Calderón–Zygmund operators are of weak type (1,1)

Assume first $f\in L^1\cap L^2$ and $B>0$, and apply the decomposition [F2] at height $\mu =\gamma \lambda$ with $\gamma :=2^{-(n+1)}{B}^{-1}$; write $f=g+b$, $b={\sum }_j{b}_j$. The series ${\sum }_j{b}_j$ converges in $L^2$: the $b_j$ are supported on the pairwise disjoint cubes $Q_j$, and $\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_2^2\le 2{\int }_{Q_j}\mathrm{\mid }f{\mathrm{\mid }}^2+2(2^n\gamma \lambda {)}^2\mathrm{\mid }{Q}_j\mathrm{\mid }$ by $b_j=(f-\mathrm{\mid }{Q}_j{\mathrm{\mid }}^{-1}{\int }_{Q_j}f){\mathbf{1}}_{Q_j}$ and $\mid \mathrm{\mid }{Q}_j{\mathrm{\mid }}^{-1}{\int }_{Q_j}f\mid \le 2^n\gamma \lambda$ (the value of $\mathrm{\mid }g\mathrm{\mid }$ on $Q_j$), so ${\sum }_j\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+2\cdot 4^n\gamma \lambda \mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$ because $f\in L^1\cap L^2$ and ${\sum }_j\mathrm{\mid }{Q}_j\mathrm{\mid }\le (\gamma \lambda {)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$; hence $b\in L^2$ and, by linearity and $L^2$-continuity of $T$ in [F1], $Tf=Tg+{\sum }_{j}T{b}_j$ as $L^2$ classes, so $\mathrm{\mid }Tf\mathrm{\mid }\le \mathrm{\mid }Tg\mathrm{\mid }+{\sum }_j\mathrm{\mid }T{b}_j\mathrm{\mid }$ almost everywhere: choose image partial sums whose squared $L^2$ errors relative to $Tb$ are at most $2^{-3r}$. By Chebyshev the sets where the errors exceed $2^{-r}$ have measures at most $2^{-r}$; their tail unions have measures tending to zero, so this subsequence converges almost everywhere, and the finite triangle inequalities pass to the limit. The good part is controlled at the target level $\lambda \mathrm{/}2$ directly: Chebyshev's inequality for $\mathrm{\mid }Tg{\mathrm{\mid }}^2$ at level $(\lambda \mathrm{/}2{)}^2$, the $L^2$ bound $\mathrm{\parallel }Tg{\mathrm{\parallel }}_2\le B\mathrm{\parallel }g{\mathrm{\parallel }}_2$ of [F1] and the decomposition bound $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\le 2^n\gamma \lambda \mathrm{\parallel }f{\mathrm{\parallel }}_1$ at height $\mu =\gamma \lambda$ from [F2] give $\mathrm{\mid }\{\mathrm{\mid }Tg\mathrm{\mid }>\lambda \mathrm{/}2\}\mathrm{\mid }\le \frac{4}{{\lambda }^2}\mathrm{\parallel }Tg{\mathrm{\parallel }}_2^2\le \frac{4B^2}{{\lambda }^2}{2}^n\gamma \lambda \mathrm{\parallel }f{\mathrm{\parallel }}_1=4B^2{2}^n\gamma {\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1=2B{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ the last equality by the choice $\gamma =2^{-(n+1)}{B}^{-1}$.

The dilated cubes satisfy $\mathrm{\mid }{Q}_j^{\ast }\mathrm{\mid }=(2\sqrt{n}{)}^n\mathrm{\mid }{Q}_j\mathrm{\mid }$ by the dilation identity [F3] applied to the concentric dilation of $Q_j$, so the union bound and the decomposition's summability give $\mathrm{\mid }{\bigcup }_j{Q}_j^{\ast }\mathrm{\mid }\le {\sum }_j\mathrm{\mid }{Q}_j^{\ast }\mathrm{\mid }\le (2\sqrt{n}{)}^n(\gamma \lambda {)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1=2^{n+1}{n}^{n\mathrm{/}2}{2}^{n}B{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{.}$

