Calderon-Zygmund operators map L-infinity to BMO

The local term obeys $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }T(b{\mathbf{1}}_{{\lambda }_{n}Q})\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}2}\mathrm{\parallel }T(b{\mathbf{1}}_{{\lambda }_{n}Q}){\mathrm{\parallel }}_2\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}2}B\mathrm{\parallel }b{\mathbf{1}}_{{\lambda }_{n}Q}{\mathrm{\parallel }}_2\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}2}B\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}\mathrm{\mid }{\lambda }_{n}Q{\mathrm{\mid }}^{1\mathrm{/}2}={\lambda }_n^{n\mathrm{/}2}B\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$, by Cauchy-Schwarz [F4], the $L^2$ bound [F1] and $\mathrm{\mid }{\lambda }_{n}Q\mathrm{\mid }={\lambda }_n^n\mathrm{\mid }Q\mathrm{\mid }$ [F3].

For $x,a\in Q$ one has $\mathrm{\mid }x-a\mathrm{\mid }\le \sqrt{n}\mathrm{\ell }(Q)$; for $y\mathrm{\notin }{\lambda }_{n}Q$ the coordinatewise distance from $a$ to the complement of the concentric cube ${\lambda }_{n}Q$ is at least $\frac{{\lambda }_n-1}{2}\mathrm{\ell }(Q)=2\sqrt{n}\mathrm{\ell }(Q)$, so $\mathrm{\mid }a-y\mathrm{\mid }\ge 2\sqrt{n}\mathrm{\ell }(Q)\ge 2\mathrm{\mid }x-a\mathrm{\mid }$; and if $Q\subseteq R$, $a\in Q\subseteq R$ and $y\mathrm{\notin }{\lambda }_{n}R$, the same computation with $\mathrm{\ell }(R)$ gives $\mathrm{\mid }a-y\mathrm{\mid }\ge 2\sqrt{n}\mathrm{\ell }(R)\ge 2\mathrm{\mid }x-a\mathrm{\mid }$.

The tail integral converges absolutely for every $x\in Q$. If $x=a$, its integrand is zero. Otherwise, by [F2] and step 1.2, $\mathrm{\mid }k(x-y)-k(a-y)\mathrm{\mid }\le A_2^{\mathrm{\prime }}\mathrm{\mid }x-a{\mathrm{\mid }}^{\delta }\mathrm{\mid }a-y{\mathrm{\mid }}^{-n-\delta }\le A_2^{\mathrm{\prime }}(\sqrt{n}\mathrm{\ell }(Q){)}^{\delta }\mathrm{\mid }a-y{\mathrm{\mid }}^{-n-\delta }$ for $y\mathrm{\notin }{\lambda }_{n}Q$, and by [F5] the tail power integral is ${\int }_{{\mathbb{R}}^n\setminus {\lambda }_{n}Q}\mathrm{\mid }a-y{\mathrm{\mid }}^{-n-\delta }dy\le {\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 2\sqrt{n}\mathrm{\ell }(Q)}\mathrm{\mid }z{\mathrm{\mid }}^{-n-\delta }dz=\sigma (S^{n-1}){\delta }^{-1}(2\sqrt{n}\mathrm{\ell }(Q){)}^{-\delta }$. Hence $\mid {\int }_{{\mathbb{R}}^n\setminus {\lambda }_{n}Q}[k(x-y)-k(a-y)]b(y)dy\mid \le A_2^{\mathrm{\prime }}\sigma (S^{n-1}){\delta }^{-1}{2}^{-\delta }\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$, a bound independent of $x$ and $Q$; this is (i), and the tail is a bounded measurable function of $x$: [F2] implies continuity of $k$ away from zero, while its displayed bound gives an integrable majorant uniformly on $Q$, so dominated convergence gives continuity of the tail there.

