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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!

Daniele Tampieri
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Feng
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1 Answers1

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Let $R=|y|, r=|y| / 2$, define $$ \Omega=B(0,R)\cap B(y,r). $$ Then $|\Omega| \sim r^n \sim R^n$, $$ \begin{aligned} &\int_{|x|>2|y|}|K(x-y)-K(x)| \mathrm{d} x \\ &\leq \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x+\int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x \end{aligned} $$ Averaging with respect to $y^{\prime}$ on $\Omega$, $$ \begin{aligned} \int_{|x|>2|y|}|K(x-y)-K(x)| \mathrm{d} x \\ & \leq \frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x \\ &+\frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x \end{aligned} $$ While $R=|y|$ and $\Omega \subset B(0, R)$, $$ \frac{1}{|\Omega|} \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x \leq \frac{C}{R^n} \int_{\left|y^{\prime}\right|<R} \mathrm{~d} y^{\prime} \int_{|x|>2 R}\left|K\left(x-y^{\prime}\right)-K(x)\right| \mathrm{d} x $$ On the other hand, $$ \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{|x|>2|y|}\left|K\left(x-y^{\prime}\right)-K(x-y)\right| \mathrm{d} x $$ $$ \begin{aligned} &=\int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{\left|x^{\prime}+y\right|>2|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d}\left(x^{\prime}+y\right) \\ &\leq \int_{y^{\prime} \in \Omega} \mathrm{d} y^{\prime} \int_{\left|x^{\prime}\right|>|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime} \\ &\leq \int_{\left|y^{\prime}-y\right|<r} \mathrm{~d}\left(y^{\prime}-y\right) \int_{\left|x^{\prime}\right|>|y|}\left|K\left(x^{\prime}+y-y^{\prime}\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime} \\ &\leq \int_{\left|y^{\prime}-y\right|<r} \mathrm{~d}\left(y^{\prime}-y\right) \int_{\left|x^{\prime}\right|>2 r}\left|K\left(x^{\prime}-\left(y^{\prime}-y\right)\right)-K\left(x^{\prime}\right)\right| \mathrm{d} x^{\prime} \end{aligned} $$

Taking the supremum in $y'$ firstly and in $y$ secondly results the desired conclusion.

IMOS
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