Extending a discrete singular kernel

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.
• 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". Nov 19 '21 at 13:27
• @ 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. Nov 19 '21 at 14:18
• @Iosif. I have edited the question. There is no need to mention Calderon-Zygmund kernel in the question. Nov 19 '21 at 14:21
• Can you also remove Calderon-Zygmund from the title? Nov 19 '21 at 14:26
• @Iosif. It's done. Nov 19 '21 at 14:52

$$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}{\lfloor#1\rfloor}$$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 $$N\le\ep. Then $$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.

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.

• In the definition of $j$ should the $|x|$ be replaced by $x$? Nov 22 '21 at 9:11
• Also how do you conclude your last statement Next, by the linear in..... So, (2) follows."? Nov 22 '21 at 9:17
• @Abeginnermathmatician : (i) No, the definition of $j$ is as intended. (ii) I have added further details on "(2) follows". Nov 22 '21 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 $$|\phi(n) - \phi(n - k)| \leqslant \sum_{j = 1}^k |\phi(n - j) - \phi(n - j + 1)| \leqslant \sum_{j = 1}^k \frac{C_2}{|n - j|^2} \leqslant C_2 \frac{k}{(|n| - |k|)^2} .$$

• 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. Nov 22 '21 at 5:19