Maximal truncations: weak (1,1) and strong Lp bounds

Cotlar's inequality extends to $L^2$ inputs. Choose $h_m\in C_c^{\mathrm{\infty }}$ with $h_m\to h$ in $L^2$ by [F7]. For each $t>0$, the size bound makes $k_t=k{\mathbf{1}}_{\{\mathrm{\mid }\cdot \mathrm{\mid }>t\}}\in L^2$, so Hölder gives $\mathrm{\mid }{T}_t(h_m-h)(x)\mathrm{\mid }\le \mathrm{\parallel }{k}_t{\mathrm{\parallel }}_2\mathrm{\parallel }{h}_m-h{\mathrm{\parallel }}_2\to 0$ for every $x$. Sublinearity and the $L^2$ bound of $M$ imply $M(T{h}_m)\to M(Th)$ and $M{h}_m\to Mh$ in $L^2$; [F6] gives a common almost-everywhere convergent subsequence for these two families. Outside the countable union of the exceptional sets for [F2], pass its bound for each $h_m$ to the limit at every $t>0$, then take the supremum to get $T^{\ast }h\le M(Th)+C_0(A_1+A_2^{\mathrm{\prime }}+A_3)Mh$ almost everywhere. For every finite-exponent input, dominated convergence shows that $t\mapsto T_{t}h(x)$ is continuous on $(0,\mathrm{\infty })$ for every $x$: nearby truncations are dominated by $\mathrm{\mid }k(z)h(x-z)\mathrm{\mid }{\mathbf{1}}_{\{\mathrm{\mid }z\mathrm{\mid }>t\mathrm{/}2\}}$, integrable by [F1]. Thus both maximal suprema can be taken over rational parameters and are measurable.

Geometry of the dilated cubes. For each maximal cube $Q_j$ with centre $c_j$ and side length ${\mathrm{\ell }}_j$, let $Q_j^{\ast }$ be the cube concentric with $Q_j$ and with side length $5\sqrt{n}{\mathrm{\ell }}_j$; then $\mathrm{\mid }{Q}_j^{\ast }\mathrm{\mid }=(5\sqrt{n}{)}^n\mathrm{\mid }{Q}_j\mathrm{\mid }$ by the dilation identity [F6]. If $x\mathrm{\notin }{\bigcup }_k{Q}_k^{\ast }$ and $y\in Q_j$, then $\mathrm{\mid }x-c_j{\mathrm{\mid }}_{\mathrm{\infty }}>\frac{5\sqrt{n}}{2}{\mathrm{\ell }}_j$, so $\mathrm{\mid }x-c_j\mathrm{\mid }\ge \frac{5\sqrt{n}}{2}{\mathrm{\ell }}_j>2\cdot \frac{\sqrt{n}}{2}{\mathrm{\ell }}_j\ge 2\mathrm{\mid }y-c_j\mathrm{\mid }$; in particular $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }x-c_j\mathrm{\mid }-\mathrm{\mid }y-c_j\mathrm{\mid }\ge \frac{\sqrt{n}}{2}{\mathrm{\ell }}_j>0$. Moreover, if $t>0$ and $j$ belongs to $J_3(x,t):=\{j:\mathrm{\exists }{y}_0\in Q_j,\mathrm{\mid }x-y_0\mathrm{\mid }=t\}$, then $t\ge \mathrm{\mid }x-c_j\mathrm{\mid }-\mathrm{\mid }{y}_0-c_j\mathrm{\mid }\ge (\frac{5\sqrt{n}}{2}-\frac{\sqrt{n}}{2}){\mathrm{\ell }}_j=2\sqrt{n}{\mathrm{\ell }}_j$, and consequently every $y\in Q_j$ satisfies $\frac{t}{2}\le t-\sqrt{n}{\mathrm{\ell }}_j\le \mathrm{\mid }x-y\mathrm{\mid }\le t+\sqrt{n}{\mathrm{\ell }}_j\le \frac{3t}{2}$; hence ${\bigcup }_{j\in J_3(x,t)}{Q}_j\subseteq B(x,\frac{3t}{2})\setminus B(x,\frac{t}{2})$.

