Calderón–Zygmund operators are bounded on Lp

The adjoint kernel $k^{\ast }(x)=\overline{k(-x)}$ satisfies the annular bound and Hörmander's condition with the constants $A_1,A_2$: the annular integral of $\mathrm{\mid }{k}^{\ast }\mathrm{\mid }$ is that of $\mathrm{\mid }k\mathrm{\mid }$ under the reflection $x\mapsto -x$, and $\mathrm{\mid }{k}^{\ast }(x-y)-k^{\ast }(x)\mathrm{\mid }=\mathrm{\mid }k(y-x)-k(-x)\mathrm{\mid }=\mathrm{\mid }\overline{k(y-x)}-\overline{k(-x)}\mathrm{\mid }$, whose integral over $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }$ equals ${\int }_{\mathrm{\mid }u\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }k(u+y)-k(u)\mathrm{\mid }du\le A_2$ by the substitution $u=-x$ and the Hörmander condition applied to $-y$. The off-support representation for $T^{\ast }$ also follows from that of $T$. For compactly supported $h\in L^2$ and a compact set $E$ disjoint from its support, $u(y)=\int \overline{k(x-y)}h(x)dx$ is absolutely convergent for almost every $y\in E$: integrating its absolute majorant over $E$ is bounded by $\mathrm{\parallel }h{\mathrm{\parallel }}_1{\int }_{E-\mathrm{supp}h}\mathrm{\mid }k(-z)\mathrm{\mid }dz<\mathrm{\infty }$, since the difference set is compact and avoids zero. For a bounded test $g$ supported in $E$, Fubini in the kernel formula for $Tg$ gives $⟨Tg,h⟩=⟨g,u⟩$; absolute integrability follows from the same majorant times $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. The adjoint identity then says $T^{\ast }h=u$ almost everywhere on $E$, since both are locally integrable and agree against all such tests. Exhausting the complement of the support by countably many compact sets proves exactly the required representation with $k^{\ast }$. Hence $T^{\ast }$, which is $L^2$-bounded with norm $B$ by [F4], is again a Calderón–Zygmund operator with constants $A_1,A_2,B$, and by [F2] both $T$ and $T^{\ast }$ are weak $(1,1)$ with constant $A:=C_n(A_2+B)$.

Sharp two-level interpolation. Let $S$ be any linear operator defined on $L^1+L^2$, weak $(1,1)$ with constant $A$ and $L^2$-bounded with constant $B$, and let $1<q<2$. For $f\in L^q$ and any $\delta >0$ put $f_0:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>\delta t\}}$ and $f_1:=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le \delta t\}}$ at height $t>0$; then $f_0\in L^1$ and $f_1\in L^2$, since $\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_1\le (\delta t{)}^{1-q}\mathrm{\parallel }f{\mathrm{\parallel }}_q^q$ and $\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_2^2\le (\delta t{)}^{2-q}\mathrm{\parallel }f{\mathrm{\parallel }}_q^q$. Hence $\{\mathrm{\mid }Sf\mathrm{\mid }>t\}\subseteq \{\mathrm{\mid }S{f}_0\mathrm{\mid }>t\mathrm{/}2\}\cup \{\mathrm{\mid }S{f}_1\mathrm{\mid }>t\mathrm{/}2\}$ and the weak $(1,1)$ bound for $S{f}_0$ together with Chebyshev and the $L^2$ bound for $S{f}_1$ give $\mathrm{\mid }\{\mathrm{\mid }Sf\mathrm{\mid }>t\}\mathrm{\mid }\le \frac{2A}{t}{\int }_{\{\mathrm{\mid }f\mathrm{\mid }>\delta t\}}\mathrm{\mid }f\mathrm{\mid }+\frac{4B^2}{t^2}{\int }_{\{\mathrm{\mid }f\mathrm{\mid }\le \delta t\}}\mathrm{\mid }f{\mathrm{\mid }}^2$. Integrating $t^{q-1}\mathrm{\mid }\{\mathrm{\mid }Sf\mathrm{\mid }>t\}\mathrm{\mid }$ over $(0,\mathrm{\infty })$ and exchanging the integrals by Tonelli gives $\mathrm{\parallel }Sf{\mathrm{\parallel }}_q^q\le q[\frac{2A{\delta }^{1-q}}{q-1}+\frac{4B^2{\delta }^{2-q}}{2-q}]\mathrm{\parallel }f{\mathrm{\parallel }}_q^q$, because ${\int }_0^{\mathrm{\mid }f\mathrm{\mid }\mathrm{/}\delta }{t}^{q-2}dt=\frac{(\mathrm{\mid }f\mathrm{\mid }\mathrm{/}\delta {)}^{q-1}}{q-1}$ and ${\int }_{\mathrm{\mid }f\mathrm{\mid }\mathrm{/}\delta }^{\mathrm{\infty }}{t}^{q-3}dt=\frac{(\mathrm{\mid }f\mathrm{\mid }\mathrm{/}\delta {)}^{q-2}}{2-q}$; choosing $\delta :=A\mathrm{/}(2B^2)$ when $A,B>0$ balances the two terms at a constant multiple of $A^{2-q}{B}^{2q-2}$, so that $\parallel Sf{\parallel }_q\le C_q{A}^{\frac{2}{q}-1}{B}^{2-\frac{2}{q}}\mathrm{\parallel }f{\mathrm{\parallel }}_q\le C_q(A+B)\mathrm{\parallel }f{\mathrm{\parallel }}_q,$ the last inequality because the exponents $\frac{2}{q}-1$ and $2-\frac{2}{q}$ are nonnegative and sum to one, so the weighted geometric mean is at most the sum; if $A=0$, let $\delta \downarrow 0$, and if $B=0$, let $\delta \to \mathrm{\infty }$ in the preceding inequality, obtaining $Sf=0$ in either case.

