[proofplan] The "only if" direction packages standard facts: if $T$ is bounded on $L^2$, the weak boundedness property follows from Cauchy–Schwarz on bump functions, and $T(1), T^*(1) \in \mathrm{BMO}$ follow from the [Fefferman–Stein result](/theorems/???) that $L^2$-bounded Calderón–Zygmund operators map $L^\infty \to \mathrm{BMO}$. For the "if" direction, we use $b_1 := T(1)$ and $b_2 := T^*(1)$ to build two paraproducts $\Pi_{b_1}$ and $\Pi_{b_2}$, each bounded on $L^2$ by the [Carleson Embedding Theorem](/theorems/???), and decompose $T = \Pi_{b_1} + \Pi_{b_2}^* + R$ where the residual operator $R$ has kernel of standard type and *cancellation* $R(1) = R^*(1) = 0$. The residual $R$ is $L^2$-bounded by the [Cotlar–Stein Lemma](/theorems/3166) applied to its dyadic Littlewood–Paley decomposition: the cancellation conditions $R(1) = R^*(1) = 0$ together with the standard kernel estimates give the almost-orthogonality bounds that drive the lemma. [/proofplan] [step:Establish necessity of WBP and $T(1), T^*(1) \in \mathrm{BMO}$] Assume $T: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ is bounded with norm $\|T\|_{\mathcal{L}(L^2)} \le M < \infty$. **WBP.** For any normalised bump function $\varphi \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp} \varphi \subseteq B(x_0, r)$, $\|\varphi\|_{L^\infty} \le 1$, $\|\nabla \varphi\|_{L^\infty} \le r^{-1}$, and similarly for $\psi$, we estimate \begin{align*} |T\varphi(\psi)| = |(T\varphi, \overline{\psi})_{L^2}| \le \|T\varphi\|_{L^2}\,\|\psi\|_{L^2} \le M\,\|\varphi\|_{L^2}\,\|\psi\|_{L^2} \le M\,r^n, \end{align*} using $\|\varphi\|_{L^2} \le \|\varphi\|_{L^\infty}\,\mathcal{L}^n(B(x_0,r))^{1/2} \le \alpha_n^{1/2}\,r^{n/2}$ where $\alpha_n := \mathcal{L}^n(B(0,1))$. The bound $|T\varphi(\psi)| \le M\alpha_n\,r^n$ is the WBP normalisation, so WBP holds. **$T(1), T^*(1) \in \mathrm{BMO}$.** This is the [Fefferman–Stein $L^\infty \to \mathrm{BMO}$ Boundedness of Calderón–Zygmund Operators](/theorems/???): a Calderón–Zygmund operator that is bounded on $L^2$ extends to a bounded operator $L^\infty(\mathbb{R}^n) \to \mathrm{BMO}(\mathbb{R}^n)$. Since the constant function $1 \in L^\infty(\mathbb{R}^n)$, both $T(1)$ and $T^*(1)$ are well-defined elements of $\mathrm{BMO}(\mathbb{R}^n)$. [/step] [step:Construct the paraproducts $\Pi_{b_1}, \Pi_{b_2}$ from $b_1 := T(1)$ and $b_2 := T^*(1)$] Henceforth assume the three conditions. Set $b_1 := T(1) \in \mathrm{BMO}$ and $b_2 := T^*(1) \in \mathrm{BMO}$. Fix a Littlewood–Paley decomposition: choose $\eta \in C_c^\infty(\mathbb{R}^n)$ radial, non-negative, supported in $\{1/2 \le |\xi| \le 2\}$, with $\sum_{j \in \mathbb{Z}} \eta(2^{-j} \xi)^2 = 1$ for $\xi \neq 0$. The associated frequency-localised operators are \begin{align*} \Delta_j: L^2(\mathbb{R}^n) &\to L^2(\mathbb{R}^n), \\ f &\mapsto \mathcal{F}^{-1}\big(\eta(2^{-j} \cdot)\,\hat{f}\big), \end{align*} and the low-frequency cutoffs are $S_j := \sum_{k < j} \Delta_k^2$ (telescoping to a Schwartz convolution operator). For $b \in \mathrm{BMO}(\mathbb{R}^n)$, define the **paraproduct** \begin{align*} \Pi_b: L^2(\mathbb{R}^n) &\to L^2(\mathbb{R}^n), \\ f &\mapsto \sum_{j \in \mathbb{Z}} \Delta_j(b)\,S_{j-1}(\Delta_j f). \end{align*} By the [Carleson Embedding Theorem applied to the BMO Carleson measure](/theorems/???) — which uses that $|\Delta_j(b)|^2\,\frac{d\mathcal{L}^n(x)\,dj}{1}$ is a Carleson measure with constant $\lesssim \|b\|_{\mathrm{BMO}}^2$ — together with [Plancherel](/theorems/???), the operator $\Pi_b$ is bounded on $L^2(\mathbb{R}^n)$ with \begin{align*} \|\Pi_b\|_{\mathcal{L}(L^2)} \le C(n)\,\|b\|_{\mathrm{BMO}}. \end{align*} Apply this with $b = b_1$ and $b = b_2$: \begin{align*} \|\Pi_{b_1}\|_{\mathcal{L}(L^2)} \le C(n)\,\|T(1)\|_{\mathrm{BMO}}, \qquad \|\Pi_{b_2}\|_{\mathcal{L}(L^2)} \le C(n)\,\|T^*(1)\|_{\mathrm{BMO}}. \end{align*} The key algebraic properties used below are \begin{align*} \Pi_b(1)(\varphi) = b(\varphi) \qquad \text{and} \qquad \Pi_b^*(1)(\varphi) = 0 \quad \text{for all } \varphi \in C_c^\infty(\mathbb{R}^n) \text{ with } \int \varphi\,d\mathcal{L}^n = 0. \end{align*} We verify both identities. Throughout, $1 \in \mathcal{S}'(\mathbb{R}^n)$ is interpreted as the constant tempered distribution, and $\Delta_j$, $S_{j-1}$ are extended to $\mathcal{S}'$ by duality in the standard way; the same applies to their formal $L^2$-adjoints $\Delta_j^*$ and $S_{j-1}^*$. \medskip \noindent \textbf{Identity 1: $\Pi_b(1) = b$ on mean-zero test functions.} For $\varphi \in C_c^\infty(\mathbb{R}^n)$ with $\int \varphi\,d\mathcal{L}^n = 0$, \begin{align*} \Pi_b(1)(\varphi) = \sum_{j \in \mathbb{Z}} \big(\Delta_j(b)\,S_{j-1}(\Delta_j(1))\big)(\varphi). \end{align*} Apply $\Delta_j$ to the constant function $1$: since $\widehat{1} = (2\pi)^n \delta_0$ and the multiplier $\eta(2^{-j}\xi)$ vanishes in a neighbourhood of $\xi = 0$, we have $\Delta_j(1) = 0$ as a tempered distribution for every $j \in \mathbb{Z}$. With $\Delta_j(1) = 0$ the formula above does not directly yield $b$; the correct algebraic identity is that the partial-sum telescoping of the Littlewood–Paley resolution gives \begin{align*} \sum_{|j| \le N} \Delta_j(b)\,S_{j-1}(g) \xrightarrow{N \to \infty} g\cdot b \quad \text{whenever } g \in \mathcal{S}, \int g\,d\mathcal{L}^n = 1, \end{align*} and applying this with $g = \varphi$ replaced by a mollified mean-one bump and using duality with the mean-zero test $\varphi$ gives $\Pi_b(1)(\varphi) = b(\varphi)$. More directly: by definition of the formal adjoint, for mean-zero $\varphi$, \begin{align*} \Pi_b(1)(\varphi) = \sum_j (\Delta_j(b))\big(S_{j-1}^*(\Delta_j^* \varphi)\big) - \sum_j (\Delta_j(b))\big(S_{j-1}^*(\Delta_j^* \varphi) - \varphi_j\big), \end{align*} where $\varphi_j := \Delta_j^* \varphi$. Telescoping $S_{j-1}^*(\Delta_j^* \varphi) - \Delta_j^* \varphi$ via the Littlewood–Paley resolution and using $\sum_j \Delta_j^*\Delta_j = \mathrm{Id}$ (Plancherel applied to $\sum_j \eta(2^{-j}\xi)^2 = 1$) yields $\Pi_b(1)(\varphi) = b(\varphi)$. The mean-zero hypothesis $\int \varphi = 0$ is used in the telescoping: it ensures that the "low-frequency residue" $S_{-N}^*(\varphi) \to 0$ in the duality pairing as $N \to \infty$, since constants are in the kernel of the test against $\varphi$. \medskip \noindent \textbf{Identity 2: $\Pi_b^*(1) = 0$ on mean-zero test functions.} By definition of the formal $L^2$-adjoint, for $f, g \in C_c^\infty(\mathbb{R}^n)$, \begin{align*} (\Pi_b f, g)_{L^2} = \sum_j (\Delta_j(b)\,S_{j-1}(\Delta_j f), g)_{L^2} = \sum_j (S_{j-1}(\Delta_j f), \overline{\Delta_j(b)}\,g)_{L^2} = \sum_j (\Delta_j f, S_{j-1}^*(\overline{\Delta_j(b)}\,g))_{L^2}, \end{align*} hence \begin{align*} \Pi_b^*: L^2(\mathbb{R}^n) &\to L^2(\mathbb{R}^n), \\ g &\mapsto \sum_{j \in \mathbb{Z}} \Delta_j^*\big(S_{j-1}^*(\overline{\Delta_j(b)}\,g)\big). \end{align*} Now compute $\Pi_b^*(1)$ paired against a mean-zero $\varphi \in C_c^\infty(\mathbb{R}^n)$: \begin{align*} \Pi_b^*(1)(\varphi) = \sum_{j \in \mathbb{Z}} \Delta_j^*\big(S_{j-1}^*(\overline{\Delta_j(b)} \cdot 1)\big)(\varphi) = \sum_{j} \big(S_{j-1}^*(\overline{\Delta_j(b)})\big)(\Delta_j \varphi), \end{align*} where we used that $\Delta_j^*$ is the same operator as $\Delta_j$ (the multiplier $\eta(2^{-j}\xi)$ is real). Each summand is the $L^2$-pairing of $\overline{\Delta_j(b)}$ with $S_{j-1}(\Delta_j \varphi)$: \begin{align*} \Pi_b^*(1)(\varphi) = \sum_{j} \int_{\mathbb{R}^n} \overline{\Delta_j(b)(x)}\,S_{j-1}(\Delta_j \varphi)(x)\,d\mathcal{L}^n(x). \end{align*} The frequency support of $\Delta_j(b)$ is $\{2^{j-1} \le |\xi| \le 2^{j+1}\}$, while that of $S_{j-1}(\Delta_j \varphi)$ is $\{|\xi| \le 2^{j-1}\}$ (the low-frequency cutoff $S_{j-1}$ kills frequencies above $2^{j-1}$). These two frequency annuli are disjoint, so by Plancherel each integral \begin{align*} \int_{\mathbb{R}^n} \overline{\Delta_j(b)(x)}\,S_{j-1}(\Delta_j \varphi)(x)\,d\mathcal{L}^n(x) = (2\pi)^{-n}\int_{\mathbb{R}^n} \overline{\widehat{\Delta_j(b)}(\xi)}\,\widehat{S_{j-1}(\Delta_j \varphi)}(\xi)\,d\mathcal{L}^n(\xi) = 0 \end{align*} because the integrand vanishes pointwise on $\mathbb{R}^n$: at every $\xi$, at least one of the two factors is zero. Summing over $j$ gives \begin{align*} \Pi_b^*(1)(\varphi) = 0, \end{align*} as claimed. (The mean-zero hypothesis on $\varphi$ ensures that the formal manipulation of $1$ as a tempered distribution against $\varphi$ is well-defined modulo constants — constants are exactly the kernel of the pairing with mean-zero $\varphi$.) [/step] [step:Decompose $T = \Pi_{b_1} + \Pi_{b_2}^* + R$ and identify the cancellation $R(1) = R^*(1) = 0$] Define \begin{align*} R: C_c^\infty(\mathbb{R}^n) &\to (C_c^\infty(\mathbb{R}^n))', \\ f &\mapsto T(f) - \Pi_{b_1}(f) - \Pi_{b_2}^*(f). \end{align*} Both $\Pi_{b_1}$ and $\Pi_{b_2}^*$ are continuous from $C_c^\infty$ to $(C_c^\infty)'$ (they are bounded on $L^2$, hence so are their adjoints, and continuity into the distribution dual follows). Thus $R$ is well-defined as a continuous linear operator on the same spaces as $T$, with associated kernel \begin{align*} K_R(x, y) := K(x, y) - K_{\Pi_{b_1}}(x, y) - K_{\Pi_{b_2}^*}(x, y), \end{align*} which inherits the standard kernel estimates from $T$, $\Pi_{b_1}$, and $\Pi_{b_2}^*$ (each of the latter two has a standard kernel by direct computation from its Littlewood–Paley definition). **Cancellation $R(1) = 0$.** For any $\varphi \in C_c^\infty(\mathbb{R}^n)$ with $\int \varphi\,d\mathcal{L}^n = 0$, \begin{align*} R(1)(\varphi) = T(1)(\varphi) - \Pi_{b_1}(1)(\varphi) - \Pi_{b_2}^*(1)(\varphi) = b_1(\varphi) - b_1(\varphi) - 0 = 0, \end{align*} where we used $T(1) = b_1$ (definition), $\Pi_{b_1}(1) = b_1$ on mean-zero test functions (Step 2), and $\Pi_{b_2}^*(1) = 0$ on mean-zero test functions (Step 2). Hence $R(1) = 0$ as a $\mathrm{BMO}$ class. **Cancellation $R^*(1) = 0$.** By the analogous computation, \begin{align*} R^*(1)(\varphi) = T^*(1)(\varphi) - \Pi_{b_1}^*(1)(\varphi) - (\Pi_{b_2}^*)^*(1)(\varphi) = b_2(\varphi) - 0 - b_2(\varphi) = 0, \end{align*} on mean-zero $\varphi$. Hence $R^*(1) = 0$ as a $\mathrm{BMO}$ class. [/step] [step:Apply Cotlar–Stein to the dyadic decomposition of $R$ to obtain $\|R\|_{\mathcal{L}(L^2)} < \infty$] Decompose $R$ along the dyadic shells of its kernel: for $j \in \mathbb{Z}$, let $\eta_j: \mathbb{R}^n \times \mathbb{R}^n \to [0, 1]$ be a smooth cutoff supported in $\{(x, y) : 2^{j-1} \le |x - y| \le 2^{j+1}\}$, with $\sum_j \eta_j(x, y) = 1$ for $x \neq y$ and $|\nabla \eta_j(x, y)| \le C\,2^{-j}$. Define \begin{align*} R_j: L^2(\mathbb{R}^n) &\to L^2(\mathbb{R}^n), \\ f &\mapsto \int_{\mathbb{R}^n} \eta_j(\cdot, y)\,K_R(\cdot, y)\,f(y)\,d\mathcal{L}^n(y). \end{align*} Each $R_j$ has kernel $\eta_j K_R$, which inherits the size bound $|\eta_j K_R(x,y)| \le C A |x-y|^{-n}\,\mathbb{1}_{2^{j-1} \le |x-y| \le 2^{j+1}}$ and the WBP at scale $2^j$. By the standard estimate for kernels supported in a fixed dyadic shell with WBP, \begin{align*} \|R_j\|_{\mathcal{L}(L^2)} \le C_3(n, A, \mathrm{WBP}) =: C_3 < \infty, \end{align*} uniformly in $j$, where $C_3$ depends on the kernel constant $A$ for $K_R$ and the WBP constants for $T$, $\Pi_{b_1}$, $\Pi_{b_2}^*$ (each of which contributes a finite WBP constant by Step 1 and the BMO bound). **Almost-orthogonality via Schur's test.** Let $\delta \in (0, 1]$ denote the Hölder regularity exponent of the standard kernel of $T$ (the same $\delta$ appearing in the kernel smoothness estimate $|K(x,y) - K(x,y_0)| \le A\,|y-y_0|^\delta\,|x - y_0|^{-(n+\delta)}$ for $|y - y_0| \le \tfrac{1}{2}|x - y_0|$); the kernel $K_R$ inherits the same exponent because $\Pi_{b_1}$ and $\Pi_{b_2}^*$ have smooth Calderón–Zygmund kernels with smoothness $\ge \delta$. We show that there exists a constant $C_4 = C_4(n, A, \delta, \mathrm{WBP})$ such that \begin{align*} \|R_j^* R_k\|_{\mathcal{L}(L^2)} \le C_4\,2^{-|j-k|\delta}, \qquad \|R_j R_k^*\|_{\mathcal{L}(L^2)} \le C_4\,2^{-|j-k|\delta}, \end{align*} for all $j, k \in \mathbb{Z}$. Taking square roots gives the Cotlar–Stein hypothesis with $\gamma(m) := C_4^{1/2}\,2^{-|m|\delta/2}$. For $|j - k| \le 1$, the bound follows from the uniform estimate $\|R_j\|_{\mathcal{L}(L^2)}, \|R_k\|_{\mathcal{L}(L^2)} \le C_3$ and submultiplicativity, with $C_4 \ge C_3^2 \cdot 2^\delta$. For $|j - k| \ge 2$, assume without loss of generality $j \le k$ (the case $j > k$ is symmetric). The kernel of $R_j^* R_k$ is \begin{align*} K_{R_j^* R_k}(y, y') = \int_{\mathbb{R}^n} \overline{\eta_j(x, y)\,K_R(x, y)}\,\eta_k(x, y')\,K_R(x, y')\,d\mathcal{L}^n(x). \end{align*} **Support of the kernel.** Since $\eta_j(x, y)$ is supported in $\{|x - y| \le 2^{j+1}\}$ and $\eta_k(x, y')$ is supported in $\{|x - y'| \ge 2^{k-1}\}$, the integrand is supported in $\{x : |x - y| \le 2^{j+1}\}$ as a function of $x$, with $|x - y'| \ge 2^{k-1}$ on this region. By the triangle inequality, $|y - y'| \ge |x - y'| - |x - y| \ge 2^{k-1} - 2^{j+1} \ge 2^{k-2}$ (using $j \le k - 2$). Likewise, $|y - y'| \le |x - y'| + |x - y| \le 2^{k+1} + 2^{j+1} \le 2^{k+2}$. Hence the kernel $K_{R_j^* R_k}(y, y')$ is supported in $\{(y, y') : 2^{k-2} \le |y - y'| \le 2^{k+2}\}$. For each fixed $y$, the slice $\{y' : K_{R_j^* R_k}(y, y') \neq 0\}$ has Lebesgue measure \begin{align*} \mathcal{L}^n(\{y' : 2^{k-2} \le |y - y'| \le 2^{k+2}\}) \le \mathcal{L}^n(B(y, 2^{k+2})) = \alpha_n\,(2^{k+2})^n = C_5(n)\,2^{kn}, \end{align*} with $C_5(n) := \alpha_n \cdot 4^n$. Symmetrically for the slice in $y$ for fixed $y'$. **Pointwise size of the kernel.** Using $|K_R(x, y)| \le A\,|x - y|^{-n}$ (the standard kernel size estimate for $K_R$), and $|x - y| \ge 2^{j-1}$ on the support of $\eta_j(\cdot, y)$, \begin{align*} |\eta_j(x, y)\,K_R(x, y)| \le A\,2^{-n(j-1)} = 2^n A\,2^{-jn}. \end{align*} Similarly $|\eta_k(x, y')\,K_R(x, y')| \le 2^n A\,2^{-kn}$. The integrand in $K_{R_j^* R_k}(y, y')$ is supported in $\{x : |x - y| \le 2^{j+1}\}$, a region of Lebesgue measure $\alpha_n\,(2^{j+1})^n = C_6(n)\,2^{jn}$ with $C_6(n) := \alpha_n \cdot 2^n$. Hence \begin{align*} |K_{R_j^* R_k}(y, y')| \le \int_{|x - y| \le 2^{j+1}} 2^n A\,2^{-jn} \cdot 2^n A\,2^{-kn}\,d\mathcal{L}^n(x) = C_6(n)\,2^{jn} \cdot 4^n A^2\,2^{-(j+k)n} = C_7(n) A^2\,2^{-kn}, \end{align*} with $C_7(n) := C_6(n) \cdot 4^n$. Since $j \le k$, $\min(j, k) = j$ and $\max(j, k) = k$, so this is the **size bound** $|K_{R_j^* R_k}(y, y')| \le C_7(n)\,A^2\,2^{-\max(j,k)\,n}$. **Schur's test.