The [Cauchy integral formula](/theorems/345) recovers a holomorphic function from its [boundary](/page/Boundary) values: if $f$ is holomorphic inside a simple closed contour $\gamma$ and [continuous](/page/Continuity) up to the boundary, then \begin{align*} f(z) = \frac{1}{2\pi i} \oint_\gamma \frac{f(w)}{w - z} \, dw \qquad \text{for every } z \text{ inside } \gamma. \end{align*} The right-hand side makes sense even when $f$ is not holomorphic — one can feed any reasonable function or measure into the kernel $1/(w - z)$ and obtain a new function of $z$. This observation leads to the **Cauchy transform**: given a compactly supported finite measure $\mu$ on $\mathbb{C}$, the Cauchy transform $\mathcal{C}\mu$ is a holomorphic function defined away from the support of $\mu$ that encodes analytic information about $\mu$. The Cauchy transform sits at a crossroads of several branches of analysis. In complex analysis, it provides a tool for constructing holomorphic [functions](/page/Function) with prescribed singularities. In harmonic analysis, its boundary behaviour is governed by a singular integral operator closely related to the [Hilbert transform](/page/Hilbert%20Transform). In partial differential equations, it inverts the Cauchy–Riemann operator $\bar{\partial}$, yielding solutions to the inhomogeneous equation $\bar{\partial} u = f$. And in geometric measure theory, the $L^2$-boundedness of the Cauchy transform characterises rectifiability of [sets](/page/Set) — a deep connection discovered by Mattila, Melnikov, and Verdera in the 1990s. ## Motivation [motivation] ### From the Cauchy Integral Formula to the Cauchy Transform The classical Cauchy integral formula applies when the function $f$ is holomorphic inside the contour and continuous up to the boundary. But in many situations — solving the $\bar{\partial}$-equation, studying boundary behaviour from one side of a curve, constructing holomorphic functions with jumps — one needs to integrate a non-holomorphic function or a measure against the Cauchy kernel. The question is: what properties does the resulting function inherit from the kernel $1/(w - z)$ alone, independent of any holomorphicity assumption on the input? ### The Kernel $1/(w - z)$ The function $w \mapsto 1/(w - z)$, for fixed $z \in \mathbb{C}$, is holomorphic in $w$ on $\mathbb{C} \setminus \{z\}$ and has a simple pole at $w = z$. As a function of $z$ for fixed $w$, it is likewise holomorphic on $\mathbb{C} \setminus \{w\}$. This double holomorphicity is the engine behind the Cauchy transform: integrating against this kernel in $w$ produces a function of $z$ that is holomorphic wherever the integration does not "see" the pole — that is, away from the support of the integrand. The kernel is not locally integrable in two real dimensions: $|1/(w - z)| = 1/|w - z|$, and $\int_{|w| < 1} |w|^{-1} \, d\mathcal{L}^2(w) = 2\pi \int_0^1 r^{-1} \cdot r \, dr = 2\pi < \infty$. So the kernel *is* locally integrable with respect to area measure $\mathcal{L}^2$, which is why the Cauchy transform of an $L^p$ function (with respect to $\mathcal{L}^2$) is well-defined pointwise almost everywhere. This should be contrasted with one dimension, where the kernel $1/(x - t)$ is *not* locally [integrable](/page/Integral) with respect to $\mathcal{L}^1$, and the corresponding transform (the Hilbert transform) requires a principal value. [/motivation] ## Definition The Cauchy transform can be defined for several classes of input: finite Borel measures, $L^p$ functions with respect to area measure, or functions on a curve. We begin with the most general setting — measures — and specialise afterward. Given a compactly supported finite Borel measure $\mu$ on $\mathbb{C}$, the integrand $w \mapsto 1/(w - z)$ is $\mu$-integrable for every $z \notin \operatorname{supp}(\mu)$, since $|w - z|^{-1}$ is bounded on $\operatorname{supp}(\mu)$ when $z$ is at positive distance from the support. The Cauchy transform is simply the integral of $\mu$ against this kernel. [definition: Cauchy Transform Of A Measure] Let $\mu$ be a compactly supported finite Borel measure on $\mathbb{C}$. The **Cauchy transform** of $\mu$ is the function \begin{align*} \mathcal{C}\mu : \mathbb{C} \setminus \operatorname{supp}(\mu) &\to \mathbb{C} \\ z &\mapsto \frac{1}{\pi} \int_{\mathbb{C}} \frac{1}{z - w} \, d\mu(w). \end{align*} [/definition] The normalisation factor $1/\pi$ is a convention that varies across the literature. Some authors use $1/(2\pi i)$ (matching the Cauchy integral formula), others use $1/\pi$ (matching the natural normalisation for the $\bar{\partial}$-operator), and some omit the constant entirely. We adopt the $1/\pi$ convention because it gives the cleanest statements for the $\bar{\partial}$-inversion and the Plemelj–Sokhotski formulas in the area-measure setting. When working with contour integrals or the Cauchy integral formula, the $1/(2\pi i)$ convention is more natural; we will