Nested consistency. Let $Q\subseteq R$ be cubes with points $a_Q\in Q$ and $a_R\in R$, and let $x\in Q$. Since ${\lambda }_{n}Q\subseteq {\lambda }_{n}R$, splitting the complement of ${\lambda }_{n}Q$ into ${\lambda }_{n}R\setminus {\lambda }_{n}Q$ and ${\mathbb{R}}^n\setminus {\lambda }_{n}R$ and using linearity of $T$ gives $(Tb{)}_Q(x)-(Tb{)}_R(x)=-T(b{\mathbf{1}}_{{\lambda }_{n}R\setminus {\lambda }_{n}Q})(x)+{\int }_{{\lambda }_{n}R\setminus {\lambda }_{n}Q}[k(x-y)-k(a_Q-y)]b(y)dy+{\int }_{{\mathbb{R}}^n\setminus {\lambda }_{n}R}[k(a_R-y)-k(a_Q-y)]b(y)dy$. The function $b{\mathbf{1}}_{{\lambda }_{n}R\setminus {\lambda }_{n}Q}$ is compactly supported $L^2$ with support disjoint from $Q$, so the off-support representation [F1] gives $T(b{\mathbf{1}}_{{\lambda }_{n}R\setminus {\lambda }_{n}Q})(x)={\int }_{{\lambda }_{n}R\setminus {\lambda }_{n}Q}k(x-y)b(y)dy$ for almost every $x\in Q$; the first integral over ${\lambda }_{n}R\setminus {\lambda }_{n}Q$ converges absolutely because that region lies in a bounded annulus about $a_Q$ on which the annular bound of [F1] controls $k$. If $a_R=a_Q$, the last integral is zero; otherwise it converges absolutely by Hormander's condition [F1] applied at centre $a_R$ with nonzero translation $v=a_R-a_Q$, since $\mathrm{\mid }{a}_R-y\mathrm{\mid }\ge 2\sqrt{n}\mathrm{\ell }(R)\ge 2\mathrm{\mid }{a}_R-a_Q\mathrm{\mid }$ for $y\mathrm{\notin }{\lambda }_{n}R$ by step 1.2. The resulting expression is independent of $x$, so $(Tb{)}_Q-(Tb{)}_R$ is almost everywhere constant on $Q$, which is (ii).

Combining steps 1.1 and 2.1, $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }(Tb{)}_Q\mathrm{\mid }\le ({\lambda }_n^{n\mathrm{/}2}B+A_2^{\mathrm{\prime }}\sigma (S^{n-1}){\delta }^{-1}{2}^{-\delta })\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ for every cube $Q$, hence by the optimal-constant bound of [F8] with $c=0$ the mean oscillation of $(Tb{)}_Q$ over $Q$ is at most $2({\lambda }_n^{n\mathrm{/}2}B+A_2^{\mathrm{\prime }}\sigma (S^{n-1}){\delta }^{-1}{2}^{-\delta })\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$.

Coherence and gluing. If $Q_1,Q_2$ are cubes, choose a cube $R$ containing both; step 2.2 shows that each $(Tb{)}_{Q_i}-(Tb{)}_R$ is constant almost everywhere on $Q_i$, so $(Tb{)}_{Q_1}-(Tb{)}_{Q_2}$ is constant almost everywhere on $Q_1\cap Q_2$. Let $Q_k=[-k,k{]}^n$ for $k\ge 1$ and choose representatives of the countably many localisations, which Countable Choice permits [F7]. Define constants $c_k$ inductively by $c_1=0$ and $c_{k+1}=c_k-[(Tb{)}_{Q_{k+1}}-(Tb{)}_{Q_k}]$, the bracket being the constant of step 2.2 on $Q_k$; then $u_k:=(Tb{)}_{Q_k}+c_k$ satisfies $u_{k+1}=u_k$ almost everywhere on $Q_k$. Removing the countable union of the exceptional null sets, which is null by [F7], define $u(x):=u_k(x)$ for $x\in Q_k$ outside that null set, and set $u=0$ on the null set; this is well defined and locally integrable, and for every cube $Q$, choosing $k$ with $Q\subseteq Q_k$, the function $u-(Tb{)}_Q=[u-(Tb{)}_{Q_k}]+[(Tb{)}_{Q_k}-(Tb{)}_Q]$ is almost everywhere constant on $Q$ by the construction and step 2.2. Two such global representatives differ by constants on the nested $Q_k$; these constants agree on their positive-measure overlaps, so the global class is unique. Linearity follows from linearity of each localisation and this uniqueness.