On the complement, Tonelli's theorem for the nonnegative series and the bad-part bound of [F2] give ${\int }_{{\mathbb{R}}^n\setminus {\bigcup }_j{Q}_j^{\ast }}{\sum }_j\mathrm{\mid }T{b}_j\mathrm{\mid }dx={\sum }_j{\int }_{{\mathbb{R}}^n\setminus {\bigcup }_k{Q}_k^{\ast }}\mathrm{\mid }T{b}_j\mathrm{\mid }dx\le {\sum }_j{\int }_{{\mathbb{R}}^n\setminus Q_j^{\ast }}\mathrm{\mid }T{b}_j\mathrm{\mid }dx\le A_2{\sum }_j\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_1,$ and the decomposition's bounds $\mathrm{\parallel }{b}_j{\mathrm{\parallel }}_1\le 2^{n+1}\gamma \lambda \mathrm{\mid }{Q}_j\mathrm{\mid }$ and ${\sum }_j\mathrm{\mid }{Q}_j\mathrm{\mid }\le (\gamma \lambda {)}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$ show this is at most $2^{n+1}\gamma \lambda (\gamma \lambda {)}^{-1}{A}_2\mathrm{\parallel }f{\mathrm{\parallel }}_1=2^{n+1}{A}_2\mathrm{\parallel }f{\mathrm{\parallel }}_1$; hence Chebyshev [F3] at level $\lambda \mathrm{/}2$ yields $\mathrm{\mid }\{x\mathrm{\notin }{\bigcup }_j{Q}_j^{\ast }:{\sum }_j\mathrm{\mid }T{b}_j\mathrm{\mid }>\lambda \mathrm{/}2\}\mathrm{\mid }\le 2^{n+2}{A}_2{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$.

Combining step 1.1 (which supplies the almost-everywhere inequality and the good-part estimate), step 1.2 and step 1.3, $\mathrm{\mid }\{\mathrm{\mid }Tf\mathrm{\mid }>\lambda \}\mathrm{\mid }\le \mathrm{\mid }\{\mathrm{\mid }Tg\mathrm{\mid }>\lambda \mathrm{/}2\}\mathrm{\mid }+\mid {\bigcup }_j{Q}_j^{\ast }\mid +\mid \{x\text{ outside the cubes}:{\sum }_j\mathrm{\mid }T{b}_j\mathrm{\mid }>\lambda \mathrm{/}2\}\mid \le C_n(A_2+B){\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1,$ where $C_n:=\max \{2+2^{2n+1}{n}^{n\mathrm{/}2},2^{n+2}\}$ is the maximum of the three dimensional constants collected from steps 1.1–1.3; this is the assertion for $f\in L^1\cap L^2$.

Extension to all $f\in L^1$ and $B\ge 0$. If $B=0$, extend the zero operator on $L^2$ by zero on $L^1$. Otherwise put $D=C_n(A_2+B)$ and $f_m=f{\mathbf{1}}_{B(0,m)}{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le m\}}\in L^1\cap L^2$. The integrable tails show $\mathrm{\parallel }{f}_m-f{\mathrm{\parallel }}_1\to 0$, so step 2.1 applied to differences makes $u_m=T{f}_m$ Cauchy in measure. Select increasing $m_r$ such that $\mathrm{\mid }\{\mathrm{\mid }{u}_{m_{r+1}}-u_{m_r}\mathrm{\mid }>2^{-r}\}\mathrm{\mid }\le 2^{-r}$. The measure of the union of these exceptional sets for $r\ge R$ is at most ${\sum }_{r\ge R}{2}^{-r}\to 0$; outside their null limsup the successive differences are eventually bounded by $2^{-r}$, so $u_{m_r}$ converges to a finite measurable limit, denoted $Tf$. For each $\lambda >0$, ${\mathbf{1}}_{\{\mathrm{\mid }Tf\mathrm{\mid }>\lambda \}}\le {\mathrm{lim\hspace{0.17em}inf}}_r{\mathbf{1}}_{\{\mathrm{\mid }T{f}_{m_r}\mathrm{\mid }>\lambda \}}$ almost everywhere. Fatou's lemma [F4] and step 2.1 yield $\mathrm{\mid }\{\mathrm{\mid }Tf\mathrm{\mid }>\lambda \}\mathrm{\mid }\le {\mathrm{lim\hspace{0.17em}inf}}_r\mathrm{\mid }\{\mathrm{\mid }T{f}_{m_r}\mathrm{\mid }>\lambda \}\mathrm{\mid }\le D{\lambda }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1$. The same difference estimate implies uniqueness of limits in measure and independence of the chosen $L^1\cap L^2$ approximants; it also proves linearity by approximating two inputs and their linear combination. For $f\in L^1\cap L^2$ these truncations converge in $L^2$, so $Tf$ agrees with the original operator. Thus the compatible linear extension satisfies the required weak $(1,1)$ bound on all of $L^1$.