note this when it arises. The sign convention also deserves comment: we write $1/(z - w)$ rather than $1/(w - z)$. This is equivalent to the opposite convention up to a global sign, and again varies by author. Our choice makes the Cauchy transform the [convolution](/page/Convolution) $\mu * K$ where $K(z) = 1/(\pi z)$, which is the standard normalisation for the fundamental solution of $\bar{\partial}$. When $\mu$ is absolutely continuous with respect to area measure — that is, $d\mu(w) = f(w) \, d\mathcal{L}^2(w)$ for some function $f$ — the Cauchy transform becomes a singular integral against $f$. The key difference from the measure case is that the integral is now taken over all of $\mathbb{C}$, and the kernel $1/(z - w)$ has a non-integrable singularity at $w = z$ when combined with a general $L^p$ function. However, the kernel is locally integrable with respect to $\mathcal{L}^2$ (as computed in the motivation), so the integral converges absolutely for $f \in L^p_c(\mathbb{C})$ (compactly supported $L^p$ functions) with $p > 2$, and converges in a principal-value sense for $p \le 2$. [definition: Cauchy Transform Of An Integrable Function] Let $f \in L^1_c(\mathbb{C}, \mathcal{L}^2)$ (compactly supported and integrable with respect to area measure). The **Cauchy transform** of $f$ is \begin{align*} \mathcal{C}f : \mathbb{C} &\to \mathbb{C} \\ z &\mapsto \frac{1}{\pi} \int_{\mathbb{C}} \frac{f(w)}{z - w} \, d\mathcal{L}^2(w). \end{align*} The integral converges absolutely for $\mathcal{L}^2$-almost every $z \in \mathbb{C}$. [/definition] The almost-everywhere convergence follows from [Young's convolution inequality](/theorems/463): $1/(\pi z)$ lies in $L^q_{\text{loc}}(\mathbb{C}, \mathcal{L}^2)$ for every $q < 2$ (since $\int_{|z|<R} |z|^{-q} \, d\mathcal{L}^2(z) = 2\pi \int_0^R r^{1-q} \, dr < \infty$ when $q < 2$), and $f \in L^1_c$, so by Young's inequality $\mathcal{C}f \in L^q_{\text{loc}}$ for every $q < 2$. ## Basic Properties The most fundamental property of the Cauchy transform is holomorphicity away from the support. This is inherited directly from the kernel: [differentiation](/page/Derivative) in $z$ passes under the integral sign and produces $\partial_z [1/(z-w)] = -1/(z-w)^2$, which is still integrable against $\mu$ when $z \notin \operatorname{supp}(\mu)$. [theorem: Holomorphicity Of The Cauchy Transform] Let $\mu$ be a compactly supported finite Borel measure on $\mathbb{C}$. Then $\mathcal{C}\mu$ is holomorphic on $\mathbb{C} \setminus \operatorname{supp}(\mu)$, with derivatives of all orders given by \begin{align*} \frac{d^n}{dz^n} \mathcal{C}\mu(z) = \frac{(-1)^n n!}{\pi} \int_{\mathbb{C}} \frac{1}{(z - w)^{n+1}} \, d\mu(w) \qquad \text{for all } z \notin \operatorname{supp}(\mu), \, n \ge 0. \end{align*} [/theorem] The differentiation under the integral sign is justified by dominated convergence: for $z$ at distance $\delta > 0$ from $\operatorname{supp}(\mu)$, the integrand $(z - w)^{-(n+1)}$ is bounded by $\delta^{-(n+1)}$ on the support, which is $\mu$-integrable since $\mu$ is finite. The holomorphicity then follows from Morera's theorem or directly from the [power series](/page/Power%20Series) expansion of $1/(z - w)$. The Cauchy transform also has characteristic behaviour at infinity: it decays like $1/z$ and the leading coefficient encodes the total mass of $\mu$. This asymptotic expansion is obtained by expanding $1/(z - w) = (1/z) \cdot 1/(1 - w/z)$ as a geometric series for $|z| > \sup_{w \in \operatorname{supp}(\mu)} |w|$. [theorem: Behaviour At Infinity] Let $\mu$ be a compactly supported finite Borel measure on $\mathbb{C}$ with $\operatorname{supp}(\mu) \subseteq \overline{B}(0, R)$. For $|z| > R$, the Cauchy transform admits the Laurent expansion \begin{align*} \mathcal{C}\mu(z) = \sum_{n=0}^{\infty} \frac{c_n}{z^{n+1}}, \qquad c_n = \frac{1}{\pi} \int_{\mathbb{C}} w^n \, d\mu(w). \end{align*} In particular, $\mathcal{C}\mu(z) = \frac{\mu(\mathbb{C})}{\pi z} + O(|z|^{-2})$ as $|z| \to \infty$, and $\mathcal{C}\mu(z) \to 0$ as $|z| \to \infty$. [/theorem] The coefficient $c_0 = \mu(\mathbb{C})/\pi$ is the total mass (up to the normalisation factor), and $c_1 = \frac{1}{\pi}\int w \, d\mu(w)$ is related to the centre of mass. These coefficients are the "moments" of $\mu$ as seen from infinity. An important consequence is that the Cauchy transform determines the moments of $\mu$: if $\mathcal{C}\mu_1 = \mathcal{C}\mu_2$ on the complement of a large disc, then $\mu_1$ and $\mu_2$ have the same moments. Under mild growth conditions (e.g., compact support), this implies $\mu_1 = \mu_2$ — the Cauchy transform is injective on compactly supported measures. [example: Cauchy Transform Of A Point Mass] Let $\mu = a \delta_{z_0}$ be a point mass of weight $a \in \mathbb{C}$ at the point $z_0 \in \mathbb{C}$. The Cauchy transform is \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_{\mathbb{C}} \frac{1}{z - w} \, d(a\delta_{z_0})(w) = \frac{a}{\pi(z - z_0)} \qquad \text{for } z \neq z_0. \end{align*} This is a scalar multiple of the Cauchy kernel itself: $\mathcal{C}(a\delta_{z_0})(z) = \frac{a}{\pi} \cdot \frac{1}{z - z_0}$. The function has a simple pole at $z = z_0$ with residue $a/\pi$, and decays like $a/(\pi z)$ at infinity. By linearity, the Cauchy transform of a finite sum of point masses $\mu = \sum_{k=1}^N a_k \delta_{z_k}$ is a rational function: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \sum_{k=1}^N \frac{a_k}{z - z_k}. \end{align*} This is a partial fraction decomposition — conversely, every rational function vanishing at infinity with only simple poles is the Cauchy transform of a weighted sum of point masses (with weights equal to $\pi$ times the residues). [/example] [example: Cauchy Transform Of Arc Length On The Unit Circle] Let $\mu = \mathcal{H}^1 \mres \partial B(0,1)$ be the arc-length measure on the unit circle. Parametrising $w = e^{i\theta}$ for $\theta \in [0, 2\pi)$, we have $d\mu(w) = d\theta$ and: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_0^{2\pi} \frac{1}{z - e^{i\theta}} \, d\theta. \end{align*} We evaluate this by converting to a contour integral. Setting $w = e^{i\theta}$, $d\theta = dw/(iw)$: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \oint_{|w|=1} \frac{1}{z - w} \cdot \frac{dw}{iw} = \frac{1}{\pi i} \oint_{|w|=1} \frac{1}{w(z - w)} \, dw. \end{align*} The partial fraction decomposition is \begin{align*} \frac{1}{w(z - w)} = \frac{1}{z} \left(\frac{1}{w} + \frac{1}{z - w}\right), \end{align*} valid when $z \neq 0$. By the residue theorem: **Case 1: $|z| > 1$.** Both poles $w = 0$ and $w = z$ are relevant, but only $w = 0$ lies inside the unit circle (since $|z| > 1$). The integral of $1/w$ around the unit circle is $2\pi i$. The integral of $1/(z - w)$ around the unit circle is $-2\pi i$ if $z$ is inside and $0$ if $z$ is outside; since $|z| > 1$, it is $0$. Therefore: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi i} \cdot \frac{1}{z} \cdot 2\pi i = \frac{2}{z}. \end{align*} **Case 2: $|z| < 1$.** Now $w = 0$ is inside the unit circle and $w = z$ is also inside (since $|z| < 1$). The integral of $1/w$ gives $2\pi i$. For $1/(z - w) = -1/(w - z)$, the pole at $w = z$ is inside, giving $-2\pi i$. Therefore: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi i} \cdot \frac{1}{z} \cdot (2\pi i - 2\pi i) = 0. \end{align*} Summarising: \begin{align*} \mathcal{C}\mu(z) = \begin{cases} 2/z & \text{if } |z| > 1, \\ 0 & \text{if } |z| < 1. \end{cases} \end{align*} This result illustrates a striking phenomenon: the Cauchy transform can be identically zero on one side of the support and nontrivial on the other. The function jumps from $0$ to $2/z$ as one crosses the unit circle from inside to outside. Note also that the exterior value $2/z$ decays like $2/z$ at infinity, consistent with the Laurent expansion theorem: $c_0 = \mu(\mathbb{C})/\pi = 2\pi/\pi = 2$, confirming $\mathcal{C}\mu(z) \sim 2/z$. [/example] [example: Cauchy Transform Of Area Measure On The Unit Disc] Let $\mu = \mathcal{L}^2 \mres B(0,1)$ be the area (Lebesgue) measure restricted to the open unit disc. For $z \in \mathbb{C}$: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_{B(0,1)} \frac{1}{z - w} \, d\mathcal{L}^2(w). \end{align*} We compute this using polar coordinates centered at $z$. However, it is easier to use the [distributional](/page/Distribution) identity $\bar{\partial}(1/z) = \pi \delta_0$ (which we develop in detail in the section on the $\bar{\partial}$-equation below) and the fact that $\mathcal{C}\mu$ should solve $\bar{\partial}(\mathcal{C}\mu) = \mathbb{1}_{B(0,1)}$ in a distributional sense. Instead, we compute directly. **For $|z| > 1$ (exterior):** Switch to polar coordinates $w = re^{i\theta}$ with $r \in [0,1)$ and $\theta \in [0, 2\pi)$: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_0^1 \int_0^{2\pi} \frac{r}{z - re^{i\theta}} \, d\theta \, dr. \end{align*} The inner integral is a contour integral: setting $w = re^{i\theta}$, $d\theta = dw/(iw)$: \begin{align*} \int_0^{2\pi} \frac{r}{z - re^{i\theta}} \, d\theta = \oint_{|w|=r} \frac{1}{z - w} \cdot \frac{dw}{i} = \frac{1}{i} \cdot 0 = 0 \end{align*} since for $|z| > 1 > r$, the pole $w = z$ lies outside the circle $|w| = r$. Therefore: \begin{align*} \mathcal{C}\mu(z) = 0 \qquad \text{for } |z| > 1. \end{align*} **For $|z| < 1$ (interior):** We split the radial integral at $|z|$. For $0 \le r < |z|$, the inner angular integral vanishes as before (the pole is outside). For $|z| < r < 1$, the pole $w = z$ lies inside $|w| = r$, and: \begin{align*} \oint_{|w|=r} \frac{1}{z - w} \cdot \frac{dw}{i} = \frac{1}{i} \cdot (-2\pi i) = -2\pi. \end{align*} Therefore: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_{|z|}^1 (-2\pi) \, dr = \frac{1}{\pi} \cdot (-2\pi)(1 - |z|) = -2(1 - |z|). \end{align*} Wait — this is not quite right, because the $r$ in the Jacobian must be accounted