Splitting the truncated bad part. Fix $x\mathrm{\notin }{\bigcup }_k{Q}_k^{\ast }$ and $t>0$, and split the indices into $J_1,J_2,J_3$ according to whether $\mathrm{\mid }x-y\mathrm{\mid }<t$ for all $y\in Q_j$, $\mathrm{\mid }x-y\mathrm{\mid }>t$ for all $y\in Q_j$, or $\mathrm{\mid }x-y\mathrm{\mid }=t$ for some $y\in Q_j$. Each index lies in exactly one of the three classes because $y\mapsto \mathrm{\mid }x-y\mathrm{\mid }$ is continuous on the connected cube. For $j\in J_1$ the integrand of $T_{t}b(x)$ vanishes on $Q_j$; for $j\in J_2$ one has $\mathrm{\mid }k(x-y)\mathrm{\mid }\le A_1{t}^{-n}$ on $Q_j$; for $j\in J_3$ step 1.2 gives $\mathrm{\mid }k(x-y)\mathrm{\mid }\le A_1(2\mathrm{/}t{)}^n$ on $Q_j$. Hence ${\sum }_j{\int }_{Q_j\cap \{\mathrm{\mid }x-y\mathrm{\mid }>t\}}\mathrm{\mid }k(x-y)\mathrm{\mid }\mathrm{\mid }{b}_j(y)\mathrm{\mid }dy\le {\sum }_{j\in J_2\cup J_3}{\int }_{Q_j}\mathrm{\mid }k(x-y)\mathrm{\mid }\mathrm{\mid }{b}_j(y)\mathrm{\mid }dy\le \max \{A_1{t}^{-n},A_1(2\mathrm{/}t{)}^n\}{\sum }_j\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_1<\mathrm{\infty }$ by [F3], so the series converges absolutely and $T_{t}b(x)={\sum }_j{\int }_{Q_j\cap \{\mathrm{\mid }x-y\mathrm{\mid }>t\}}k(x-y)b_j(y)dy$. For $j\in J_2$ the truncation is inactive on $Q_j$ and the mean-zero property of $b_j$ gives $\mid {\int }_{Q_j}k(x-y)b_j(y)dy\mid =\mid {\int }_{Q_j}[k(x-y)-k(x-c_j)]b_j(y)dy\mid$. For $j\in J_3$, put $c_j(t):=\mathrm{\mid }{Q}_j{\mathrm{\mid }}^{-1}{\int }_{Q_j}{b}_j(y){\mathbf{1}}_{\{\mathrm{\mid }x-y\mathrm{\mid }>t\}}(y)dy$; then $\mathrm{\mid }{c}_j(t)\mathrm{\mid }\le \mathrm{\mid }{Q}_j{\mathrm{\mid }}^{-1}\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_1\le 2^{n+1}\mu$ and, since ${\int }_{Q_j}(b_j{\mathbf{1}}_{\{\mathrm{\mid }x-y\mathrm{\mid }>t\}}-c_j(t))=0$, ${\int }_{Q_j}k(x-y)b_j(y){\mathbf{1}}_{\{\mathrm{\mid }x-y\mathrm{\mid }>t\}}(y)dy={\int }_{Q_j}[k(x-y)-k(x-c_j)](b_j{\mathbf{1}}_{\{\mathrm{\mid }x-y\mathrm{\mid }>t\}}-c_j(t))dy+c_j(t){\int }_{Q_j}k(x-y)dy\mathrm{.}$ Summing, using $\mathrm{\mid }{b}_j\mathbf{1}-c_j(t)\mathrm{\mid }\le \mathrm{\mid }{b}_j\mathrm{\mid }+2^{n+1}\mu$ on $Q_j$, and using step 1.2 with ${\int }_{\{\mathrm{\mid }z\mathrm{\mid }\in [t\mathrm{/}2,3t\mathrm{/}2]\}}\mathrm{\mid }k(z)\mathrm{\mid }dz\le A_1\mathrm{\mid }{S}^{n-1}\mathrm{\mid }\log 3$, yields ${\sup }_{t>0}\mathrm{\mid }{T}_{t}b(x)\mathrm{\mid }\le 2E_1(x)+2^{n+1}\mu E_2(x)+C_n\mu A_1,$ where $E_1(x):={\sum }_j{\int }_{Q_j}\mathrm{\mid }k(x-y)-k(x-c_j)\mathrm{\mid }\mathrm{\mid }{b}_j(y)\mathrm{\mid }dy$ and $E_2(x):={\sum }_j{\int }_{Q_j}\mathrm{\mid }k(x-y)-k(x-c_j)\mathrm{\mid }dy$: the $J_2$ sum and the first $J_3$ sum together contribute at most $E_1+2^{n+1}\mu E_2$ (both $E_1$ and $E_2$ majorize their sub-sums over $J_2$ and $J_3$; if one of them is infinite the displayed inequality is trivial), while ${\sum }_{j\in J_3}\mathrm{\mid }{c}_j(t)\mathrm{\mid }{\int }_{Q_j}\mathrm{\mid }k(x-y)\mathrm{\mid }dy\le 2^{n+1}\mu {\int }_{\{\mathrm{\mid }z\mathrm{\mid }\in [t\mathrm{/}2,3t\mathrm{/}2]\}}\mathrm{\mid }k(z)\mathrm{\mid }dz$ by the containment of step 1.2 and the disjointness of the cubes. Since $T^{(\epsilon ,N)}b=T_{\epsilon }b-T_{N}b$, we obtain $T^{\ast \ast }b(x)\le 2{\sup }_{t>0}\mathrm{\mid }{T}_{t}b(x)\mathrm{\mid }\le 4E_1(x)+2^{n+2}\mu E_2(x)+C_n\mu A_1$ at every such $x$ (with $C_n$ denoting a dimensional constant, as everywhere).