The case $1<p<2$: by step 1.1 the operator $T$ has weak $(1,1)$ constant $A=C_n(A_2+B)$ and $L^2$ norm $B$, so step 1.2 with $q=p$ gives $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p\le C_p{A}^{2\mathrm{/}p-1}{B}^{2-2\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_p\le C_{n,p}(A_2+B)\mathrm{\parallel }f{\mathrm{\parallel }}_p\le C_{n,p}(A_2+B)\max (p,(p-1{)}^{-1})\mathrm{\parallel }f{\mathrm{\parallel }}_p$ for every $f\in L^p$ (which lies in $L^1+L^2$ by the canonical split of step 1.2, so $Tf$ is defined).

The case $2<p<\mathrm{\infty }$: apply step 1.2 with $q=p^{\mathrm{\prime }}$ to the adjoint $T^{\ast }$, which by step 1.1 has weak $(1,1)$ constant $A$ and $L^2$ norm $B$: for every $g\in L^{p^{\mathrm{\prime }}}$, $\mathrm{\parallel }{T}^{\ast }g{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le C_{p^{\mathrm{\prime }}}(A_2+B)\mathrm{\parallel }g{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$. Then for $f\in L^p\cap L^2$ and $g\in L^{p^{\mathrm{\prime }}}\cap L^2$ with $\mathrm{\parallel }g{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le 1$, the adjoint identity and Hölder's inequality give $\mathrm{\mid }\int (Tf)g\mathrm{\mid }=\mathrm{\mid }⟨f,T^{\ast }\bar{g}⟩\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }{T}^{\ast }\bar{g}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le C_{p^{\mathrm{\prime }}}(A_2+B)\mathrm{\parallel }f{\mathrm{\parallel }}_p$, To establish $Tf\in L^p$ before using norm recovery, put $u=Tf\in L^2$, $E_N=B(0,N)\cap \{\mathrm{\mid }u\mathrm{\mid }\le N\}$ and $a_N=({\int }_{E_N}\mathrm{\mid }u{\mathrm{\mid }}^p{)}^{1\mathrm{/}p}$. If $a_N>0$, take $g_N={\mathbf{1}}_{E_N}\mathrm{\mid }u{\mathrm{\mid }}^{p-1}{\theta }_u\mathrm{/}{a}_N^{p-1}$, with ${\theta }_u=\bar{u}\mathrm{/}\mathrm{\mid }u\mathrm{\mid }$ on $u\mathrm{\ne }0$ and zero otherwise. This test is bounded on a finite-measure set, hence lies in $L^{p^{\mathrm{\prime }}}\cap L^2$, and satisfies $\mathrm{\parallel }{g}_N{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}=1$, $\int u{g}_N=a_N$. The preceding pairing bound gives $a_N\le C_{p^{\mathrm{\prime }}}(A_2+B)\mathrm{\parallel }f{\mathrm{\parallel }}_p$; if $a_N=0$ the same inequality is immediate. Since $E_N$ increases to a full-measure set, monotone convergence gives $\mathrm{\parallel }Tf{\mathrm{\parallel }}_p\le C_{p^{\mathrm{\prime }}}(A_2+B)\mathrm{\parallel }f{\mathrm{\parallel }}_p$; since $C_c^{\mathrm{\infty }}\subseteq L^p\cap L^2$ is dense in $L^p$, this bound extends uniquely to all of $L^p$, and $C_{p^{\mathrm{\prime }}}(A_2+B)\le C_{n,p}(A_2+B)\max (p,(p-1{)}^{-1})$.