for. Let us redo the computation more carefully. In polar coordinates $w = re^{i\theta}$, $d\mathcal{L}^2(w) = r \, dr \, d\theta$, so: \begin{align*} \mathcal{C}\mu(z) &= \frac{1}{\pi} \int_0^1 \left(\int_0^{2\pi} \frac{r}{z - re^{i\theta}} \, d\theta\right) dr. \end{align*} The inner integral for fixed $r > 0$: \begin{align*} \int_0^{2\pi} \frac{r}{z - re^{i\theta}} \, d\theta. \end{align*} Set $w = re^{i\theta}$, so $dw = ire^{i\theta} \, d\theta = iw \, d\theta/1$, hence $d\theta = dw/(iw)$ and $re^{i\theta} = w$: \begin{align*} \int_0^{2\pi} \frac{r}{z - re^{i\theta}} \, d\theta = \oint_{|w|=r} \frac{r}{z - w} \cdot \frac{dw}{iw} = \frac{r}{i} \oint_{|w|=r} \frac{dw}{w(z-w)}. \end{align*} Using the partial fraction $\frac{1}{w(z-w)} = \frac{1}{z}\left(\frac{1}{w} + \frac{1}{z-w}\right)$: For $r < |z|$: only $w = 0$ is inside $|w| = r$, giving $\oint \frac{dw}{w} = 2\pi i$ and $\oint \frac{dw}{z - w} = 0$. So the inner integral is $\frac{r}{i} \cdot \frac{1}{z} \cdot 2\pi i = \frac{2\pi r}{z}$. For $r > |z|$: both $w = 0$ and $w = z$ are inside, giving $\oint \frac{dw}{w} = 2\pi i$ and $\oint \frac{dw}{z - w} = -2\pi i$. So the inner integral is $\frac{r}{i} \cdot \frac{1}{z} \cdot (2\pi i - 2\pi i) = 0$. Therefore, for $|z| < 1$: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_0^{|z|} \frac{2\pi r}{z} \, dr = \frac{2}{z} \cdot \frac{|z|^2}{2} = \frac{|z|^2}{z} = \bar{z}. \end{align*} Summarising: \begin{align*} \mathcal{C}(\mathcal{L}^2 \mres B(0,1))(z) = \begin{cases} \bar{z} & \text{if } |z| < 1, \\ 0 & \text{if } |z| > 1. \end{cases} \end{align*} Several features of this result are noteworthy. First, $\mathcal{C}\mu$ is *not* holomorphic on $B(0,1)$: the function $z \mapsto \bar{z}$ fails the Cauchy–Riemann equations (it is antiholomorphic). This is expected — the Cauchy transform of a function that is not identically zero cannot be holomorphic on its support, since it must solve $\bar{\partial}(\mathcal{C}\mu) = f$ there. Second, $\mathcal{C}\mu = 0$ on the exterior, so the Cauchy transform is identically zero outside the unit disc even though $\mu$ is nonzero. Third, $\mathcal{C}\mu$ is continuous across the boundary $\partial B(0,1)$: as $|z| \to 1^-$, $\bar{z} \to \bar{z}$ with $|z| = 1$, while $\mathcal{C}\mu = 0$ on the exterior, and $|\bar{z}| = 1 \neq 0$ on the boundary — so in fact $\mathcal{C}\mu$ is *not* continuous across the boundary. The exterior value is $0$ while the interior limiting value on $|z| = 1$ is $\bar{z}$ with $|\bar{z}| = 1$. The jump is $\bar{z} - 0 = \bar{z}$ on the unit circle. [/example] ## Inversion of the $\bar{\partial}$-Operator The deepest structural property of the Cauchy transform is that it provides a right inverse for the $\bar{\partial}$-operator. This connection arises because the Cauchy kernel $1/(\pi z)$ is the fundamental solution of $\bar{\partial}$ in two dimensions, playing the same role that $1/(2\pi) \ln|z|$ plays for the Laplacian. Recall that the Wirtinger derivatives are defined by \begin{align*} \partial := \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right), \qquad \bar{\partial} := \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right), \end{align*} where $z = x + iy$. A $C^1$ function $f$ is holomorphic if and only if $\bar{\partial} f = 0$ (this is a restatement of the Cauchy–Riemann equations). The $\bar{\partial}$-operator is thus the obstruction to holomorphicity, and solving $\bar{\partial} u = f$ amounts to finding a function whose "failure to be holomorphic" is prescribed by $f$. The fundamental solution of $\bar{\partial}$ is the distribution $K(z) = 1/(\pi z)$, in the sense that $\bar{\partial} K = \delta_0$. This identity, which requires distributional calculus to make rigorous, is the complex-variable analogue of the identity $\Delta \left(\frac{1}{2\pi}\ln|z|\right) = \delta_0$ for the Laplacian. The key consequence is that convolving any compactly supported function with $K$ should solve the $\bar{\partial}$-equation. [theorem: Fundamental Solution Of Dbar] Let $K : \mathbb{C} \setminus \{0\} \to \mathbb{C}$ be the function $K(z) = 1/(\pi z)$. Then, in the sense of distributions on $\mathbb{C}$: \begin{align*} \bar{\partial} K = \delta_0. \end{align*} That is, for every [test function](/page/Test%20Function) $\varphi \in C^\infty_c(\mathbb{C})$: \begin{align*} -\int_{\mathbb{C}} K(z) \bar{\partial}\varphi(z) \, d\mathcal{L}^2(z) = \varphi(0). \end{align*} [/theorem] The proof uses a standard limiting argument. For $\epsilon > 0$, write \begin{align*} -\int_{\mathbb{C}} K(z) \bar{\partial}\varphi(z) \, d\mathcal{L}^2(z) = -\lim_{\epsilon \to 0^+} \int_{|z| > \epsilon} \frac{1}{\pi z} \bar{\partial}\varphi(z) \, d\mathcal{L}^2(z). \end{align*} On the region $|z| > \epsilon$, the function $K$ is smooth and $\bar{\partial} K = 0$ (since $1/z$ is holomorphic away from the origin). Integrating by parts (using the identity $\bar{\partial}(K\varphi) = (\bar{\partial}K)\varphi + K\bar{\partial}\varphi = K\bar{\partial}\varphi$ on this region) and applying Stokes' theorem, the volume integral reduces to a boundary integral over the circle $|z| = \epsilon$. Parametrising $z = \epsilon e^{i\theta}$: \begin{align*} -\int_{|z| > \epsilon} K \bar{\partial}\varphi \, d\mathcal{L}^2 = \frac{1}{2i} \oint_{|z|=\epsilon} K(z) \varphi(z) \, dz = \frac{1}{2\pi i} \oint_{|z|=\epsilon} \frac{\varphi(z)}{z} \, dz. \end{align*} As $\epsilon \to 0$, $\varphi(z) \to \varphi(0)$ uniformly on $|z| = \epsilon$ (since $\varphi$ is smooth), and $\frac{1}{2\pi i}\oint_{|z|=\epsilon} \frac{dz}{z} = 1$, so the limit is $\varphi(0)$. The theorem immediately yields the solvability of the inhomogeneous $\bar{\partial}$-equation: if $f$ is a compactly supported smooth function, then $u = K * f = \mathcal{C}f$ satisfies $\bar{\partial} u = f$. The following theorem makes this precise and extends it to $L^p$ functions. [theorem: Dbar Inversion By The Cauchy Transform] Let $f \in L^p_c(\mathbb{C}, \mathcal{L}^2)$ with $1 < p < \infty$. Then $u := \mathcal{C}f$ satisfies $\bar{\partial} u = f$ in the distributional sense. If $p > 2$, then $u$ is continuous and satisfies $\bar{\partial} u = f$ pointwise almost everywhere. Moreover, $u$ is the unique solution in $L^q_{\text{loc}}(\mathbb{C})$ (for appropriate $q$) that vanishes at infinity. [/theorem] The equation $\bar{\partial}(\mathcal{C}f) = f$ is sometimes called the **Pompeiu formula** or the **generalised Cauchy integral formula**. It says that the Cauchy transform is a right inverse for $\bar{\partial}$ — applying $\bar{\partial}$ to the Cauchy transform of $f$ recovers $f$. This is the two-dimensional analogue of the [fundamental theorem of calculus](/theorems/632): $\frac{d}{dx}\int_{-\infty}^x f(t) \, dt = f(x)$. [example: Solving Dbar U Equals A Constant On A Disc] Find a continuous function $u : \mathbb{C} \to \mathbb{C}$ satisfying $\bar{\partial} u = \mathbb{1}_{B(0,R)}$ in the distributional sense, where $R > 0$. By the $\bar{\partial}$-inversion theorem, $u = \mathcal{C}(\mathbb{1}_{B(0,R)})$. This is a rescaled version of the unit disc computation from the previous section. Substituting $w = Rw'$ in the integral and simplifying (or repeating the polar coordinate argument with radius $R$), one finds: \begin{align*} u(z) = \mathcal{C}(\mathbb{1}_{B(0,R)})(z) = \begin{cases} \bar{z} & \text{if } |z| < R, \\ R^2/z & \text{if } |z| > R. \end{cases} \end{align*} Let us verify the interior case: for $|z| < R$, the same polar-coordinate computation as before (with the outer radius changed from $1$ to $R$) gives \begin{align*} u(z) = \frac{1}{\pi} \int_0^{|z|} \frac{2\pi r}{z} \, dr = \frac{|z|^2}{z} = \bar{z}. \end{align*} And indeed $\bar{\partial}(\bar{z}) = \bar{\partial}\left(\frac{1}{2}(x + iy) + \frac{1}{2}(x - iy)\right)$. Writing $\bar{z} = x - iy$ and computing: $\bar{\partial}(\bar{z}) = \frac{1}{2}(\partial_x + i\partial_y)(x - iy) = \frac{1}{2}(1 + i \cdot (-i)) = \frac{1}{2}(1 + 1) = 1 = \mathbb{1}_{B(0,R)}(z)$ for $|z| < R$. For $|z| > R$: the exterior formula $R^2/z$ is holomorphic, so $\bar{\partial}(R^2/z) = 0 = \mathbb{1}_{B(0,R)}(z)$ for $|z| > R$. The function $u$ is *not* smooth across the boundary $|z| = R$: the interior formula gives $\bar{z}$ and the exterior gives $R^2/z$. On $|z| = R$, both agree: $\bar{z} = R^2/z$ when $|z| = R$ (since $z\bar{z} = |z|^2 = R^2$). So $u$ is continuous but has a kink at the boundary — its first derivatives have a jump discontinuity, reflecting the jump in $f = \mathbb{1}_{B(0,R)}$ at $|z| = R$. [/example] ## The Cauchy Transform on Curves and the Plemelj–Sokhotski Formulas When the measure $\mu$ is supported on a smooth curve $\Gamma$, the Cauchy transform exhibits a characteristic jump discontinuity as one crosses $\Gamma$. This phenomenon is governed by the Plemelj–Sokhotski formulas, which express the boundary values of $\mathcal{C}\mu$ from either side of $\Gamma$ in terms of a principal-value integral and the density of $\mu$. Let $\Gamma$ be a smooth oriented Jordan curve in $\mathbb{C}$ and let $\varphi : \Gamma \to \mathbb{C}$ be a Hölder-continuous function. Consider the Cauchy-type integral (here using the $1/(2\pi i)$ convention matching the Cauchy integral formula): \begin{align*} F(z) := \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z} \, dw, \qquad z \notin \Gamma. \end{align*} Note the sign convention: $1/(w - z)$ rather than $1/(z - w)$, and the $1/(2\pi i)$ normalisation. This is the standard convention for Cauchy-type integrals on curves and differs from our earlier $1/\pi$ convention for area integrals by a factor and a sign. The function $F$ is holomorphic on $\mathbb{C} \setminus \Gamma$ and has well-defined non-tangential boundary values $F^+(z_0)$ (limit from the left of $\Gamma$, relative to the orientation) and $F^-(z_0)$ (limit from the right) at each point $z_0 \in \Gamma$. These boundary values are not equal — their difference is the density $\varphi$, and their average is the principal-value integral. To state the formulas precisely, we need the principal-value Cauchy integral. At a point $z_0 \in \Gamma$, the integrand $\varphi(w)/(w - z_0)$ has a non-integrable singularity at $w = z_0$. The principal value is defined by excising a symmetric neighbourhood: [definition: Principal Value Cauchy Integral] Let $\Gamma$ be a smooth oriented Jordan curve and let $\varphi : \Gamma \to \mathbb{C}$ be Hölder continuous. The **principal-value Cauchy integral** at $z_0 \in \Gamma$ is \begin{align*} \operatorname{p.v.