Maximal truncated singular integrals

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Fix an integer $n\ge 1$ and let $k:{\mathbb{R}}^n\setminus \{0\}\to \mathbb{C}$ be a measurable function that is integrable on compact subsets of ${\mathbb{R}}^n\setminus \{0\}$ and satisfies the pointwise size bound $\mathrm{\mid }k(x)\mathrm{\mid }\le A_1\mathrm{\mid }x{\mathrm{\mid }}^{-n}(x\mathrm{\ne }0)(1)$ for some finite constant $0\le A_1<\mathrm{\infty }$. With the complex $L^p$ conventions of Complex Lp classes and Euclidean test-function conventions, fix $1\le p<\mathrm{\infty }$ and $f\in L^p({\mathbb{R}}^n)$.

For $0<\epsilon <\mathrm{\infty }$ define the truncated singular integral $T_{\epsilon }f(x):={\int }_{\mathrm{\mid }y\mathrm{\mid }>\epsilon }k(y)f(x-y)dy,x\in {\mathbb{R}}^n,$ and for $0<\epsilon <N<\mathrm{\infty }$ define the doubly truncated singular integral $T^{(\epsilon ,N)}f(x):={\int }_{\epsilon <\mathrm{\mid }y\mathrm{\mid }<N}k(y)f(x-y)dy,x\in {\mathbb{R}}^n\mathrm{.}$ Both integrals converge absolutely for every $x$. Indeed, the truncated kernel $k_{\epsilon }:=k{\mathbf{1}}_{\{\mathrm{\mid }\cdot \mathrm{\mid }>\epsilon \}}$ satisfies $\mathrm{\parallel }{k}_{\epsilon }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}^{p^{\mathrm{\prime }}}={\int }_{\mathrm{\mid }y\mathrm{\mid }>\epsilon }\mathrm{\mid }k(y){\mathrm{\mid }}^{p^{\mathrm{\prime }}}dy\le A_1^{p^{\mathrm{\prime }}}{\int }_{\mathrm{\mid }y\mathrm{\mid }>\epsilon }\mathrm{\mid }y{\mathrm{\mid }}^{-n{p}^{\mathrm{\prime }}}dy=A_1^{p^{\mathrm{\prime }}}\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\frac{{\epsilon }^{n-n{p}^{\mathrm{\prime }}}}{n{p}^{\mathrm{\prime }}-n}<\mathrm{\infty }$ by Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma for $1<p<\mathrm{\infty }$ (the case $p=1$, $p^{\mathrm{\prime }}=\mathrm{\infty }$, uses $\mathrm{\parallel }{k}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}\le A_1{\epsilon }^{-n}$), and likewise $\mathrm{\parallel }k{\mathbf{1}}_{\{\epsilon <\mathrm{\mid }\cdot \mathrm{\mid }<N\}}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}<\mathrm{\infty }$; Hölder's inequality (Complex Holder, Minkowski, and the quotient norm) therefore gives $\mathrm{\mid }{T}_{\epsilon }f(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }{k}_{\epsilon }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}<\mathrm{\infty }$ and $\mathrm{\mid }{T}^{(\epsilon ,N)}f(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }k{\mathbf{1}}_{\{\epsilon <\mathrm{\mid }\cdot \mathrm{\mid }<N\}}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}<\mathrm{\infty }$ at every $x$. The maximal truncated singular integral and the doubly truncated maximal singular integral are $T^{\ast }f(x):={\sup }_{\epsilon >0}\mathrm{\mid }{T}_{\epsilon }f(x)\mathrm{\mid },T^{\ast \ast }f(x):={\sup }_{0<\epsilon <N<\mathrm{\infty }}\mathrm{\mid }{T}^{(\epsilon ,N)}f(x)\mathrm{\mid },$ with values in $[0,\mathrm{\infty }]$.

Under the size bound (1) the two maximal operators are pointwise comparable: $T^{\ast }f\le T^{\ast \ast }f\le 2T^{\ast }f\text{pointwise on }{\mathbb{R}}^n\mathrm{.}(2)$ For the left inequality, fix $\epsilon >0$ and $x$: dominated convergence (Dominated convergence) applied to the absolutely convergent integral defining $T_{\epsilon }f(x)$ gives $T_{\epsilon }f(x)={\lim }_{N\to \mathrm{\infty }}{T}^{(\epsilon ,N)}f(x)$, so $\mathrm{\mid }{T}_{\epsilon }f(x)\mathrm{\mid }\le T^{\ast \ast }f(x)$; take the supremum in $\epsilon$. For the right inequality, split the defining integral of $T^{(\epsilon ,N)}$ at $N$ to get $T^{(\epsilon ,N)}f(x)=T_{\epsilon }f(x)-T_{N}f(x)$ and hence $\mathrm{\mid }{T}^{(\epsilon ,N)}f(x)\mathrm{\mid }\le \mathrm{\mid }{T}_{\epsilon }f(x)\mathrm{\mid }+\mathrm{\mid }{T}_{N}f(x)\mathrm{\mid }\le 2T^{\ast }f(x)$; take the supremum in $\epsilon <N$. Thus $T^{\ast }$ and $T^{\ast \ast }$ have the same finiteness set and the same boundedness properties. The definition itself asserts neither an upper truncation for $T^{\ast }$ nor the existence of a principal-value limit ${\lim }_{\epsilon \downarrow 0}{T}_{\epsilon }f(x)$; the doubly truncated form is the primitive object, because it is defined from the kernel alone. Here the pointwise size bound (1) is a stronger hypothesis than the annular size condition of the Calderón–Zygmund kernel definition: (1) implies ${\int }_{R\le \mathrm{\mid }x\mathrm{\mid }\le 2R}\mathrm{\mid }k\mathrm{\mid }\le A_1\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log 2$, while the annular condition does not by itself prevent pointwise spikes. Countable Choice is inherited from the polar-coordinate evaluation; the truncations themselves require no selection.