The exponent dependence can be made dimension-only. Put $D=A_2+B$. If $D=0$, then $B=0$ and $T=0$. Otherwise normalize $S=T\mathrm{/}D$. Step 1.1 and the weak endpoint give weak $(1,1)$ constants at most $a_n=C_n$ for both $S$ and $S^{\ast }$, and their $L^2$ norms are at most one. The interpolation-and-duality lemma [F4] at $q=3\mathrm{/}2$ and its conjugate $3$ gives $\mathrm{\parallel }S{\mathrm{\parallel }}_{3\to 3},\mathrm{\parallel }{S}^{\ast }{\mathrm{\parallel }}_{3\to 3}\le d_n$ with $d_n$ depending only on $n$. For $1<p\le 2$, split $f=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>t\}}+f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le t\}}$. Weak $(1,1)$ on the first part and Chebyshev with the strong $(3,3)$ bound on the second yield $\mathrm{\mid }\{\mathrm{\mid }Sf\mathrm{\mid }>t\}\mathrm{\mid }\le 2a_n{t}^{-1}{\int }_{\mathrm{\mid }f\mathrm{\mid }>t}\mathrm{\mid }f\mathrm{\mid }+8d_n^3{t}^{-3}{\int }_{\mathrm{\mid }f\mathrm{\mid }\le t}\mathrm{\mid }f{\mathrm{\mid }}^3$. Layer cake and Tonelli, as in step 1.2, give $\mathrm{\parallel }Sf{\mathrm{\parallel }}_p^p\le p[2a_n\mathrm{/}(p-1)+8d_n^3\mathrm{/}(3-p)]\mathrm{\parallel }f{\mathrm{\parallel }}_p^p\le K_n(p-1{)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_p^p$, with $K_n\ge 1$ independent of $p$. The compatible $L^1$ and $L^3$ actions agree on their intersection by approximation with bounded compact-support functions in both norms. Thus $\mathrm{\parallel }S{\mathrm{\parallel }}_{p\to p}\le K_n\mathrm{/}(p-1)$ for $1<p\le 2$. Apply this estimate to $S^{\ast }$ at $p^{\mathrm{\prime }}=p\mathrm{/}(p-1)$ and use the finite-support tests and density argument of step 2.2 to get $\mathrm{\parallel }S{\mathrm{\parallel }}_{p\to p}\le K_n(p-1)\le K_{n}p$ for $p>2$. Consequently $\mathrm{\parallel }T{\mathrm{\parallel }}_{p\to p}\le K_{n}D\max (p,(p-1{)}^{-1})$ throughout the strict range.

Steps 2.1 and 2.2 prove boundedness and uniqueness in the two open ranges; the given $L^2$ bound handles $p=2$. Step 3.1 also proves the stronger bound with a dimensional constant independent of $p$, and hence the stated estimate.