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.

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

1 Answer 1


$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$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.)

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    $\begingroup$ @Abeginnermathmatician : I have added more on this. $\endgroup$ Nov 16, 2021 at 15:13
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    $\begingroup$ @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. $\endgroup$ Nov 16, 2021 at 18:23
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    $\begingroup$ @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 $x>3/2>x−y$. $\endgroup$ Nov 16, 2021 at 18:25
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    $\begingroup$ @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 $n-1/2$. $\endgroup$ Nov 17, 2021 at 16:25
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    $\begingroup$ @Abeginnermathmatician : My first name is Iosif (pronounced like Yosef), which is similar to Joseph, Josef, Giuseppe, José, Yusef, etc. $\endgroup$ Nov 17, 2021 at 16:30

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