By step 3.2, $u-(Tb{)}_Q$ is constant almost everywhere on every cube $Q$, so the mean oscillation of $u$ over $Q$ equals that of $(Tb{)}_Q$; by step 3.1, $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }u-u_Q\mathrm{\mid }\le 2({\lambda }_n^{n\mathrm{/}2}B+A_2^{\mathrm{\prime }}\sigma (S^{n-1}){\delta }^{-1}{2}^{-\delta })\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ for every cube. Hence $u\in \mathrm{B}\mathrm{M}\mathrm{O}({\mathbb{R}}^n)$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}\le C_{n,\delta }(A_2^{\mathrm{\prime }}+B)\mathrm{\parallel }b{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ for $C_{n,\delta }:=2\max ({\lambda }_n^{n\mathrm{/}2},\sigma (S^{n-1}){\delta }^{-1}{2}^{-\delta })$, and its class modulo constants is the class $Tb$ of the statement, which is (iii).

Suppose $b\in L^2({\mathbb{R}}^n)$; fix a cube $Q$ and $a\in Q$ and write $c:={\mathbb{R}}^n\setminus {\lambda }_{n}Q$ and $\mathrm{\Psi }(x):={\int }_c[k(x-y)-k(a-y)]b(y)dy$ for the tail of the localisation. Let $b_N:=b{\mathbf{1}}_{c\cap B(0,N)}$ for $N\ge 1$. Each $b_N$ is compactly supported $L^2$ with support disjoint from $Q$, so [F1] gives $T{b}_N(x)={\int }_{c}k(x-y)b_N(y)dy$ for almost every $x\in Q$; since $b_N\to b{\mathbf{1}}_c$ in $L^2$, boundedness of $T$ gives $T{b}_N\to T(b{\mathbf{1}}_c)$ in $L^2$. By [F9] choose a subsequence whose representatives converge almost everywhere to a representative of $T(b{\mathbf{1}}_c)$, and intersect this full-measure set with the full-measure set where the countably many off-support identities hold. For $x=x^{\mathrm{\prime }}$ the kernel difference below is zero. For distinct $x,x^{\mathrm{\prime }}\in Q$ outside the exceptional null set and $y\in c$, one has $\mathrm{\mid }{x}^{\mathrm{\prime }}-y\mathrm{\mid }\ge 2\sqrt{n}\mathrm{\ell }(Q)\ge 2\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$ by step 1.2, so $\mathrm{\mid }k(x-y)-k(x^{\mathrm{\prime }}-y)\mathrm{\mid }\le A_2^{\mathrm{\prime }}\mathrm{\mid }x-x^{\mathrm{\prime }}{\mathrm{\mid }}^{\delta }\mathrm{\mid }{x}^{\mathrm{\prime }}-y{\mathrm{\mid }}^{-n-\delta }$ by [F2]; the function $y\mapsto A_2^{\mathrm{\prime }}\mathrm{\mid }x-x^{\mathrm{\prime }}{\mathrm{\mid }}^{\delta }\mathrm{\mid }{x}^{\mathrm{\prime }}-y{\mathrm{\mid }}^{-n-\delta }\mathrm{\mid }b(y)\mathrm{\mid }$ is integrable on $c$ by Cauchy-Schwarz [F4] and [F5], so dominated convergence [F6] gives along that subsequence ${\lim }_j[T{b}_{N_j}(x)-T{b}_{N_j}(x^{\mathrm{\prime }})]={\int }_c[k(x-y)-k(x^{\mathrm{\prime }}-y)]b(y)dy=\mathrm{\Psi }(x)-\mathrm{\Psi }(x^{\mathrm{\prime }})$. Hence $T(b{\mathbf{1}}_c)-\mathrm{\Psi }$ has equal values at almost every pair of points of $Q$, so by Fubini [F6] it is almost everywhere constant on $Q$. Since $Tb=T(b{\mathbf{1}}_{{\lambda }_{n}Q})+T(b{\mathbf{1}}_c)$ in $L^2$ and $(Tb{)}_Q=T(b{\mathbf{1}}_{{\lambda }_{n}Q})+\mathrm{\Psi }$, the difference $Tb-(Tb{)}_Q$ is almost everywhere constant on $Q$; that is (iv), and it identifies the $L^2$ class of $Tb$ with the localisation class of step 3.2 modulo constants.

Steps 2.1, 2.2, 4.1 and 4.2 prove (i), (ii), (iii) and (iv) respectively, and steps 3.1 and 3.2 supply the global representative $u$ used in (iii). The argument uses Countable Choice exactly in the countably many applications of the off-support representation and in the selection of the representatives and subsequence in steps 3.2 and 4.2, and it uses no other choice principle.