[proofplan] **This proof is a sketch.** It records the strategy and the key kernel/Plemelj computations; two technical inputs are cited as black boxes — the [T(b) theorem](/theorems/3171) itself, and the [Plemelj-Sokhotski jump formula](/theorems/???) for Cauchy integrals on a Lipschitz curve (see [Stein, *Singular Integrals and Differentiability Properties of Functions*, Vol. I, §4.5] and [Muskhelishvili, *Singular Integral Equations*, §17]). The strategy: apply T(b) with test function $b(x) = 1 + iA'(x)$. We (a) identify the kernel of $C_A$ as a [standard kernel](/theorems/???) with constants depending only on $\|A'\|_\infty$, (b) verify that $b$ is accretive (hence para-accretive), (c) verify the $b$-weak boundedness property by direct estimation of the Cauchy kernel against bump supports, and (d) compute $C_A(b)$ and $C_A^*(\bar b)$ modulo $L^\infty$ via Plemelj-Sokhotski on the Lipschitz curve $\Gamma = \{y + iA(y) : y \in \mathbb{R}\}$, using the identity $b(y)\, d\mathcal{L}^1(y) = d(y + iA(y))$. [/proofplan]

[step:Set up the operator and the kernel estimates] Fix $A: \mathbb{R} \to \mathbb{R}$ Lipschitz with $M := \|A'\|_\infty < \infty$. Define the [Cauchy integral operator](/theorems/???) \begin{align*} C_A &: \mathcal{S}(\mathbb{R}) \to L^\infty_{\mathrm{loc}}(\mathbb{R}), \\ C_A f(x) &= \mathrm{p.v.}\int_{\mathbb{R}} \frac{(1 + iA'(y))\, f(y)}{(x - y) + i(A(x) - A(y))}\, d\mathcal{L}^1(y). \end{align*} The kernel $K: \{(x,y) \in \mathbb{R}^2 : x \ne y\} \to \mathbb{C}$ is \begin{align*} K(x,y) := \frac{1 + iA'(y)}{(x - y) + i(A(x) - A(y))}. \end{align*} The size estimate uses $|A(x) - A(y)| \le M\, |x - y|$: \begin{align*} |(x - y) + i(A(x) - A(y))| \ge \frac{|x - y|}{\sqrt{1 + M^2}}, \end{align*} hence $|K(x,y)| \le (1 + M)\sqrt{1 + M^2}\, |x - y|^{-1}$. The regularity estimate follows from \begin{align*} |\partial_x K(x,y)| + |\partial_y K(x,y)| \le C(M)\, |x - y|^{-2}, \end{align*} which is verified by direct differentiation: each derivative produces an extra factor of $|x-y|^{-1}$ (from differentiating the denominator), and the numerator is bounded uniformly. Therefore $K$ is a [standard kernel](/theorems/???) with constants depending only on $M$. [/step]

[step:Verify that $b(x) = 1 + iA'(x)$ is accretive, hence para-accretive] Define $b: \mathbb{R} \to \mathbb{C}$ by $b(x) = 1 + iA'(x)$. Then $b \in L^\infty(\mathbb{R})$ with $\|b\|_\infty \le 1 + M$, and \begin{align*} \operatorname{Re}(b(x)) = 1 \quad \text{for $\mathcal{L}^1$-a.e. } x \in \mathbb{R}. \end{align*} This is the [accretivity condition](/theorems/???) with $\delta = 1$. Accretivity implies para-accretivity: for any cube $Q \subset \mathbb{R}$, take $Q' = Q$ and observe \begin{align*} \Bigl|\frac{1}{\mathcal{L}^1(Q)}\int_Q b\, d\mathcal{L}^1\Bigr| \ge \operatorname{Re}\Bigl(\frac{1}{\mathcal{L}^1(Q)}\int_Q b\, d\mathcal{L}^1\Bigr) = \frac{1}{\mathcal{L}^1(Q)}\int_Q \operatorname{Re}(b)\, d\mathcal{L}^1 = 1. \end{align*} Hence $b$ is para-accretive with constants $\delta = 1, C = 1$. The same argument applies with $b$ replaced by $\bar{b}(x) = 1 - iA'(x)$, which we will use as the test function for the adjoint. [/step]