} \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z_0} \, dw := \lim_{\epsilon \to 0^+} \frac{1}{2\pi i} \int_{\Gamma \setminus B(z_0, \epsilon)} \frac{\varphi(w)}{w - z_0} \, dw. \end{align*} [/definition] The existence of this limit for Hölder-continuous $\varphi$ on a smooth curve is a classical result. The key point is that the Hölder condition $|\varphi(w) - \varphi(z_0)| \le C|w - z_0|^\alpha$ for some $\alpha > 0$ makes the integrand of $(\varphi(w) - \varphi(z_0))/(w - z_0)$ integrable near $z_0$, and the principal value of $\int \frac{\varphi(z_0)}{w - z_0} \, dw$ exists by symmetry of the excision. With the principal-value integral in hand, the Plemelj–Sokhotski formulas express the boundary values as: [theorem: Plemelj Sokhotski Formulas] Let $\Gamma$ be a smooth oriented Jordan curve in $\mathbb{C}$ and let $\varphi : \Gamma \to \mathbb{C}$ be Hölder continuous with exponent $\alpha \in (0, 1]$. Define \begin{align*} F(z) = \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z} \, dw, \qquad z \notin \Gamma. \end{align*} Then for every $z_0 \in \Gamma$, the non-tangential [limits](/page/Limit) $F^+(z_0)$ (from the left of $\Gamma$) and $F^-(z_0)$ (from the right) exist and satisfy: \begin{align*} F^+(z_0) &= \frac{1}{2}\varphi(z_0) + \operatorname{p.v.} \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z_0} \, dw, \\ F^-(z_0) &= -\frac{1}{2}\varphi(z_0) + \operatorname{p.v.} \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z_0} \, dw. \end{align*} Equivalently: \begin{align*} F^+(z_0) - F^-(z_0) &= \varphi(z_0), \\ \frac{1}{2}\left(F^+(z_0) + F^-(z_0)\right) &= \operatorname{p.v.} \frac{1}{2\pi i} \int_\Gamma \frac{\varphi(w)}{w - z_0} \, dw. \end{align*} [/theorem] The first identity — the jump formula — says that the Cauchy transform has a jump of exactly $\varphi(z_0)$ across $\Gamma$ at $z_0$. This is the fundamental reason the Cauchy transform is useful for constructing functions with prescribed discontinuities: to build a function that is holomorphic on $\mathbb{C} \setminus \Gamma$ and jumps by $\varphi$ across $\Gamma$, simply take the Cauchy-type integral of $\varphi$. The second identity — the average formula — says that the principal-value integral is the arithmetic mean of the two boundary values. This is the analogue of the fact that for the Poisson integral on the disc, the boundary value is the average of the harmonic function from the interior and the harmonic conjugate from the exterior. The proof proceeds by writing $F(z) = \frac{1}{2\pi i}\int_\Gamma \frac{\varphi(w) - \varphi(z_0)}{w - z} \, dw + \frac{\varphi(z_0)}{2\pi i}\int_\Gamma \frac{dw}{w - z}$. The first integral has a removable singularity at $z = z_0$ (by the Hölder condition) and is continuous across $\Gamma$. The second integral is an explicit function of $z$ that can be evaluated by the residue theorem: for $z$ inside $\Gamma$, $\frac{1}{2\pi i}\oint_\Gamma \frac{dw}{w-z} = 1$, and for $z$ outside, it equals $0$. As $z$ approaches $z_0 \in \Gamma$, the integral approaches $1/2$ (by a symmetry argument involving the tangent line at $z_0$). [example: Plemelj Sokhotski For A Constant Density On The Unit Circle] Let $\Gamma = \partial B(0,1)$ (oriented counterclockwise) and $\varphi \equiv 1$. Then: \begin{align*} F(z) = \frac{1}{2\pi i} \oint_{|w|=1} \frac{1}{w - z} \, dw = \begin{cases} 1 & \text{if } |z| < 1, \\ 0 & \text{if } |z| > 1, \end{cases} \end{align*} by the Cauchy integral formula (the function $1$ is holomorphic everywhere). The Plemelj–Sokhotski formulas predict: \begin{align*} F^+(z_0) - F^-(z_0) &= \varphi(z_0) = 1, \\ \frac{1}{2}(F^+(z_0) + F^-(z_0)) &= \operatorname{p.v.