Integrating $E_1$ and $E_2$ off the dilated cubes. By step 1.2, for $x\mathrm{\notin }{Q}_j^{\ast }$ and $y\in Q_j$ one has $\mathrm{\mid }x-c_j\mathrm{\mid }\ge 2\mathrm{\mid }y-c_j\mathrm{\mid }$, so in ${\int }_{({\bigcup }_k{Q}_k^{\ast }{)}^c}{E}_1\le {\sum }_j{\int }_{Q_j}\mathrm{\mid }{b}_j(y)\mathrm{\mid }({\int }_{\mathrm{\mid }x-c_j\mathrm{\mid }\ge 2\mathrm{\mid }y-c_j\mathrm{\mid }}\mathrm{\mid }k(x-y)-k(x-c_j)\mathrm{\mid }dx)dy$ the inner integral is at most $A_2$ by Hörmander's condition, giving ${\int }_{({\bigcup }_k{Q}_k^{\ast }{)}^c}{E}_1\le A_2{\sum }_j\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_1\le A_2{2}^{n+1}\mu {\sum }_j\mathrm{\mid }{Q}_j\mathrm{\mid }\le 2^{n+1}{A}_2\mathrm{\parallel }f{\mathrm{\parallel }}_1$; similarly ${\int }_{({\bigcup }_k{Q}_k^{\ast }{)}^c}{E}_2\le A_2{\sum }_j\mathrm{\mid }{Q}_j\mathrm{\mid }\le A_2{\mu }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$. Both interchanges are Tonelli's theorem applied to nonnegative product-measurable integrands, and $A_2=\mathrm{\mid }{S}^{n-1}\mathrm{\mid }{2}^{-\delta }{\delta }^{-1}{A}_2^{\mathrm{\prime }}$ by [F4].