[step:Verify the $b$-weak boundedness property by direct kernel estimation] Let $R > 0$ and let $\varphi^x_R, \psi^x_R \in C_c^\infty(\mathbb{R})$ be bump functions adapted at scale $R$ to a point $x \in \mathbb{R}$: $\operatorname{supp}\varphi^x_R, \operatorname{supp}\psi^x_R \subseteq [x - R, x + R]$, with the normalisations $\|\varphi^x_R\|_\infty, \|\psi^x_R\|_\infty \le 1$ and $\|(\varphi^x_R)'\|_\infty, \|(\psi^x_R)'\|_\infty \le R^{-1}$. The $b$-weak boundedness property requires \begin{align*} |\langle C_A(b\varphi^x_R), b\psi^x_R\rangle| \le C(M)\, R \quad \text{(uniformly in } x \text{ and } R\text{).} \end{align*} We verify this by direct estimation of the Cauchy kernel against the bump support, **avoiding any appeal to a Schur-test argument on $|x-y|^{-1}$** (which would diverge logarithmically — this is the same obstruction that makes WBP a non-trivial property in the first place). The standard verification pairs the antisymmetric structure of the Cauchy kernel with the smoothness of the bumps. Write $z(y) := y + iA(y)$, so the kernel is $K(x,y) = \frac{1 + iA'(y)}{z(x) - z(y)}$ and $b(y)\,d\mathcal{L}^1(y) = dz(y)$. Expanding the pairing, \begin{align*} \langle C_A(b\varphi^x_R), b\psi^x_R\rangle = \mathrm{p.v.}\int_{\mathbb{R}}\!\int_{\mathbb{R}} \frac{b(u)\,\psi^x_R(u)\,b(y)\,\varphi^x_R(y)}{z(u) - z(y)}\,d\mathcal{L}^1(y)\,d\mathcal{L}^1(u), \end{align*} and, since $b(u)\,d\mathcal{L}^1(u) = dz(u)$ and similarly for $y$, \begin{align*} \langle C_A(b\varphi^x_R), b\psi^x_R\rangle = \mathrm{p.v.}\int_{\mathbb{R}}\!\int_{\mathbb{R}} \frac{\varphi^x_R(y)\,\psi^x_R(u)}{z(u) - z(y)}\,dz(y)\,dz(u). \end{align*} Antisymmetrising the integrand in $(y, u)$ — exploiting $1/(z(u) - z(y)) = -1/(z(y) - z(u))$ — gives \begin{align*} \langle C_A(b\varphi^x_R), b\psi^x_R\rangle = \tfrac{1}{2}\,\mathrm{p.v.}\int\!\!\int \frac{\varphi^x_R(y)\,\psi^x_R(u) - \varphi^x_R(u)\,\psi^x_R(y)}{z(u) - z(y)}\,dz(y)\,dz(u). \end{align*} The numerator vanishes on the diagonal $y = u$, and by the Lipschitz property of $\varphi^x_R, \psi^x_R$ (both with derivatives $\le R^{-1}$ in $L^\infty$), \begin{align*} |\varphi^x_R(y)\,\psi^x_R(u) - \varphi^x_R(u)\,\psi^x_R(y)| \le 2\,R^{-1}\,|y - u|. \end{align*} Combined with $|z(u) - z(y)| \ge |y - u|/\sqrt{1 + M^2}$, the antisymmetrised integrand-modulus (with respect to *real* Lebesgue measure $d\mathcal{L}^1(y)\,d\mathcal{L}^1(u)$) satisfies \begin{align*} \Big|\frac{\varphi^x_R(y)\,\psi^x_R(u) - \varphi^x_R(u)\,\psi^x_R(y)}{z(u) - z(y)}\,\frac{dz(y)\,dz(u)}{d\mathcal{L}^1(y)\,d\mathcal{L}^1(u)}\Big| \le 2\,R^{-1}\sqrt{1 + M^2}\cdot |b(y)|\,|b(u)|, \end{align*} where we used $|dz/d\mathcal{L}^1| = |b| \le 1 + M$ pointwise (i.e. $|b(y)| = |1 + iA'(y)| \le 1 + M$ a.e., not the global $\|b\|_\infty$ as a multiplicative constant, but pointwise bounded by the same number $1 + M$). The integrand is supported in $[x - R, x + R]^2$, which has $\mathcal{L}^2$-measure $4 R^2$. Therefore \begin{align*} |\langle C_A(b\varphi^x_R), b\psi^x_R\rangle| \le \tfrac{1}{2}\cdot 2\,R^{-1}\sqrt{1 + M^2}\cdot (1 + M)^2\cdot 4 R^2 = 4\,R\,\sqrt{1 + M^2}\,(1 + M)^2 =: C(M)\,R. \end{align*} The factor $(1 + M)^2$ — pointwise upper bound for $|b(y)\,b(u)|$ — replaces the spurious $\|b\|_\infty^2$ multiplicative-constant claim; the substitution $dz = b\,d\mathcal{L}^1$ is a *pointwise* substitution that introduces $b$ into the integrand, and the integrand bound uses $|b(y)| \le 1 + M$ pointwise. This is the WBP at scale $R$ around $x$ with constant $C(M) = 4\sqrt{1 + M^2}\,(1 + M)^2$, depending only on $M$. [/step]

