[proofplan] We invoke the [Fefferman duality theorem](/theorems/???), which identifies $\mathrm{BMO}(\mathbb{R}^n)$ with the dual of the Hardy space $H^1(\mathbb{R}^n)$ via the pairing $g \mapsto \Lambda_g(f) = \int_{\mathbb{R}^n} f g \, d\mathcal{L}^n$. Concretely, $g \in \mathrm{BMO}$ if and only if the linear functional $\Lambda_g$ defined on $H^1$-atoms is uniformly bounded, with $\|g\|_{\mathrm{BMO}} \asymp \sup_{a \text{ atom}} |\Lambda_g(a)|$. Thus to control $\|Tf\|_{\mathrm{BMO}}$ for $f \in L^\infty$, it suffices to bound $|\int (Tf) a \, d\mathcal{L}^n|$ uniformly over $H^1$-atoms $a$. We achieve this by passing to the formal adjoint $T^*$ via $L^2$ duality and then invoking the $H^1 \to L^1$ boundedness of Calderón–Zygmund operators on atoms. [/proofplan] [step:Reduce $L^\infty \to \mathrm{BMO}$ boundedness to a uniform pairing bound on $H^1$-atoms] Recall that an [$H^1$-atom](/page/Atomic%20Hardy%20Space) is a measurable function $a: \mathbb{R}^n \to \mathbb{C}$ supported in some ball $B = B(x_0, r) \subset \mathbb{R}^n$ with $\|a\|_{L^\infty(\mathbb{R}^n)} \le |B|^{-1}$ and $\int_{\mathbb{R}^n} a \, d\mathcal{L}^n = 0$. By the [atomic decomposition of $H^1$](/theorems/???), every $f \in H^1(\mathbb{R}^n)$ admits a representation $f = \sum_j \lambda_j a_j$ with $a_j$ atoms and $\sum_j |\lambda_j| \asymp \|f\|_{H^1}$. By the [Fefferman duality theorem](/theorems/???), a locally integrable function $g: \mathbb{R}^n \to \mathbb{C}$ belongs to $\mathrm{BMO}(\mathbb{R}^n)$ if and only if the linear functional \begin{align*} \Lambda_g: H^1(\mathbb{R}^n) &\to \mathbb{C} \\ f &\mapsto \int_{\mathbb{R}^n} f(x) g(x) \, d\mathcal{L}^n(x) \end{align*} (initially defined on a dense subset of $H^1$ where the integral is absolutely convergent) extends to a bounded linear functional on $H^1$, with \begin{align*} \|g\|_{\mathrm{BMO}(\mathbb{R}^n)} \asymp \|\Lambda_g\|_{(H^1)^*} \asymp \sup_{a \text{ $H^1$-atom}} \left| \int_{\mathbb{R}^n} a(x) g(x) \, d\mathcal{L}^n(x) \right|. \end{align*} The implicit constants depend only on $n$. To prove $Tf \in \mathrm{BMO}(\mathbb{R}^n)$ with the asserted bound, it therefore suffices to show that there exists a constant $C_1 = C_1(n, T) > 0$ such that for every $H^1$-atom $a$ supported in a ball $B \subset \mathbb{R}^n$, \begin{align*} \left| \int_{\mathbb{R}^n} (Tf)(x) \, a(x) \, d\mathcal{L}^n(x) \right| \le C_1 \|f\|_{L^\infty(\mathbb{R}^n)}. \end{align*} [/step] [step:Move $T$ to the test function via the formal adjoint $T^*$] The Calderón–Zygmund operator $T$ has a [formal adjoint](/page/Adjoint%20Operator) $T^*: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$, defined by the relation \begin{align*} \int_{\mathbb{R}^n} (Th)(x) \, k(x) \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} h(x) \, (T^* k)(x) \, d\mathcal{L}^n(x) \quad \text{for all } h, k \in L^2(\mathbb{R}^n). \end{align*} Since $T$ is bounded on $L^2$, so is $T^*$, with $\|T^*\|_{\mathcal{L}(L^2)} = \|T\|_{\mathcal{L}(L^2)}$. The kernel of $T^*$ is $K^*(x, y) = \overline{K(y, x)}$, and inherits the size and Hörmander conditions from $K$ (with the same constants up to interchanging the roles of $x$ and $y$). Hence $T^*$ is itself a Calderón–Zygmund operator with the same quantitative kernel constants as $T$. Fix an $H^1$-atom $a$ supported in a ball $B = B(x_0, r) \subset \mathbb{R}^n$. Since $a \in L^\infty(\mathbb{R}^n)$ with compact support, $a \in L^2(\mathbb{R}^n)$. The bound $\|a\|_{L^\infty} \le |B|^{-1}$ together with $\operatorname{supp}(a) \subseteq B$ gives \begin{align*} \|a\|_{L^2(\mathbb{R}^n)}^2 = \int_B |a(x)|^2 \, d\mathcal{L}^n(x) \le |B|^{-2} \cdot |B| = |B|^{-1}. \end{align*} Approximate $f \in L^\infty(\mathbb{R}^n)$ by truncations $f_R := f \cdot \mathbb{1}_{B(0,R)}$, which lie in $L^\infty \cap L^2$. For each $R > 0$ the $L^2$ duality of $T$ applies: \begin{align*} \int_{\mathbb{R}^n} (T f_R)(x) \, a(x) \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} f_R(x) \, (T^* a)(x) \, d\mathcal{L}^n(x). \end{align*} We will show below (next step) that $T^* a \in L^1(\mathbb{R}^n)$ with $\|T^* a\|_{L^1} \le C_2(n, T)$. Granting this, both integrands are dominated independently of $R$ — the right-hand side by the $L^\infty$–$L^1$ pairing, the left-hand side by the $L^2$–$L^2$ pairing — and the [dominated convergence theorem](/theorems/???) applies as $R \to \infty$ (with $f_R \to f$ pointwise a.e. and uniformly bounded by $\|f\|_{L^\infty}$ on the right, and $T f_R \to T f$ in the appropriate sense on the left, which we will not need to invoke directly because we only need the right-hand side to interpret $\int (Tf) a \, d\mathcal{L}^n$). Concretely, define $\int_{\mathbb{R}^n} (Tf)(x) a(x) \, d\mathcal{L}^n(x)$ as the limit of the truncated pairings, equivalently as \begin{align*} \int_{\mathbb{R}^n} (Tf)(x) \, a(x) \, d\mathcal{L}^n(x) := \int_{\mathbb{R}^n} f(x) \, (T^* a)(x) \, d\mathcal{L}^n(x), \end{align*} which is the well-defined definition of $T$ on $L^\infty$ paired against an atom. [/step] [step:Bound $\|T^* a\|_{L^1}$ uniformly over $H^1$-atoms] We show that for every $H^1$-atom $a$ supported in a ball $B = B(x_0, r) \subset \mathbb{R}^n$, the image $T^* a$ lies in $L^1(\mathbb{R}^n)$ with \begin{align*} \|T^* a\|_{L^1(\mathbb{R}^n)} \le C_2 = C_2(n, \|T\|_{\mathcal{L}(L^2)}, \text{kernel constants of } T), \end{align*} independently of $a$. Split the integral $\int_{\mathbb{R}^n} |T^* a| \, d\mathcal{L}^n$ at the doubled ball $2B = B(x_0, 2r)$: \begin{align*} \|T^* a\|_{L^1(\mathbb{R}^n)} = \int_{2B} |T^* a(x)| \, d\mathcal{L}^n(x) + \int_{\mathbb{R}^n \setminus 2B} |T^* a(x)| \, d\mathcal{L}^n(x). \end{align*} [claim:The local part is bounded: $\int_{2B} |T^* a| \, d\mathcal{L}^n \le C \|T\|_{\mathcal{L}(L^2)}$] [proof] By the [Cauchy–Schwarz inequality](/theorems/???) applied to $T^* a \cdot \mathbb{1}_{2B}$, \begin{align*} \int_{2B} |T^* a| \, d\mathcal{L}^n &\le |2B|^{1/2} \|T^* a\|_{L^2(\mathbb{R}^n)} \\ &\le |2B|^{1/2} \|T^*\|_{\mathcal{L}(L^2)} \|a\|_{L^2(\mathbb{R}^n)} \\ &\le |2B|^{1/2} \|T\|_{\mathcal{L}(L^2)} |B|^{-1/2}. \end{align*} The first inequality is Cauchy–Schwarz with the $L^2$-pair $(T^* a \cdot \mathbb{1}_{2B}, \mathbb{1}_{2B})$. The second inequality is the $L^2$-boundedness of $T^*$, which has the same operator norm as $T$. The third inequality uses $\|a\|_{L^2}^2 \le |B|^{-1}$ from the atomic normalisation. Since $|2B| = 2^n |B|$, we obtain \begin{align*} \int_{2B} |T^* a| \, d\mathcal{L}^n \le 2^{n/2} \|T\|_{\mathcal{L}(L^2)}, \end{align*} which is the claimed bound. [/proof] [/claim] [claim:The far-field part is bounded: $\int_{\mathbb{R}^n \setminus 2B} |T^* a| \, d\mathcal{L}^n \le C(n, \text{kernel constants})$] [proof] For $x \in \mathbb{R}^n \setminus 2B$ we have $|x - x_0| \ge 2r$, while $\operatorname{supp}(a) \subseteq B$ so $|y - x_0| < r$ for all $y$ in the support of $a$. Then $|x - y| \ge |x - x_0| - |y - x_0| > |x - x_0|/2 \ge r$, so the kernel $K^*(x, y)$ is well-defined (off the diagonal). Use the mean-zero property $\int_{\mathbb{R}^n} a(y) \, d\mathcal{L}^n(y) = 0$ to write, for $x \in \mathbb{R}^n \setminus 2B$, \begin{align*} T^* a(x) = \int_B K^*(x, y) a(y) \, d\mathcal{L}^n(y) = \int_B [K^*(x, y) - K^*(x, x_0)] a(y) \, d\mathcal{L}^n(y). \end{align*} The substitution of $K^*(x, y) - K^*(x, x_0)$ for $K^*(x, y)$ is justified because $\int_B a(y) \, d\mathcal{L}^n(y) = 0$, so the term $K^*(x, x_0) \int_B a(y) \, d\mathcal{L}^n(y)$ vanishes. Taking absolute values and using $\|a\|_{L^\infty} \le |B|^{-1}$, \begin{align*} |T^* a(x)| \le |B|^{-1} \int_B |K^*(x, y) - K^*(x, x_0)| \, d\mathcal{L}^n(y). \end{align*} Integrating in $x$ over $\mathbb{R}^n \setminus 2B$ and applying [Tonelli's theorem](/theorems/???) (legal because the integrand is non-negative), \begin{align*} \int_{\mathbb{R}^n \setminus 2B} |T^* a(x)| \, d\mathcal{L}^n(x) &\le |B|^{-1} \int_B \left( \int_{\mathbb{R}^n \setminus 2B} |K^*(x, y) - K^*(x, x_0)| \, d\mathcal{L}^n(x) \right) d\mathcal{L}^n(y). \end{align*} For each $y \in B$, $|y - x_0| < r$ and $\mathbb{R}^n \setminus 2B \subseteq \{x: |x - x_0| > 2r\} \subseteq \{x: |x - x_0| > 2|y - x_0|\}$. The [Hörmander condition](/theorems/???) on the kernel $K^*$ states that there exists a constant $C_H = C_H(n, T) > 0$ such that for all $y, x_0 \in \mathbb{R}^n$ with $y \ne x_0$, \begin{align*} \int_{|x - x_0| > 2|y - x_0|} |K^*(x, y) - K^*(x, x_0)| \, d\mathcal{L}^n(x) \le C_H. \end{align*} Thus the inner integral above is bounded by $C_H$ uniformly in $y \in B$, giving \begin{align*} \int_{\mathbb{R}^n \setminus 2B} |T^* a(x)| \, d\mathcal{L}^n(x) \le |B|^{-1} \cdot |B| \cdot C_H = C_H, \end{align*} which proves the claim. [/proof] [/claim] Adding the two claims: \begin{align*} \|T^* a\|_{L^1(\mathbb{R}^n)} \le 2^{n/2} \|T\|_{\mathcal{L}(L^2)} + C_H =: C_2. \end{align*} [/step] [step:Combine the pairing bounds and conclude $L^\infty \to \mathrm{BMO}$ boundedness] For an arbitrary $H^1$-atom $a$ supported in a ball $B \subset \mathbb{R}^n$ and arbitrary $f \in L^\infty(\mathbb{R}^n)$, combining the previous two steps, \begin{align*} \left| \int_{\mathbb{R}^n} (Tf)(x) \, a(x) \, d\mathcal{L}^n(x) \right| &= \left| \int_{\mathbb{R}^n} f(x) \, (T^* a)(x) \, d\mathcal{L}^n(x) \right| \\ &\le \|f\|_{L^\infty(\mathbb{R}^n)} \|T^* a\|_{L^1(\mathbb{R}^n)} \\ &\le C_2 \|f\|_{L^\infty(\mathbb{R}^n)}, \end{align*} where $C_2 = 2^{n/2} \|T\|_{\mathcal{L}(L^2)} + C_H$ depends only on $n$ and the structural constants of $T$. This is exactly the uniform pairing bound on atoms required by Step 1. By the Fefferman duality characterisation, this shows $Tf \in \mathrm{BMO}(\mathbb{R}^n)$ with \begin{align*} \|Tf\|_{\mathrm{BMO}(\mathbb{R}^n)} \le C_3 \cdot C_2 \|f\|_{L^\infty(\mathbb{R}^n)} =: C \|f\|_{L^\infty(\mathbb{R}^n)}, \end{align*} where $C_3 = C_3(n)$ is the implicit constant from the duality identification $\|g\|_{\mathrm{BMO}} \asymp \sup_a |\Lambda_g(a)|$. The constant $C$ depends only on $n$, $\|T\|_{\mathcal{L}(L^2)}$, and the Hörmander kernel constant $C_H$ of $T$. This proves the asserted boundedness $T: L^\infty(\mathbb{R}^n) \to \mathrm{BMO}(\mathbb{R}^n)$. [/step]