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 $N\to\infty.$


Denote
$$
R(x):=
\begin{cases}
1-|x| & \text{for }|x|<1\\
0 &\text{otherwise}
\end{cases}.
$$
and consider
$$
K(x)=\sum\_{n\in\mathbb Z}\phi(n)R(x-n),
$$ which exists as a function on $\mathbb R.$


* Can anyone prove that
$$
\lim\_{\epsilon\to \infty}\int\limits\_{\frac{1}{\epsilon}<|x|<\epsilon}K(x)\,\mathrm{d}x
$$ exists?
* Also are the following true?
$$
\begin{align}
|K(x)-K(x-y)| &\leq C\_3\frac{|y|}{|x|^2} &\text{for }|x|>2|y|>0\\ 
|K(x)| & \leq C\_4|x|^{-1}&\text{for }x\neq 0.
\end{align}
$$ Also it is okay to have the above conditions true almost everywhere.



Extending a discrete singular kernel
$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$The answer is yes to each of your two questions.


Let $a\_n:=\phi(n)$. Then
\begin{equation\*}
 K(x)=\sum\_{n\in\Z}a\_n R(x-n). 
\end{equation\*}
Note that for all $j\in\Z$ we have $K(j)=a\_j$ and $K$ linear (or, more exactly, affine) on the interval $[j,j+1]$. Also, $K$ is continuous. So, $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a\_j$. In particular, $K$ is bounded.


So, for real $\ep>0$,
\begin{equation\*}
 \int\_{1/\ep<|x|<\ep}K(x)\,dx=I\_\ep+O(1/\ep),
\end{equation\*}
where
\begin{equation\*}
 I\_\ep:=\int\_{|x|<\ep}K(x)\,dx
 =\sum\_{n\in\Z}a\_n J\_n,
\end{equation\*}
\begin{equation\*}
 J\_n:=\int\_{-\ep}^\ep dx\,R(x-n). 
\end{equation\*}


Let now $N:=\lfloor\ep\rfloor$, so that $N\le\ep<N+1$. Then $J\_n=\int\_\R dx\,R(x-n)=1$ if $|n|\le N-1$ and $J\_n=0$ if $|n|\ge N+2$. 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}|+|a\_{N+1}|+|a\_{-N-1}|). 
\end{equation\*}
So, $I\_\ep$ converges, since $N\to\infty$ (as $\ep\to\infty$), $\sum\_{|n|\le N-1}a\_n$ converges, and $|a\_N|+|a\_{-N}|+|a\_{N+1}|+|a\_{-N-1}|=O(1/N)\to0$. So, $\int\_{1/\ep<|x|<\ep}K(x)\,dx$ converges.


This provides the positive answer to your first question.


The answer to your second question is also positive, that is, for some real $C\_3$ and $C\_4$,
\begin{equation\*}
 |K(x)|\le C\_3|x|^{-1} \text{ if }x\ne 0 \tag{1}
\end{equation\*}
and
\begin{equation\*}
 |K(x)-K(x-y)| \le C\_4\frac{|y|}{|x|^2} \text{ if }|x|>2|y|>0. \tag{2}
\end{equation\*}


Indeed, since $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a\_j$ and the $a\_j$'s are bounded, we see that the function $K$ is bounded and Lipschitz, so that without loss of generality $|x|>8$ in (1) and (2).


Now take indeed any real $x$ with $|x|>8$ and any real $y$ as in (2). Let
\begin{equation}
 j:=\fl{|x|},\quad m:=\fl{|x-y|},
\end{equation}
so that $|j|\ge|m|\ge1+|x|/4$ and also $|j+1|\ge|x|/4$.


So, by the linear interpolation observation and the condition $|a\_n|\le C\_1/|n|$ for $n\ne0$,
\begin{equation\*}
 |K(x)|\le|a\_j|+|a\_{j+1}|\le C\_1(|j|^{-1}+|j+1|^{-1})\le8C\_1|x|^{-1}, 
\end{equation\*}
which verifies (1).


Next, by the linear interpolation observation and the condition $|a\_{n+1}-a\_n|\le C\_2/n^2$ for $n\ne0$, the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $C\_2/(m-1)^2\le4C\_2/x^2$.
Also, $|x|\ge j\ge m$, $|x-y|\ge m$, and, by the condition $|x|>2|y|>0$ in (2), $x$ and $x-y$ are of the same sign. So, either both $x$ and $x-y$ are in $[m,\infty)$ or they are both in $(-\infty,-m]$.


Therefore and because the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $4C\_2/x^2$, (2) follows.