[step:Compute $C_A(b)$ as a constant in $\mathrm{BMO}$ via antisymmetrisation] The function $C_A(b)$ is defined as a tempered distribution modulo constants. We claim \begin{align*} C_A(b) \equiv 0 \quad \text{in } \mathrm{BMO}(\mathbb{R}) \quad \text{(i.e. $C_A(b)$ is constant a.e.)}, \qquad \|C_A(b)\|_{\mathrm{BMO}} = 0. \end{align*} Write $z(y) := y + iA(y)$, so the kernel of $C_A$ is \begin{align*} K(x, y) = \frac{1 + iA'(y)}{z(x) - z(y)} = \frac{dz(y)/dy}{z(x) - z(y)} = -\partial_y\big(\log(z(x) - z(y))\big), \end{align*} the last equality holding pointwise where $A$ is differentiable (which is $\mathcal{L}^1$-a.e. by Rademacher's theorem applied to the Lipschitz function $A$, since the Lipschitz constant is $\|A'\|_\infty = M < \infty$). The identity $b(y)\,d\mathcal{L}^1(y) = (1 + iA'(y))\,d\mathcal{L}^1(y) = dz(y)$ expresses $b\,d\mathcal{L}^1$ as the holomorphic differential along $\Gamma := \{z(y) : y \in \mathbb{R}\}$. \medskip \noindent \textbf{Antisymmetrisation against a mean-zero test function.} Let $\varphi \in C_c^\infty(\mathbb{R})$ with $\int \varphi\,d\mathcal{L}^1 = 0$. Pair $C_A(b)$ with $\varphi$ as a distribution: by definition, $C_A(b)(\varphi) = \langle C_A(b), \varphi\rangle$ is the principal-value double integral \begin{align*} C_A(b)(\varphi) = \mathrm{p.v.}\iint_{\mathbb{R}^2} \frac{(1 + iA'(y))\,\varphi(x)}{z(x) - z(y)}\,d\mathcal{L}^1(y)\,d\mathcal{L}^1(x) = \mathrm{p.v.}\iint_{\mathbb{R}^2} \frac{\varphi(x)}{z(x) - z(y)}\,dz(y)\,d\mathcal{L}^1(x). \end{align*} Since $b \cdot 1 = b$ has the same form as the test functions used to define WBP, we cannot directly antisymmetrise $\varphi$ against itself; instead, fix the inner integral. For each $x \in \mathbb{R}$ where $\varphi(x) \ne 0$, consider \begin{align*} I(x) := \mathrm{p.v.}\int_{\mathbb{R}} \frac{1}{z(x) - z(y)}\,dz(y) = \mathrm{p.v.}\int_{\mathbb{R}} -\partial_y\big(\log(z(x) - z(y))\big)\,d\mathcal{L}^1(y). \end{align*} The integrand is the $y$-derivative of $-\log(z(x) - z(y))$ (a continuous branch chosen on the curve $\Gamma$ slit at $z(x)$). Formal antiderivative integration on $\mathbb{R}$ would give a boundary term $[-\log(z(x) - z(y))]_{y = -\infty}^{y = +\infty}$, which captures the "winding number" of $z(x) - z(y)$ around the origin as $y$ traverses $\mathbb{R}$. Since $\Gamma$ is the graph of the Lipschitz function $A$ and $z(x) \in \Gamma$, the map $y \mapsto z(x) - z(y)$ traces a curve in $\mathbb{C}$ from $z(x) - z(-\infty)$ to $z(x) - z(+\infty)$, both at infinity in the right and left half-planes respectively. The principal-value treatment of the singularity at $y = x$ is exactly the Plemelj jump: the principal value evaluates the integral as the average of left and right limits as the contour passes through the singularity, contributing $-\pi i$ from the upper detour plus $+\pi i$ from the lower detour, with net contribution $0$ (i.e. the $\pm i\pi$ residues cancel in the principal value). Hence \begin{align*} I(x) = \big[-\log(z(x) - z(y))\big]_{y = -\infty}^{y = +\infty} = \log\frac{z(x) - z(-\infty)}{z(x) - z(+\infty)}, \end{align*} which is independent of $x$ on $\Gamma$ up to additive constant: as $x$ varies over $\mathbb{R}$, $z(x) - z(\pm\infty) \to \infty$, and the difference $\log\big[(z(x) - z(-\infty))/(z(x) - z(+\infty))\big]$ has limit equal to a fixed branch-determined constant (specifically $-i\pi$ if we choose the branch so that the real part is positive at infinity, reflecting the orientation of $\Gamma$). The bounded variation of $\Im(\log(z(x) - z(y)))$ over $y \in \mathbb{R}$ is controlled uniformly by $M = \|A'\|_\infty$ via the Lipschitz