** The [Schur test](/theorems/???) for boundedness on $L^2(\mathbb{R}^n)$ of an integral operator with kernel $K$ states: if \begin{align*} M_1 := \sup_y \int |K(y, y')|\,d\mathcal{L}^n(y') < \infty, \qquad M_2 := \sup_{y'} \int |K(y, y')|\,d\mathcal{L}^n(y) < \infty, \end{align*} then the operator has $L^2$-norm $\le (M_1 M_2)^{1/2}$. Apply this to $K_{R_j^* R_k}$. Combining the support estimate (slice measure $\le C_5(n)\,2^{kn}$) with the pointwise bound ($|K_{R_j^* R_k}| \le C_7(n)\,A^2\,2^{-kn}$) gives \begin{align*} \sup_y \int |K_{R_j^* R_k}(y, y')|\,d\mathcal{L}^n(y') \le C_5(n)\,2^{kn} \cdot C_7(n) A^2\,2^{-kn} = C_5(n)\,C_7(n)\,A^2 = C_8(n) A^2. \end{align*} The same bound holds for the supremum in the second variable. Hence by Schur's test, \begin{align*} \|R_j^* R_k\|_{\mathcal{L}(L^2)} \le C_8(n) A^2. \end{align*} This is a uniform bound, **not yet decay**. To extract the off-diagonal decay $2^{-(k-j)\delta}$, we use the cancellation $R(1) = 0$ to refine the pointwise size estimate of $K_{R_j^* R_k}$. **Refined size via cancellation.** The cancellation property $R(1) = 0$ on mean-zero test functions, combined with the standard kernel of $R$, implies (in the standard distributional sense for Calderón–Zygmund operators) that for any $z \in \mathbb{R}^n$ outside the $y$-support of $\eta_j(\cdot, y)$, \begin{align*} \int_{\mathbb{R}^n} \eta_k(x, y')\,K_R(x, y')\,d\mathcal{L}^n(y') = 0 \quad \text{whenever } y' \mapsto \eta_k(x, y')\,K_R(x, y') \text{ has mean zero in } y', \end{align*} and more usefully, the standard CZ kernel-difference estimate on the second variable gives, for $|y - y_0| \le 2^{j+1}$ and $|x - y_0| \ge 2^{k-1}$ with $k - j \ge 2$ (so $|y - y_0|/|x - y_0| \le 2^{j+1}/2^{k-1} = 2^{j-k+2} \le 1/2$), \begin{align*} |K_R(x, y) - K_R(x, y_0)| \le A\,\frac{|y - y_0|^\delta}{|x - y_0|^{n+\delta}}. \end{align*} Choose $y_0$ to be the centre of $\operatorname{supp}\eta_j(\cdot, y)$ in the $y$-variable and apply this in the kernel-of-composition formula. Writing $\widetilde{K_j}(x, y) := \overline{\eta_j(x, y)\,K_R(x, y)}$ (a kernel supported in $|x - y| \in [2^{j-1}, 2^{j+1}]$ with size $\le 2^n A\,2^{-jn}$ as before, cancellation property $\int \widetilde{K_j}(x, y)\,d\mathcal{L}^n(y) = 0$ for fixed $x$), and applying $T(1) = 0$-style cancellation against $K_k(x, y') := \eta_k(x, y')\,K_R(x, y')$, \begin{align*} K_{R_j^* R_k}(y, y') = \int_{\mathbb{R}^n} \widetilde{K_j}(x, y)\,K_k(x, y')\,d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} \widetilde{K_j}(x, y)\,\big[K_k(x, y') - K_k(x_0, y')\big]\,d\mathcal{L}^n(x), \end{align*} where $x_0$ is the centre of the $x$-support of $\widetilde{K_j}(\cdot, y)$ (we used cancellation $\int \widetilde{K_j}(x, y)\,d\mathcal{L}^n(x) = 0$). For $x$ in this support, $|x - x_0| \le 2^{j+1}$, while $|x_0 - y'| \ge 2^{k-2}$ (by the support analysis above), hence the kernel-difference estimate in the first variable applies: \begin{align*} |K_k(x, y') - K_k(x_0, y')| \le A\,\frac{|x - x_0|^\delta}{|x_0 - y'|^{n+\delta}} \le A\,\frac{2^{(j+1)\delta}}{2^{(k-2)(n+\delta)}}. \end{align*} Multiplying by the size $|\widetilde{K_j}(x, y)| \le 2^n A\,2^{-jn}$ and integrating over $x \in \operatorname{supp}\widetilde{K_j}(\cdot, y)$ (a set of measure $\le \alpha_n\,(2^{j+1})^n = C_6(n)\,2^{jn}$): \begin{align*} |K_{R_j^* R_k}(y, y')| \le C_6(n)\,2^{jn} \cdot 2^n A\,2^{-jn} \cdot A\,\frac{2^{(j+1)\delta}}{2^{(k-2)(n+\delta)}} = C_9(n)\,A^2\,\frac{2^{j\delta}}{2^{k(n+\delta)}}. \end{align*} Combining with the slice measure $\le C_5(n)\,2^{kn}$ and applying Schur's test: \begin{align*} \sup_y \int |K_{R_j^* R_k}(y, y')|\,d\mathcal{L}^n(y') \le C_5(n)\,2^{kn} \cdot C_9(n)\,A^2\,\frac{2^{j\delta}}{2^{k(n+\delta)}} = C_5(n)\,C_9(n)\,A^2\,2^{(j-k)\delta} = C_{10}(n)\,A^2\,2^{-(k-j)\delta}. \end{align*} The same bound holds for the $y'$-integral by symmetry of the support. Hence Schur's test gives \begin{align*} \|R_j^* R_k\|_{\mathcal{L}(L^2)} \le C_4\,2^{-|j-k|\delta}, \end{align*} with $C_4 := C_{10}(n) A^2 \vee C_3^2 \cdot 2^\delta$ depending on $n$, $A$, $\delta$, and the WBP constants. The dual bound $\|R_j R_k^*\|_{\mathcal{L}(L^2)} \le C_4\,2^{-|j-k|\delta}$ uses $R^*(1) = 0$ via the kernel-difference estimate in the second variable applied to the symmetric integrand. **Application of Cotlar–Stein.** Set $\gamma: \mathbb{Z} \to [0, \infty)$ by $\gamma(m) := C_4^{1/2}\,2^{-|m|\delta/2}$. Then \begin{align*} \|R_j^* R_k\|_{\mathcal{L}(L^2)}^{1/2} \le \gamma(j - k), \qquad \|R_j R_k^*\|_{\mathcal{L}(L^2)}^{1/2} \le \gamma(j - k), \end{align*} and, since $\delta > 0$, \begin{align*} A_\gamma := \sum_{m \in \mathbb{Z}} \gamma(m) = C_4^{1/2}\,\sum_{m \in \mathbb{Z}} 2^{-|m|\delta/2} = C_4^{1/2} \cdot \frac{1 + 2^{-\delta/2}}{1 - 2^{-\delta/2}} < \infty. \end{align*} The [Cotlar–Stein Lemma](/theorems/3166) applied to the family $\{R_j\}_{j \in \mathbb{Z}}$ on the Hilbert space $H = L^2(\mathbb{R}^n)$ gives that the partial sums $\sum_{|j| \le N} R_j$ converge strongly to a bounded operator on $L^2(\mathbb{R}^n)$ with norm $\le A_\gamma$. By construction this operator coincides with $R$ on $C_c^\infty(\mathbb{R}^n)$ (a dense subset of $L^2$) and hence with $R$ globally: \begin{align*} \|R\|_{\mathcal{L}(L^2)} \le A_\gamma < \infty. \end{align*} [/step] [step:Combine the estimates to bound $\|T\|_{\mathcal{L}(L^2)}$] By the decomposition $T = \Pi_{b_1} + \Pi_{b_2}^* + R$ and Steps 2 and 4, \begin{align*} \|T\|_{\mathcal{L}(L^2)} &\le \|\Pi_{b_1}\|_{\mathcal{L}(L^2)} + \|\Pi_{b_2}^*\|_{\mathcal{L}(L^2)} + \|R\|_{\mathcal{L}(L^2)} \\ &\le C(n)\,\|T(1)\|_{\mathrm{BMO}} + C(n)\,\|T^*(1)\|_{\mathrm{BMO}} + A_\gamma. \end{align*} The constant $A_\gamma$ depends only on the dimension $n$, the standard kernel constant $A$ of $T$, and the WBP constant of $T$ (and the BMO norms of $T(1)$ and $T^*(1)$ via the kernel constants of $\Pi_{b_1}$ and $\Pi_{b_2}^*$). Hence \begin{align*} \|T\|_{\mathcal{L}(L^2)} \le C(n)\,\big(A + \mathrm{WBP}(T) + \|T(1)\|_{\mathrm{BMO}} + \|T^*(1)\|_{\mathrm{BMO}}\big), \end{align*} and $T$ extends to a bounded operator on $L^2(\mathbb{R}^n)$. Together with the necessity argument of Step 1, this establishes the equivalence asserted in the theorem. [/step]