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.
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    $\begingroup$ Given the function $\phi$, $K$ was completely defined. But at the end of your post, you say, all of a sudden, "where $K$ is standard Calderon-Zygmund kernel". So, which of these two is $K$: $K(x)=\sum_{n\in\mathbb Z}\phi(n)R(x-n)$ or "standard Calderon-Zygmund kernel"? In fact, a Calderon-Zygmund kernel is a function of two arguments (en.wikipedia.org/wiki/…), and your $K$, everywhere in your post, is a function of one argument. You see, you got me totally confused, with this mentioning of "standard Calderon-Zygmund kernel". $\endgroup$ Nov 19, 2021 at 13:27
  • $\begingroup$ @ Iosif. The definition of Calderon-Zygmund kernel varies in the literature. What I mean is that $(x,y)\mapsto K(x-y)$ is a Calderon-Zygmund kernel. You can take the three properties that I have mentioned as the definition of Calderon-Zygmund kernel. There are many text books which follow this convention. A more general definition is what you have mentioned. $\endgroup$ Nov 19, 2021 at 14:18
  • $\begingroup$ @Iosif. I have edited the question. There is no need to mention Calderon-Zygmund kernel in the question. $\endgroup$ Nov 19, 2021 at 14:21
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    $\begingroup$ Can you also remove Calderon-Zygmund from the title? $\endgroup$ Nov 19, 2021 at 14:26
  • $\begingroup$ @Iosif. It's done. $\endgroup$ Nov 19, 2021 at 14:52

2 Answers 2


$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$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.

  • $\begingroup$ In the definition of $j$ should the $|x|$ be replaced by $x$? $\endgroup$ Nov 22, 2021 at 9:11
  • $\begingroup$ Also how do you conclude your last statement ``Next, by the linear in..... So, (2) follows."? $\endgroup$ Nov 22, 2021 at 9:17
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    $\begingroup$ @Abeginnermathmatician : (i) No, the definition of $j$ is as intended. (ii) I have added further details on "(2) follows". $\endgroup$ Nov 22, 2021 at 13:40

Come on, for the first property just use the fact that $$ \int_{-n}^n K(x) dx = \sum_{j = -n+1}^{n-1} \phi(j) + \tfrac12(\phi(-n)+\phi(n)) $$ has a finite limit as $n \to \infty$, together with convergence of $$ \biggl| \int_{-a}^a K(x) dx - \int_{-\lfloor a\rfloor}^{\lfloor a\rfloor} K(x) dx \biggr| \leqslant |\phi(-\lfloor a\rfloor-1)| + |\phi(-\lfloor a\rfloor)| + |\phi(\lfloor a\rfloor)| + |\phi(\lfloor a\rfloor+1)| $$ to zero as $a \to \infty$.

The other two properties are equally simple, once you realise that $$ |\phi(n) - \phi(n - k)| \leqslant \sum_{j = 1}^k |\phi(n - j) - \phi(n - j + 1)| \leqslant \sum_{j = 1}^k \frac{C_2}{|n - j|^2} \leqslant C_2 \frac{k}{(|n| - |k|)^2} . $$

  • $\begingroup$ Thanks! It was not difficult. But I was decoding a paper where the authors did not mention the construction of the function. So I was confused. $\endgroup$ Nov 22, 2021 at 5:19

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