} \frac{1}{2\pi i} \oint_{|w|=1} \frac{dw}{w - z_0}. \end{align*} Since $F^+ = 1$ (interior) and $F^- = 0$ (exterior), the jump is $1 - 0 = 1$ and the average is $(1 + 0)/2 = 1/2$. Both match: the principal-value integral equals $1/2$. This is a good sanity check. The "half-residue" phenomenon — the principal-value integral giving $1/2$ rather than $0$ or $1$ — is the geometric content of the Plemelj–Sokhotski formulas: the point $z_0$ on the curve "sees" half of the residue from each side. [/example] [example: Constructing A Function With A Prescribed Jump] Suppose we want a function $F : \mathbb{C} \setminus \partial B(0,1) \to \mathbb{C}$ that is holomorphic on $\mathbb{C} \setminus \partial B(0,1)$ and whose jump across the unit circle is $\varphi(w) = w + \bar{w} = 2\operatorname{Re}(w)$ for $w \in \partial B(0,1)$. On the unit circle, $\bar{w} = 1/w$, so $\varphi(w) = w + 1/w$. The Cauchy-type integral with density $\varphi$ is: \begin{align*} F(z) &= \frac{1}{2\pi i} \oint_{|w|=1} \frac{w + 1/w}{w - z} \, dw = \frac{1}{2\pi i} \oint_{|w|=1} \frac{w}{w - z} \, dw + \frac{1}{2\pi i} \oint_{|w|=1} \frac{1}{w(w - z)} \, dw. \end{align*} For the first integral, write $\frac{w}{w-z} = 1 + \frac{z}{w-z}$: \begin{align*} \frac{1}{2\pi i} \oint \frac{w}{w-z} \, dw = \frac{1}{2\pi i}\oint dw + \frac{z}{2\pi i}\oint \frac{dw}{w-z}. \end{align*} The first contour integral $\oint dw = 0$. The second is $z$ if $|z| < 1$ and $0$ if $|z| > 1$. For the second integral, use the partial fraction $\frac{1}{w(w-z)} = \frac{1}{z}\left(\frac{1}{w - z} - \frac{1}{w}\right)$ (valid for $z \neq 0$): \begin{align*} \frac{1}{2\pi i} \oint \frac{dw}{w(w-z)} = \frac{1}{z}\left(\frac{1}{2\pi i}\oint \frac{dw}{w-z} - \frac{1}{2\pi i}\oint \frac{dw}{w}\right). \end{align*} The integral $\frac{1}{2\pi i}\oint \frac{dw}{w} = 1$ always (pole at $w = 0$ inside the unit circle). The integral $\frac{1}{2\pi i}\oint \frac{dw}{w - z}$ is $1$ if $|z| < 1$ and $0$ if $|z| > 1$. Combining: **For $|z| < 1$:** $F(z) = z + \frac{1}{z}(1 - 1) = z$. **For $|z| > 1$:** $F(z) = 0 + \frac{1}{z}(0 - 1) = -1/z$. So: \begin{align*} F(z) = \begin{cases} z & \text{if } |z| < 1, \\ -1/z & \text{if } |z| > 1. \end{cases} \end{align*} Let us verify the jump: for $z_0 \in \partial B(0,1)$, $F^+(z_0) = z_0$ and $F^-(z_0) = -1/z_0 = -\bar{z}_0$ (since $|z_0| = 1$). The jump is: \begin{align*} F^+(z_0) - F^-(z_0) = z_0 + \bar{z}_0 = 2\operatorname{Re}(z_0) = \varphi(z_0). \end{align*} This confirms the Plemelj–Sokhotski jump formula. [/example] ## Connection to the Hilbert Transform When the Cauchy transform is restricted to the real line, it becomes closely related to the Hilbert transform — the prototypical singular integral operator on $\mathbb{R}$. This connection provides a bridge between the complex-analytic theory of the Cauchy transform and the real-variable theory of singular integrals. Consider a compactly supported function $f \in L^1(\mathbb{R}, \mathcal{L}^1)$. We can view $f$ as a measure on $\mathbb{C}$ supported on $\mathbb{R}$: define $\mu_f$ by $d\mu_f(w) = f(\operatorname{Re}(w)) \, d\mathcal{H}^1 \mres \mathbb{R}(w)$. The Cauchy transform of this measure, evaluated at a point $z = x + iy$ with $y > 0$ (in the upper half-plane), is: \begin{align*} \mathcal{C}\mu_f(x + iy) = \frac{1}{\pi} \int_\mathbb{R} \frac{f(t)}{(x - t) + iy} \, d\mathcal{L}^1(t). \end{align*} This is holomorphic in the upper half-plane and has boundary values as $y \to 0^+$. The real and imaginary parts of the boundary value are related to $f$ and its Hilbert transform. The Hilbert transform is the singular integral operator that emerges when one takes the principal-value limit of the Cauchy transform on the real line. It requires a principal-value prescription because the kernel $1/(x - t)$ is not integrable near $t = x$. [definition: Hilbert Transform] Let $f \in L^p(\mathbb{R}, \mathcal{L}^1)$ with $1 \le p < \infty$. The **Hilbert transform** of $f$ is \begin{align*} Hf(x) := \frac{1}{\pi} \operatorname{p.v.} \int_\mathbb{R} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t) = \frac{1}{\pi} \lim_{\epsilon \to 0^+} \int_{|x - t| > \epsilon} \frac{f(t)}{x - t} \, d\mathcal{L}^1(t). \end{align*} [/definition] The principal-value limit exists for $\mathcal{L}^1$-almost every $x$ when $f \in L^p$ with $1 \le p < \infty$. The key property of the Hilbert transform is $L^p$-boundedness: $\|Hf\|_{L^p} \le C_p \|f\|_{L^p}$ for $1 < p < \infty$, with the constant $C_p$ independent of $f$. This fails at the endpoints: $H$ is not bounded on $L^1$ (though it maps $L^1$ to $L^{1,\infty}$ — weak $L^1$) and not bounded on $L^\infty$. The connection between the Cauchy transform and the Hilbert transform is made explicit by separating real and imaginary parts. For $z = x + iy$ with $y > 0$: \begin{align*} \frac{1}{(x - t) + iy} = \frac{(x - t) - iy}{(x-t)^2 + y^2} = \frac{x - t}{(x-t)^2 + y^2} - i\frac{y}{(x-t)^2 + y^2}. \end{align*} The imaginary part $y/((x-t)^2 + y^2)$ is the Poisson kernel $P_y(x - t)$, and the real part $(x-t)/((x-t)^2 + y^2)$ is the conjugate Poisson kernel $Q_y(x - t)$. As $y \to 0^+$: The Poisson kernel $P_y$ converges to $\pi \delta_x$ (it is an approximate identity), so the imaginary part of the boundary value recovers $-f(x)$. The conjugate Poisson kernel $Q_y$ converges (in the principal-value sense) to $1/(x - t)$, so the real part of the boundary value gives the Hilbert transform $Hf(x)$. Precisely, the non-tangential boundary value of $\mathcal{C}\mu_f$ from the upper half-plane is: \begin{align*} \lim_{y \to 0^+} \mathcal{C}\mu_f(x + iy) = \frac{1}{\pi}(Hf(x) - if(x)) \qquad \text{for } \mathcal{L}^1\text{-a.e. } x \in \mathbb{R}. \end{align*} This is the Plemelj–Sokhotski formula specialised to the real line. [example: Hilbert Transform Of An Indicator Function] Let $f = \mathbb{1}_{[a,b]}$ for $a < b$. The Hilbert transform is: \begin{align*} Hf(x) = \frac{1}{\pi} \operatorname{p.v.