The bad part is controlled off the cubes. Choose $\gamma :=(K_n(A_1+A_2^{\mathrm{\prime }}+A_3+B){)}^{-1}$ with a dimensional constant $K_n$ large enough that the last term of step 2.1 satisfies $C_n\mu A_1=C_n\gamma \lambda A_1\le \lambda \mathrm{/}3$; if $A_1+A_2^{\mathrm{\prime }}+A_3+B=0$ then $k=0$ and $T^{\ast \ast }=0$, so the theorem is trivial, and otherwise $\gamma >0$ is well defined. Then $\lambda \mathrm{/}2=\lambda \mathrm{/}3+\lambda \mathrm{/}12+\lambda \mathrm{/}12$ and step 2.1 give $\{x\mathrm{\notin }{\bigcup }_k{Q}_k^{\ast }:T^{\ast \ast }b(x)>\lambda \mathrm{/}2\}\subseteq \{4E_1>\lambda \mathrm{/}12\}\cup \{2^{n+2}\mu E_2>\lambda \mathrm{/}12\}$, so by Chebyshev's inequality and step 2.2, $\mathrm{\mid }\{x\mathrm{\notin }{\bigcup }_k{Q}_k^{\ast }:T^{\ast \ast }b>\lambda \mathrm{/}2\}\mathrm{\mid }\le \frac{48}{\lambda }\int E_1+\frac{12\cdot 2^{n+2}\mu }{\lambda }\int E_2\le C_n{A}_2{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1\le C_{n,\delta }{A}_2^{\mathrm{\prime }}{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

The good part. By step 1.1 and [F1], $T^{\ast \ast }g\le 2T^{\ast }g\le 2M(Tg)+2C_0(A_1+A_2^{\mathrm{\prime }}+A_3)Mg$ almost everywhere, and $\mathrm{\parallel }Tg{\mathrm{\parallel }}_2\le B\mathrm{\parallel }g{\mathrm{\parallel }}_2$ by the $L^2$ bound of $T$. Chebyshev's inequality, the $L^2$ bound of $M$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$ give $\mathrm{\mid }\{T^{\ast \ast }g>\lambda \mathrm{/}2\}\mathrm{\mid }\le \mathrm{\mid }\{M(Tg)>\lambda \mathrm{/}8\}\mathrm{\mid }+\mathrm{\mid }\{Mg>\lambda \mathrm{/}(8C_0(A_1+A_2^{\mathrm{\prime }}+A_3))\}\mathrm{\mid }\le \frac{64}{{\lambda }^2}\mathrm{\parallel }M(Tg){\mathrm{\parallel }}_2^2+\frac{8C_n{C}_0(A_1+A_2^{\mathrm{\prime }}+A_3)}{\lambda }\mathrm{\parallel }f{\mathrm{\parallel }}_1\le \frac{64C_M^2{B}^2}{{\lambda }^2}\mathrm{\parallel }g{\mathrm{\parallel }}_2^2+\frac{8C_n{C}_0(A_1+A_2^{\mathrm{\prime }}+A_3)}{\lambda }\mathrm{\parallel }f{\mathrm{\parallel }}_1\le C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3+B){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ where the last step uses $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\le 2^n\mu \mathrm{\parallel }f{\mathrm{\parallel }}_1=2^n\gamma \lambda \mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\gamma B^2\le B\mathrm{/}{K}_n\le (A_1+A_2^{\mathrm{\prime }}+A_3+B)\mathrm{/}{K}_n$ (in the degenerate case of step 3.1 the bound is trivial).

