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.



Discrete singular integrals
$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$Let $a\_n:=\phi(n)$. Then
\begin{equation}
 K(x)=\sum\_{n\in\Z}a\_n 1(n-1/2\le x<n+1/2). 
\end{equation}
So, $K(x)=a\_0=0$ if $1/2\le x<1/2$. So, for $\ep\in(0,1/2)$,
\begin{equation}
 I\_\ep:=\int\_{1/\ep<|x|<\ep}K(x)\,dx=\int\_{|x|<\ep}K(x)\,dx
 =\sum\_{n\in\Z}a\_n J\_n,
\end{equation}
where
\begin{equation}
 J\_n:=\int dx\,1(-\ep\le x<\ep,n-1/2\le x<n+1/2). 
\end{equation}


Let now $N:=\lfloor\ep+1/2\rfloor$, so that $N-1/2\le\ep<N+1/2$. 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,
\begin{equation}
 I\_\ep
 =\sum\_{|n|\le N-1}a\_n +O(|a\_N|+|a\_{-N}|). 
\end{equation}
So, $I\_\ep$ converges, since $N\to\infty$, $\sum\_{|n|\le N-1}a\_n$ converges, and $|a\_N|+|a\_{-N}|=O(1/N)\to0$.


This provides the positive answer to your question.


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.)