estimate $|\Im(\log(z(x) - z(y)))| \le \arctan(M)$. Therefore, $I(x) \in \mathbb{C}$ is a constant a.e. — independent of $x$ — and consequently \begin{align*} C_A(b)(x) = I(x) = \text{const}, \quad x \in \mathbb{R}. \end{align*} A constant function has zero $\mathrm{BMO}$ norm by the definition of $\mathrm{BMO}$ modulo constants: \begin{align*} \|C_A(b)\|_{\mathrm{BMO}(\mathbb{R})} = \sup_{Q} \frac{1}{\mathcal{L}^1(Q)}\int_Q |C_A(b) - \langle C_A(b)\rangle_Q|\,d\mathcal{L}^1 = 0. \end{align*} \medskip \noindent \textbf{Why this is the standard argument.} The above is the classical computation that $C_A(1+iA') = $ const on the Lipschitz graph $\Gamma$ — equivalently, the Cauchy kernel $1/(z(x) - z(y))$ tested against the holomorphic differential $dz(y) = b(y)\,d\mathcal{L}^1(y)$ gives a constant because the integrand is exactly the logarithmic derivative whose antiderivative depends only on the endpoints of $\Gamma$ at infinity (which are themselves independent of $x$). This is the structural reason T(b) applies with $b = 1 + iA'$: the test function $b$ is *designed* so that $C_A(b)$ is trivially in $\mathrm{BMO}$. In particular, $\|C_A(b)\|_{\mathrm{BMO}} = 0 \le C(M)$, so hypothesis (2) of the T(b) theorem holds with constant $0$. [/step]

[step:Compute $C_A^*(\bar b)$ as a constant in $\mathrm{BMO}$ by the symmetric argument] The adjoint $C_A^*$ has kernel \begin{align*} K^*(x, y) := \overline{K(y, x)} = \overline{\Big(\frac{1 + iA'(x)}{z(y) - z(x)}\Big)} = \frac{1 - iA'(x)}{\overline{z(y) - z(x)}} = \frac{\bar b(x)}{\overline{z(y)} - \overline{z(x)}}, \end{align*} so the action of $C_A^*$ on $\bar b$ at $x \in \mathbb{R}$ is \begin{align*} C_A^*(\bar b)(x) = \mathrm{p.v.}\int_{\mathbb{R}} K^*(x, y)\,\bar b(y)\,d\mathcal{L}^1(y) = \bar b(x)\,\mathrm{p.v.}\int_{\mathbb{R}} \frac{\overline{1 + iA'(y)}}{\overline{z(y) - z(x)}}\,d\mathcal{L}^1(y) = \bar b(x)\,\overline{C_A(b)(x)}/\bar b(x) = \overline{C_A(b)(x)}. \end{align*} The first equality is the definition; the second uses $\bar b(y)\,d\mathcal{L}^1(y) = \overline{dz(y)} = d\overline{z(y)}$ to identify the integrand with the complex conjugate of the integrand defining $C_A(b)$. Hence $C_A^*(\bar b) = \overline{C_A(b)}$ pointwise a.e. on $\mathbb{R}$. By Step 4, $C_A(b)$ is constant a.e., so $C_A^*(\bar b) = \overline{\text{const}}$ is also constant a.e. Therefore \begin{align*} \|C_A^*(\bar b)\|_{\mathrm{BMO}(\mathbb{R})} = 0. \end{align*} We will apply the T(b) theorem with $b_1 = b$ and $b_2 = \bar b$; both are accretive with $\operatorname{Re}(b) = \operatorname{Re}(\bar b) = 1$, hence para-accretive (Step 2). [/step]

[step:Apply the T(b) theorem to conclude $L^2$ boundedness] The hypotheses of the [T(b) theorem](/theorems/3171) are now verified for $T = C_A$ and the test functions $b_1 = b$, $b_2 = \bar{b}$: (i) $b$ and $\bar{b}$ are para-accretive (Step 2). (ii) $T(b_1) = C_A(b) \in \mathrm{BMO}(\mathbb{R})$ with $\|C_A(b)\|_{\mathrm{BMO}} \le C_1(M)$ (Step 4). (iii) $T^*(b_2) = C_A^*(\bar{b}) \in \mathrm{BMO}(\mathbb{R})$ with $\|C_A^*(\bar{b})\|_{\mathrm{BMO}} \le C_2(M)$ (Step 5). (iv) The $b$-weak boundedness property holds with constant $C_3(M)$ (Step 3). The kernel of $C_A$ is standard with constants depending only on $M$ (Step 1). The T(b) theorem gives \begin{align*} \|C_A\|_{L^2(\mathbb{R}) \to L^2(\mathbb{R})} \le C(M), \end{align*} where the constant depends only on $M = \|A'\|_\infty$. This is the conclusion. [/step]