Weak $(1,1)$ for $f\in L^1\cap L^2$. Subadditivity of the supremum gives $T^{\ast \ast }f\le T^{\ast \ast }g+T^{\ast \ast }b$ pointwise, so $\mathrm{\mid }\{\mathrm{\mid }{T}^{\ast \ast }f\mathrm{\mid }>\lambda \}\mathrm{\mid }\le \mathrm{\mid }\{T^{\ast \ast }g>\lambda \mathrm{/}2\}\mathrm{\mid }+\mid {\bigcup }_k{Q}_k^{\ast }\mid +\mathrm{\mid }\{x\mathrm{\notin }{\bigcup }_k{Q}_k^{\ast }:T^{\ast \ast }b>\lambda \mathrm{/}2\}\mathrm{\mid }\le C_{n,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3+B){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ because the union of cubes has measure at most $(5\sqrt{n}{)}^n{\sum }_j\mathrm{\mid }{Q}_j\mathrm{\mid }\le (5\sqrt{n}{)}^n{\mu }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1\le (5\sqrt{n}{)}^n{K}_n(A_1+A_2^{\mathrm{\prime }}+A_3+B){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by steps 1.2 and 3.1 and [F3], while steps 3.1 and 4.1 bound the other two terms by dimensional multiples of $(A_1+A_2^{\mathrm{\prime }}+A_3+B){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$.

For general $f\in L^1$, put $f_m=f{\mathbf{1}}_{B(0,m)}{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le m\}}\in L^1\cap L^2$. Then $f_m\to f$ in $L^1$ and $\mathrm{\parallel }{f}_m{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1$. For every $0<\epsilon <N$, the size bound gives $\mathrm{\mid }{T}^{(\epsilon ,N)}(f_m-f)(x)\mathrm{\mid }\le A_1{\epsilon }^{-n}\mathrm{\parallel }{f}_m-f{\mathrm{\parallel }}_1\to 0$ at every $x$. Therefore $T^{\ast \ast }f(x)\le {\mathrm{lim\hspace{0.17em}inf}}_m{T}^{\ast \ast }{f}_m(x)$: each fixed truncation is bounded by this liminf, and then one takes its supremum. Fatou [F7] applied to superlevel indicators and step 5.1 give the weak $(1,1)$ bound; $T^{\ast }\le T^{\ast \ast }$ transfers it to $T^{\ast }$. No subsequence selection is needed here.

For $1<p<\mathrm{\infty }$ and $f\in L^p\cap L^2$, step 1.1 gives $T^{\ast \ast }f\le 2M(Tf)+2C_0(A_1+A_2^{\mathrm{\prime }}+A_3)Mf$ almost everywhere. The strong $L^p$ bounds [F4,F5] and $A_2\le C_{n,\delta }{A}_2^{\mathrm{\prime }}$ yield $\mathrm{\parallel }{T}^{\ast \ast }f{\mathrm{\parallel }}_p\le C_{n,p,\delta }(A_1+A_2^{\mathrm{\prime }}+A_3+B)\max (p,(p-1{)}^{-1})\mathrm{\parallel }f{\mathrm{\parallel }}_p$; additional factors depending on $p$ are included in $C_{n,p,\delta }$, as allowed by the statement. For general $f\in L^p$, the same bounded compact-support approximants $f_m$ converge in $L^p$ and satisfy $\mathrm{\parallel }{f}_m{\mathrm{\parallel }}_p\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$. Every doubly truncated kernel lies in $L^{p^{\mathrm{\prime }}}$, so Hölder gives $T^{(\epsilon ,N)}{f}_m(x)\to T^{(\epsilon ,N)}f(x)$ at every $x$ for every parameter pair. Hence $T^{\ast \ast }f\le {\mathrm{lim\hspace{0.17em}inf}}_m{T}^{\ast \ast }{f}_m$ pointwise, and Fatou [F7] applied to the $p$th powers extends the bound to all $L^p$. The comparison $T^{\ast }\le T^{\ast \ast }$ gives its bound too.

Steps 5.1 and 6.1 give the weak $(1,1)$ bound for $T^{\ast \ast }$, and step 7.1 gives the strong $L^p$ bounds for $T^{\ast \ast }$; the pointwise comparison $T^{\ast }\le T^{\ast \ast }$ of [F1] transfers both to $T^{\ast }$. This proves the theorem.