}\phi(x)dx\Bigr|=0$.
Proof. Suppose $\varepsilon>0$ is arbitrary. Pick $a_1,\cdots,a_k\in(0,a)$ such that $a_1+\cdots+a_k=I$. By hypothesis, there exist points $x_1,\cdots,x_k\in\mathbb{R}^n$ such that $|\phi(x_i)-a_i|<\varepsilon/k$ for $i=1,\cdots,k$. Then,
$$ \left|\sum_{x\in X}\phi(x)-\int_{\mathbb{R}^n}\phi(x)dx\right|=\left|\sum_{i=1}^k \big(\phi(x_i)-a_i\big)\right|\le\sum_{i=1}^k\bigl|\phi(x_i)-a_i\bigr|\le\sum_{i=1}^k\frac{\varepsilon}{k}=\varepsilon $$
for the set $X=\{x_1,\cdots,x_k\}$.