# Discrete singular integrals

Let $$\{\phi(n)\}_{n\in\mathbb Z}$$ be a sequence of complex numbers with the following properties:

1. $$\phi(0)=0$$ and $$|\phi(n)|\leq \frac{C_1}{|n|}$$ for all $$n\neq 0$$ and $$C_1>0$$ is independent of $$n.$$

2. $$|\phi(n+1)-\phi(n)|\leq \frac{C_2}{n^2}$$ for all $$n\neq 0$$ and $$C_2>0$$ is independent of $$n.$$

3. $$\sum_{-N}^N\phi(n)$$ converges as $$N\to\infty.$$

Consider $$K(x)=\sum_{n\in\mathbb Z}\phi(n)\chi_{\left[n-\frac{1}{2},n+\frac{1}{2}\right)}(x),$$ which exists as a function in $$L_p(\mathbb R).$$ Can anyone prove that $$\lim_{\epsilon\to \infty}\int\limits_{\frac{1}{\epsilon}<|x|<\epsilon}K(x)\,\mathrm{d}x$$ exists? Also is the following true? $$|K(x)-K(x-y)|\leq C_3\frac{|y|}{|x|^2}$$ for $$|x|>2|y|.$$ We also have by http://matwbn.icm.edu.pl/ksiazki/cm/cm66/cm66211.pdf that $$|K(x)|\leq C_4|x|^{-1}$$ for $$x\neq 0.$$ That is $$K$$ is standard Calderon-Zygmund kernel.

• There must be a problem with the formulation of the integral. It is $0$. Nov 16, 2021 at 8:54
• @Dieter. Then the limit exists automatically. However do you have a proof? If s please share. Nov 16, 2021 at 10:57
• The problem is $\epsilon < |x| < \epsilon$, this has to be replaced by something else. Nov 16, 2021 at 11:24
• @Dieter. Corrected now. Nov 16, 2021 at 13:11
• @ Dieter. Yes. Corrected again. Nov 16, 2021 at 13:17

$$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$$Let $$a_n:=\phi(n)$$. Then $$$$K(x)=\sum_{n\in\Z}a_n 1(n-1/2\le x So, $$K(x)=a_0=0$$ if $$1/2\le x<1/2$$. So, for $$\ep\in(0,1/2)$$, $$$$I_\ep:=\int_{1/\ep<|x|<\ep}K(x)\,dx=\int_{|x|<\ep}K(x)\,dx =\sum_{n\in\Z}a_n J_n,$$$$ where $$$$J_n:=\int dx\,1(-\ep\le x<\ep,n-1/2\le x

Let now $$N:=\lfloor\ep+1/2\rfloor$$, so that $$N-1/2\le\ep. Then $$J_n=1$$ if $$|n|\le N-1$$ and $$J_n=0$$ if $$|n|\ge N+1$$. Also, $$0\le J_n\le1$$ for all $$n\in\Z$$. So, $$$$I_\ep =\sum_{|n|\le N-1}a_n +O(|a_N|+|a_{-N}|).$$$$ So, $$I_\ep$$ converges, since $$N\to\infty$$, $$\sum_{|n|\le N-1}a_n$$ converges, and $$|a_N|+|a_{-N}|=O(1/N)\to0$$.

The answer to your second question is no, in general. Indeed, take $$x=3/2$$ and let $$y\downarrow0$$. Then $$K(x)=a_2$$ and, eventually (for all small enough $$y>0$$), $$K(x-y)=a_1$$ and $$|x|>2|y|$$. However, for any real $$C_3$$, the inequality $$|K(x)-K(x-y)|\le C_3\frac{|y|}{|x|^2}$$ will eventually fail to hold if $$a_2\ne a_1$$. However, it is not hard to see that the answer to your second question will be yes under an additional restriction such as $$|y|\ge1$$. (I will add details to this later -- now have to do something else.)
• @Abeginnermathmatician : I have taken a look at the paper you are reading, especially (K1)--(K3). I believe what they meant by "the linear extension" is the linear interpolation (en.wikipedia.org/wiki/…) Then I think their conditions (K1)--(K3) will hold. However, if you have further questions about those conditions for $K$ defined as the linear interpolation, please post them separately. Nov 16, 2021 at 18:23
• @Abeginnermathmatician : The modification of $K(x)$ that you suggested in your latter comment will not help: now let $x$ and $x−y$ be close to each other and so that $x>3/2>x−y$. Nov 16, 2021 at 18:25
• @Abeginnermathmatician : (i) I think $(1-|x|)_+$ will work. For details, you may want to post a separate question about this. (ii) With the construction as in your current post, you cannot get it almost everywhere either -- you would need to make $x$ bounded away from $n-1/2$. Nov 17, 2021 at 16:25