3
$\begingroup$

This problem has been asked in MSE, but got no answers. I guess that this exam problem may be a small lemma in some research papers, so I post it here on MathOverflow.

Let $K\in L_{\text{loc}}^1(\mathbb R^n\setminus\{0\})$. Prove that $$\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty\label{1}\tag{1}$$ if and only if $$\sup_{r>0}\frac1{r^n}\int_{B(0,r)}\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy<\infty.\label{2}\tag{2}$$

This is an old exam problem on Harmonic Analysis. Formula \eqref{1} is called the Hörmander's condition for singular integrals. The proof of \eqref{1}$\Rightarrow$\eqref{2} is quite easy: assume $$\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,\qquad \forall y\in\mathbb R^n,$$ then for $r>0$ and $y\in B(0,r)$ we have $\{x: |x|>2r\}\subset \{x: |x|>2|y|\}$, so $$\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\leq \int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,$$ hence $$\frac1{r^n}\int_{B(0,r)}\int_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy\leq \frac1{r^n}\int_{B(0,r)}M\,dy=M|B(0,1)|,\ \ \ \forall r>0.$$ This completes the proof of \eqref{1}$\Rightarrow$\eqref{2}.

However, for \eqref{2}$\Rightarrow$\eqref{1}, I don't know how to start.

Any help would be appreciated!

$\endgroup$

1 Answer 1

3
$\begingroup$

Condition \eqref{2} implies \eqref{1} with $5|y|$ instead of $2|y|$. Fix $y$, let $r=|y|$ and $I=\int_{|x| >5|y| }|K(x-y)-K(x)|\, dx$. Then $$I \leq \int_{|x| >5r }|K(x-y)-K(x-z)|\, dx+\int_{|x| >5r }|K(x-z)-K(x)|\, dx:=I_1(z)+I_2(z) $$ for every $|z| \leq r$. If $K$ is the supremum in \eqref{2}, then $r^{-n} \int_{B(0,r)} I_2(z)\, dz \leq K$. In $I_1$ we set $\xi=x-z$ so that $|\xi| \geq 4r$ and $$I_1(z) \leq \int_{|\xi| \geq 4r} |K(\xi-(y-z))-K(\xi)|\, d\xi.$$ Since $|y-z| \leq 2r$, then $$r^{-n}\int_{B(0,r)} I_1(z)\, dz \leq r^{-n} \int_{B(0,2r)}|K(\xi-w)-K(\xi)|\, dw \leq 2^n K.$$ The estimate of $I$ in terms of $K$ now follows by averaging the inequality $I \leq I_1(z)+I_2(z)$ over $B(0,r)$.

$\endgroup$
5
  • $\begingroup$ (+1)Thanks for your answer! After reading your proof, I realized that we can prove that condition (2) implies (1) with $2\delta |y|$ instead of $2|y|$ for any $\delta>1$, using a similar argument. Indeed, for fixed $\delta>1$ and fixed $y\in\mathbb R^n$, we can write $|y|=C_1r$ and then do the similar thing for every $|z|<C_2 r$, where $C_1$ and $C_2$ are constants to be determined later, depending only on $\delta>1$. Eventually, we can prove that $$\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2\delta |y|\}}|K(x-y)-K(x)|\,dx\lesssim \frac1{(\delta-1)^n},\qquad \delta>1.$$ $\endgroup$
    – Feng
    Commented Oct 12, 2022 at 2:16
  • $\begingroup$ Based on my comment above, I still believe that $(2)\Rightarrow(1)$ is true. (That is, $\delta=1$ case.) Maybe we just need some new ideas. $\endgroup$
    – Feng
    Commented Oct 12, 2022 at 2:18
  • $\begingroup$ I do not know about $\delta=1$. A related question concerns the constants in (1). Assume it holds with $2|y|$, as in the statement. Does it hold with $a|y|$ for every $a>1$? From the point of view of boundeness of singular integrals, does not matter, since any $a$ makes the proof work...do you know the answer? $\endgroup$ Commented Oct 12, 2022 at 7:34
  • $\begingroup$ You've convinced me. Yes, any $a>1$ or $\delta\geq 1$ is sufficient to prove the boundedness of singular integrals. So, I accept your answer. I don't know the answer of your question. I'll think about it. $\endgroup$
    – Feng
    Commented Oct 12, 2022 at 8:01
  • $\begingroup$ Having a bounded operator $T$, say in $L^2$, the weak estimate follows if $\int_{R^n \setminus \alpha Q} |Tf| \leq C\int_Q |f|$ holds for some $\alpha>1$, any cube $Q$ and any function with support in $Q$ and zero mean. If $T$ is given by a convolution kernel, usually one substracts the value of the kernel at the center of the cube, but any other point in the cube works and one can average. This leads to condition (2), though substantially equivalent to (1). $\endgroup$ Commented Oct 12, 2022 at 8:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .