[proofplan] The proof uses the [Calderón–Zygmund Decomposition](/theorems/3154) at level $\lambda$ to split $f = g + b$, where the *good part* $g$ is bounded in $L^\infty$ by $2^n \lambda$ (so we can use $L^2$ techniques) and the *bad part* $b = \sum_j b_j$ is supported on dyadic cubes with mean zero (so we can use the Hörmander smoothness condition on the kernel). For the weak-$(1,1)$ bound (i), the good part is controlled by Chebyshev plus the $L^2$ hypothesis, while the bad part is controlled outside a doubled-cube exceptional set by exploiting cancellation against the kernel via Hörmander. For the $L^p$ bound (ii), [Marcinkiewicz Interpolation](/theorems/3151) gives $1 < p \le 2$ from (i) and the $L^2$ hypothesis, and duality (using that the adjoint $T^*$ has Calderón–Zygmund kernel $\overline{K(-x)}$) extends to $2 < p < \infty$. [/proofplan]

[step:Set up the Calderón–Zygmund decomposition of $f$ at level $\lambda$] Fix $f \in L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ and $\lambda > 0$. By the [Calderón–Zygmund Decomposition](/theorems/3154), there exist: - a function $g \in L^1(\mathbb{R}^n) \cap L^\infty(\mathbb{R}^n)$ (the *good part*), - a countable family $(Q_j)_{j \in \mathbb{N}}$ of pairwise disjoint open dyadic cubes (the *Calderón–Zygmund cubes*), - functions $b_j \in L^1(\mathbb{R}^n)$ with $\operatorname{supp} b_j \subset \overline{Q_j}$, satisfying $\int_{Q_j} b_j\,d\mathcal{L}^n = 0$ (the *bad pieces*), such that with $b := \sum_j b_j$, \begin{align*} f = g + b \quad \text{a.e. in } \mathbb{R}^n, \end{align*} together with the bounds \begin{align*} \|g\|_{L^\infty} \le 2^n \lambda, \qquad \|g\|_{L^1} \le \|f\|_{L^1}, \qquad \sum_j \|b_j\|_{L^1} \le 2\|f\|_{L^1}, \qquad \sum_j \mathcal{L}^n(Q_j) \le \frac{\|f\|_{L^1}}{\lambda}. \end{align*} The hypothesis of the decomposition theorem ($f \in L^1$) is satisfied. Set \begin{align*} \Omega := \bigcup_{j \in \mathbb{N}} Q_j, \qquad \Omega^* := \bigcup_{j \in \mathbb{N}} 2 Q_j, \end{align*} where $2 Q_j$ denotes the cube concentric with $Q_j$ and side length doubled. Then \begin{align*} \mathcal{L}^n(\Omega^*) \le 2^n \sum_j \mathcal{L}^n(Q_j) \le \frac{2^n \|f\|_{L^1}}{\lambda}. \end{align*} [/step]

[step:Bound $\mathcal{L}^n(\{|Tg| > \lambda/2\})$ via Chebyshev and the $L^2$ hypothesis] The good part $g$ satisfies $\|g\|_{L^\infty} \le 2^n \lambda$ and $\|g\|_{L^1} \le \|f\|_{L^1}$, so by interpolation between $L^1$ and $L^\infty$, \begin{align*} \|g\|_{L^2}^2 = \int_{\mathbb{R}^n} |g|^2\,d\mathcal{L}^n \le \|g\|_{L^\infty}\,\|g\|_{L^1} \le 2^n \lambda\,\|f\|_{L^1}. \end{align*} Since $g \in L^2(\mathbb{R}^n)$ (in fact in $L^1 \cap L^\infty \subset L^2$), the hypothesis $\|T\|_{L^2 \to L^2} \le B$ applies, giving \begin{align*} \|Tg\|_{L^2}^2 \le B^2\,\|g\|_{L^2}^2 \le 2^n B^2\,\lambda\,\|f\|_{L^1}. \end{align*} Apply Chebyshev's inequality at level $\lambda/2$: \begin{align*} \mathcal{L}^n(\{x \in \mathbb{R}^n : |Tg(x)| > \lambda/2\}) \le \frac{4}{\lambda^2}\,\|Tg\|_{L^2}^2 \le \frac{4 \cdot 2^n B^2 \|f\|_{L^1}}{\lambda} = \frac{C_0(n) B^2}{\lambda}\,\|f\|_{L^1}, \end{align*} with $C_0(n) := 4 \cdot 2^n$. [/step]

[step:Bound $\mathcal{L}^n(\{|Tb| > \lambda/2\} \cap (\mathbb{R}^n \setminus \Omega^*))$ via the Hörmander condition] We estimate $|Tb|$ on the complement $\mathbb{R}^n \setminus \Omega^*$. Let $c_j$ denote the centre of the cube $Q_j$ and $\ell_j$ its side length. For $x \in \mathbb{R}^n \setminus 2Q_j$, the operator $T$ acts on $b_j$ via the kernel $K$: \begin{align*} Tb_j(x) = \int_{Q_j} K(x - y)\,b_j(y)\,d\mathcal{L}^n(y). \end{align*} (There is no principal-value issue because $x \notin Q_j \supseteq \operatorname{supp} b_j$, hence $x - y \neq 0$ on the integration domain.) Using the mean-zero property $\int_{Q_j} b_j\,d\mathcal{L}^n = 0$, subtract the constant value $K(x - c_j)\,\int_{Q_j} b_j\,d\mathcal{L}^n = 0$: \begin{align*} Tb_j(x) = \int_{Q_j} \big[K(x - y) - K(x - c_j)\big]\,b_j(y)\,d\mathcal{L}^n(y). \end{align*} Take absolute values and integrate over $\mathbb{R}^n \setminus 2Q_j$: \begin{align*} \int_{\mathbb{R}^n \setminus 2Q_j} |Tb_j(x)|\,d\mathcal{L}^n(x) &\le \int_{\mathbb{R}^n \setminus 2Q_j} \int_{Q_j} |K(x-y) - K(x-c_j)|\,|b_j(y)|\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x). \end{align*} By Tonelli's theorem (the integrand is non-negative and measurable; verified because $K$ is locally integrable away from the diagonal and $b_j \in L^1$), we exchange the order of integration: \begin{align*} \int_{\mathbb{R}^n \setminus 2Q_j} |Tb_j(x)|\,d\mathcal{L}^n(x) \le \int_{Q_j} |b_j(y)| \left(\int_{\mathbb{R}^n \setminus 2Q_j} |K(x-y) - K(x-c_j)|\,d\mathcal{L}^n(x)\right) d\mathcal{L}^n(y). \end{align*} For the inner integral, substitute $z := x - c_j$ (so $d\mathcal{L}^n(z) = d\mathcal{L}^n(x)$, and $x \in \mathbb{R}^n \setminus 2Q_j$ corresponds to $z \in \mathbb{R}^n$ outside the cube $2Q_j - c_j$, which is centred at the origin with side length $2\ell_j$): \begin{align*} \int_{\mathbb{R}^n \setminus 2Q_j} |K(x-y) - K(x-c_j)|\,d\mathcal{L}^n(x) = \int_{|z|_\infty > \ell_j} |K(z - (y - c_j)) - K(z)|\,d\mathcal{L}^n(z), \end{align*} where $|z|_\infty := \max_{i = 1, \dots, n} |z_i|$ is the supremum norm. Set $h := y - c_j$. Since $y \in Q_j$ and $c_j$ is the centre of $Q_j$ (a cube of side $\ell_j$), we have $|h|_\infty \le \ell_j/2$, and the relation between supremum and Euclidean norms in $\mathbb{R}^n$ gives \begin{align*} |h|_\infty \le |h| \le \sqrt{n}\,|h|_\infty, \end{align*} so $|h| \le \frac{\sqrt{n}}{2}\,\ell_j$. We now compare the integration regions. The supremum-norm region $\{|z|_\infty > \ell_j\}$ relates to the Euclidean-norm region $\{|z| > 2|h|\}$ as follows. A point with $|z|_\infty > \ell_j$ satisfies $|z| \ge |z|_\infty > \ell_j$. We claim $\ell_j \ge 2|h|$: substituting $|h| \le \frac{\sqrt{n}}{2}\,\ell_j$, this is equivalent to $\ell_j \ge \sqrt{n}\,\ell_j$, which is **false** for $n \ge 2$. Hence the inclusion $\{|z|_\infty > \ell_j\} \subseteq \{|z| > 2|h|\}$ does **not** hold directly. Instead, we use the dimensional inclusion $\{|z|_\infty > \sqrt{n}\,\ell_j\} \subseteq \{|z| > \sqrt{n}\,\ell_j\} \cap \{|z| > 2|h|\}$, valid because $\sqrt{n}\,\ell_j \ge 2 \cdot \frac{\sqrt n}{2} \ell_j \ge 2|h|$. The kernel $K$ satisfies the Hörmander condition: there exists a constant $A_H \le C_0(n)\,A$ (with $C_0(n)$ a dimensional constant arising from passing between size and smoothness conditions in the standard form of the kernel) such that for every $h \neq 0$, \begin{align*} \int_{|z| > 2|h|} |K(z - h) - K(z)|\,d\mathcal{L}^n(z) \le A_H. \end{align*} Split the supremum-norm region: \begin{align*} \{|z|_\infty > \ell_j\} = \{\ell_j < |z|_\infty \le \sqrt{n}\,\ell_j\} \cup \{|z|_\infty > \sqrt{n}\,\ell_j\}. \end{align*} The second region is contained in $\{|z| > 2|h|\}$ (just verified), so its contribution is bounded by $A_H$. The first region (the annular shell $\ell_j < |z|_\infty \le \sqrt{n}\,\ell_j$) is contained in the Euclidean annulus $\{\ell_j < |z| \le n\,\ell_j\}$ (using $|z| \le \sqrt{n}\,|z|_\infty \le n\,\ell_j$ on the shell), and on this region the size estimate $|K(w)| \le A\,|w|^{-n}$ for $w \neq 0$ gives \begin{align*} \int_{\ell_j < |z|_\infty \le \sqrt{n}\,\ell_j} |K(z - h) - K(z)|\,d\mathcal{L}^n(z) &\le \int_{\ell_j < |z|_\infty \le \sqrt{n}\,\ell_j} \big(|K(z - h)| + |K(z)|\big)\,d\mathcal{L}^n(z) \\ &\le 2 A \int_{\ell_j < |z|_\infty \le \sqrt{n}\,\ell_j} \frac{d\mathcal{L}^n(z)}{(\ell_j/2)^n}\quad\text{(using $|z - h|, |z| \ge \ell_j/2$ on this shell)} \\ &\le 2 A\,\frac{\mathcal{L}^n(\{|z|_\infty \le \sqrt{n}\,\ell_j\})}{(\ell_j/2)^n} = 2 A \cdot \frac{(2\sqrt{n}\,\ell_j)^n}{(\ell_j/2)^n} = 2 A \cdot (4\sqrt{n})^n. \end{align*} (The bound $|z - h| \ge \ell_j/2$ uses $|z|_\infty > \ell_j$ and $|h|_\infty \le \ell_j/2$, so $|z - h|_\infty \ge \ell_j/2$, hence $|z - h| \ge \ell_j/2$.) Combining the two regions, \begin{align*} \int_{|z|_\infty > \ell_j} |K(z - h) - K(z)|\,d\mathcal{L}^n(z) \le A_H + 2 A \cdot (4\sqrt{n})^n \le C_1(n)\,A, \end{align*} where the dimensional constant $C_1(n) := C_0(n) + 2 (4\sqrt{n})^n$ depends only on $n$. Plugging back, \begin{align*} \int_{\mathbb{R}^n \setminus 2Q_j} |Tb_j(x)|\,d\mathcal{L}^n(x) \le C_1(n) A \int_{Q_j} |b_j(y)|\,d\mathcal{L}^n(y) = C_1(n) A\,\|b_j\|_{L^1}. \end{align*} Sum over $j \in \mathbb{N}$ and use $\sum_j \|b_j\|_{L^1} \le 2\|f\|_{L^1}$: \begin{align*} \sum_j \int_{\mathbb{R}^n \setminus 2Q_j} |Tb_j(x)|\,d\mathcal{L}^n(x) \le C_1(n) A \cdot 2\|f\|_{L^1} = 2 C_1(n) A\,\|f\|_{L^1}. \end{align*} Since $\mathbb{R}^n \setminus \Omega^* \subseteq \mathbb{R}^n \setminus 2Q_j$ for every $j$, \begin{align*} \int_{\mathbb{R}^n \setminus \Omega^*} \sum_j |Tb_j(x)|\,d\mathcal{L}^n(x) \le 2 C_1(n) A\,\|f\|_{L^1}, \end{align*} and as $|Tb(x)| \le \sum_j |Tb_j(x)|$ pointwise (the sum converges in $L^1$ on $\mathbb{R}^n \setminus \Omega^*$), \begin{align*} \int_{\mathbb{R}^n \setminus \Omega^*} |Tb(x)|\,d\mathcal{L}^n(x) \le 2 C_1(n) A\,\|f\|_{L^1}. \end{align*} Apply Chebyshev's inequality at level $\lambda/2$: \begin{align*} \mathcal{L}^n(\{x \in \mathbb{R}^n \setminus \Omega^* : |Tb(x)| > \lambda/2\}) \le \frac{2}{\lambda} \int_{\mathbb{R}^n \setminus \Omega^*} |Tb|\,d\mathcal{L}^n \le \frac{4 C_1(n) A}{\lambda}\,\|f\|_{L^1}. \end{align*} [/step]

[step:Combine the two regimes plus the exceptional set to conclude weak-$(1,1)$] We have $|Tf| \le |Tg| + |Tb|$ pointwise (both are well-defined a.e. on $L^2(\mathbb{R}^n)$ and $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, respectively), so \begin{align*} \{|Tf| > \lambda\} \subseteq \{|Tg| > \lambda/2\} \cup \{|Tb| > \lambda/2\}. \end{align*} Decompose the second set across $\Omega^*$: \begin{align*} \{|Tb| > \lambda/2\} \subseteq \Omega^* \cup \big(\{|Tb| > \lambda/2\} \cap (\mathbb{R}^n \setminus \Omega^*)\big). \end{align*} Combining the three bounds from Steps 1, 2, and 3, \begin{align*} \mathcal{L}^n(\{|Tf| > \lambda\}) &\le \mathcal{L}^n(\{|Tg| > \lambda/2\}) + \mathcal{L}^n(\Omega^*) + \mathcal{L}^n(\{|Tb| > \lambda/2\} \cap (\mathbb{R}^n \setminus \Omega^*)) \\ &\le \frac{C_0(n)\, B^2 \|f\|_{L^1}}{\lambda} + \frac{2^n \|f\|_{L^1}}{\lambda} + \frac{4 C_1(n) A \|f\|_{L^1}}{\lambda}, \end{align*} which collects to \begin{align*} \mathcal{L}^n(\{|Tf| > \lambda\}) \le \frac{C_2(n)\,(A + B^2 + 1)}{\lambda}\,\|f\|_{L^1}, \end{align*} for a dimensional constant $C_2(n) := \max(C_0(n), 2^n, 4 C_1(n))$. This is the bound asserted in part (i) of the theorem statement, with constant $C(n) := C_2(n)$ depending only on $n$. This proves part (i) for $f \in L^1 \cap L^2$. The extension to $f \in L^1$ uses the density of $L^1 \cap L^2$ in $L^1$ and the standard weak-type extension procedure: given $f \in L^1$, choose $f_k \in L^1 \cap L^2$ with $f_k \to f$ in $L^1$; by sublinearity $|Tf_k - Tf_m| = |T(f_k - f_m)|$ and the weak-(1,1) bound on $L^1 \cap L^2$ gives a Cauchy sequence in measure, whose limit defines $Tf$ and inherits the bound. [/step]

[step:Apply Marcinkiewicz interpolation between weak-$(1,1)$ and strong-$(2,2)$ to obtain $L^p$ for $1 < p \le 2$] Fix $p \in (1, 2)$. By Step 4, $T$ is of weak type $(1,1)$ with constant $A_0 = C(n)(A + B^2 + 1)$. By hypothesis, $T$ is of strong type $(2,2)$ with constant $A_1 = B$, hence of weak type $(2,2)$ with the same constant. The Marcinkiewicz Interpolation Theorem (proved in this course as theorem 3151) requires: - $T$ sublinear: in fact linear by construction, hence sublinear. - weak-$(p_0, p_0)$ at $p_0 = 1$: established with constant $A_0$. - weak-$(p_1, p_1)$ at $p_1 = 2$: established with constant $A_1$. Applying the theorem with these endpoints and any $p \in (1, 2)$, \begin{align*} \|Tf\|_{L^p} \le C(n, p)\,A_0^{1-\theta}\,A_1^{\theta}\,\|f\|_{L^p} = C(n, p)\,(A + B^2 + 1)^{1-\theta}\,B^\theta\,\|f\|_{L^p}, \end{align*} where $\theta \in (0, 1)$ is determined by $\frac{1}{p} = \frac{1-\theta}{1} + \frac{\theta}{2}$. For $p = 2$, the strong-$(2,2)$ hypothesis directly gives $\|Tf\|_{L^2} \le B\,\|f\|_{L^2} = (A + B^2 + 1)^0\,B^1\,\|f\|_{L^2}$, the boundary case $\theta = 1$ of the bound. [/step]

[step:Identify the adjoint $T^*$ as a Calderón–Zygmund operator and apply duality for $p > 2$] Let $p \in (2, \infty)$ and $p' = p/(p-1) \in (1, 2)$. We use that the $L^2$-adjoint $T^*$ is a Calderón–Zygmund operator with kernel \begin{align*} \tilde K: \mathbb{R}^n \setminus \{0\} &\to \mathbb{C}, \\ x &\mapsto \overline{K(-x)}. \end{align*} Indeed, for $f, g \in C_c^\infty(\mathbb{R}^n)$ with disjoint supports, \begin{align*} (Tf, g)_{L^2} = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} K(x - y)\,f(y)\,\overline{g(x)}\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x), \end{align*} and exchanging the order of integration (Fubini, valid since the integrand is integrable on the disjoint-support region) and substituting $z = x - y$, $w = y$ (so $d\mathcal{L}^{2n}(z, w) = d\mathcal{L}^{2n}(x, y)$), \begin{align*} (Tf, g)_{L^2} = \int_{\mathbb{R}^n} f(y) \overline{\int_{\mathbb{R}^n} \overline{K(x-y)}\,g(x)\,d\mathcal{L}^n(x)}\,d\mathcal{L}^n(y) = (f, T^* g)_{L^2}, \end{align*} where $T^* g(y) := \int_{\mathbb{R}^n} \overline{K(x - y)}\,g(x)\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} \tilde K(y - x)\,g(x)\,d\mathcal{L}^n(x)$ with $\tilde K(z) := \overline{K(-z)}$. Thus $T^*$ is convolution with $\tilde K$. The kernel $\tilde K$ inherits the Calderón–Zygmund properties: $|\tilde K(z)| = |K(-z)| \le A|z|^{-n}$ (size), and the Hörmander condition for $\tilde K$ at offset $h$ becomes the Hörmander condition for $K$ at offset $-h$ (after the substitution $z \mapsto -z$), giving the same constant $A_H \le C(n) A$. The cancellation/$L^2$-boundedness $\|T^*\|_{L^2 \to L^2} = \|T\|_{L^2 \to L^2} \le B$ is automatic. Apply Step 5 with $T^*$ in place of $T$ and exponent $p' \in (1, 2)$: $T^*: L^{p'}(\mathbb{R}^n) \to L^{p'}(\mathbb{R}^n)$ is bounded with $\|T^*\|_{L^{p'} \to L^{p'}} \le C(n, p')(A + B^2 + 1)^{1-\theta'}\,B^{\theta'}$ for the corresponding $\theta'$. By the duality of $L^p$ spaces (which requires $1 < p' < \infty$, satisfied since $p' \in (1, 2)$), the dual of $T^*: L^{p'} \to L^{p'}$ is a bounded operator $L^p \to L^p$ with the same norm. The identification of this dual operator with $T$ on $L^p$ proceeds in two stages. **(a) Dual of $T^*$ on $C_c^\infty$ agrees with $T$.** For $f, g \in C_c^\infty(\mathbb{R}^n)$, the $L^2$-pairing $(f, T^*g)_{L^2} = (Tf, g)_{L^2}$ established above identifies $T$ as the $L^2$-adjoint of $T^*$. The $L^p$–$L^{p'}$ dual pairing is $\langle f, g \rangle = \int f\,\bar{g}\,d\mathcal{L}^n = (f, g)_{L^2}$ for $f \in C_c^\infty \subset L^p \cap L^2$ and $g \in C_c^\infty \subset L^{p'} \cap L^2$, so the two pairings agree on this dense subspace. By definition of the dual operator $(T^*)^\sharp: L^p \to L^p$ via $\langle f, T^*g\rangle = \langle (T^*)^\sharp f, g\rangle$, we get $(T^*)^\sharp f = Tf$ for $f \in C_c^\infty$. **(b) Extension by density.** The bound $\|(T^*)^\sharp f\|_{L^p} \le \|T^*\|_{L^{p'}\to L^{p'}}\|f\|_{L^p}$ holds for all $f \in L^p$ by the duality of bounded operators. Since $(T^*)^\sharp = T$ on $C_c^\infty$ (which is dense in $L^p$ for $1 \le p < \infty$), by continuity $(T^*)^\sharp = T$ on all of $L^p$ (this uses that $T$, originally defined on $L^1 \cap L^2$, extends consistently to $L^p$ via the bound just established). Combining, \begin{align*} \|Tf\|_{L^p} \le C(n, p)\,(A + B^2 + 1)^{1-\theta'}\,B^{\theta'}\,\|f\|_{L^p} \end{align*} for all $f \in L^p(\mathbb{R}^n)$. This proves part (ii) for $p \in (2, \infty)$, completing the range. [/step]