} \int_a^b \frac{1}{x - t} \, d\mathcal{L}^1(t). \end{align*} For $x \notin [a,b]$, the integral is not singular and evaluates directly: \begin{align*} Hf(x) = \frac{1}{\pi} \int_a^b \frac{dt}{x - t} = \frac{1}{\pi} \left[-\ln|x - t|\right]_{t=a}^{t=b} = \frac{1}{\pi} \ln\frac{|x - a|}{|x - b|}. \end{align*} For $x \in (a, b)$, the principal value gives: \begin{align*} Hf(x) &= \frac{1}{\pi} \lim_{\epsilon \to 0^+} \left(\int_a^{x - \epsilon} \frac{dt}{x - t} + \int_{x + \epsilon}^b \frac{dt}{x - t}\right) \\ &= \frac{1}{\pi} \lim_{\epsilon \to 0^+} \left(\ln(x - a) - \ln\epsilon + \ln\epsilon - \ln(b - x)\right) \\ &= \frac{1}{\pi} \ln\frac{x - a}{b - x}. \end{align*} Both cases combine into the single formula: \begin{align*} H(\mathbb{1}_{[a,b]})(x) = \frac{1}{\pi} \ln\left|\frac{x - a}{x - b}\right| \qquad \text{for } x \notin \{a, b\}. \end{align*} The Hilbert transform has logarithmic singularities at the endpoints $x = a$ and $x = b$, which is why $H(\mathbb{1}_{[a,b]}) \notin L^1(\mathbb{R})$ — it is only in $L^{1,\infty}$ (weak $L^1$). This exemplifies the failure of $L^1$-boundedness of the Hilbert transform. For $x \to +\infty$, $Hf(x) \sim \frac{b-a}{\pi x}$, giving $O(1/x)$ decay. [/example] ## The Cauchy Transform and Analytic Capacity We briefly indicate a deeper application. A compact set $K \subset \mathbb{C}$ is called **removable for bounded analytic functions** if every bounded holomorphic function on $\mathbb{C} \setminus K$ is constant. The classical problem is to characterise such sets in metric and measure-theoretic terms. The answer involves the Cauchy transform and a quantity called analytic capacity. The **analytic capacity** of a compact set $K \subset \mathbb{C}$ is \begin{align*} \gamma(K) := \sup\left\{|f'(\infty)| : f \in H^\infty(\mathbb{C} \setminus K), \, \|f\|_\infty \le 1\right\}, \end{align*} where $f'(\infty) = \lim_{z \to \infty} z(f(z) - f(\infty))$. The set $K$ is removable if and only if $\gamma(K) = 0$. The connection to the Cauchy transform is that the extremal function in the definition of $\gamma(K)$ — the *Ahlfors function* — can be represented as the Cauchy transform of a measure supported on $K$. More precisely, $\gamma(K) = \sup\{|\mu(\mathbb{C})| / \pi : \mu \text{ is supported on } K, \, \|\mathcal{C}\mu\|_\infty \le 1\}$. This reformulation shifts the problem from studying bounded analytic functions to studying the Cauchy transform of measures, which is amenable to techniques from harmonic analysis and geometric measure theory. The celebrated theorem of Tolsa (2003) characterises analytic capacity in terms of the Menger curvature of measures, confirming a conjecture of Vitushkin. The key intermediate result, proved by Mattila, Melnikov, and Verdera (1996), is that the Cauchy transform is bounded on $L^2(\mathcal{H}^1 \mres E)$ if and only if the set $E$ is rectifiable (in the sense of having finite $\mathcal{H}^1$-measure and being contained, up to a set of $\mathcal{H}^1$-measure zero, in a countable union of Lipschitz curves). This is a remarkable connection between the analytic properties of a singular integral operator and the geometric regularity of the underlying set. ## Problems [problem] Let $\mu = \mathcal{H}^1 \mres [0,1]$ (arc-length measure on the segment $[0,1] \subset \mathbb{R} \subset \mathbb{C}$). Compute $\mathcal{C}\mu(z)$ for $z \notin [0,1]$ and determine its boundary behaviour as $z$ approaches a point $x_0 \in (0,1)$ from above and below the real axis. [/problem] [solution] **Step 1: Compute $\mathcal{C}\mu(z)$ for $z \notin [0,1]$.** \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi} \int_0^1 \frac{1}{z - t} \, d\mathcal{L}^1(t) = \frac{1}{\pi} \left[\ln(z - t)\right]_{t=0}^{t=1}, \end{align*} where $\ln$ denotes a branch of the complex logarithm that is continuous on $\mathbb{C} \setminus [0,1]$. Evaluating: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi}\left(\ln(z - 1) - \ln(z)\right) = \frac{1}{\pi} \ln\frac{z - 1}{z} = \frac{1}{\pi}\ln\left(1 - \frac{1}{z}\right). \end{align*} Here $\ln$ is the principal logarithm $\operatorname{Log}$ when $z$ is in the upper half-plane and we choose the [branch cut](/page/Branch%20Cuts) to lie along $(-\infty, 0]$ for both $z - 1$ and $z$. More precisely, for $z \in \mathbb{C} \setminus [0,1]$: \begin{align*} \mathcal{C}\mu(z) = \frac{1}{\pi}\left(\operatorname{Log}(z - 1) - \operatorname{Log}(z)\right), \end{align*} which is valid when neither $z - 1$ nor $z$ is on $(-\infty, 0]$. For real $z > 1$, both $z - 1 > 0$ and $z > 0$, so the formula gives $\frac{1}{\pi}\ln\frac{z-1}{z}$ (real, negative). For real $z < 0$, both $z - 1 < 0$ and $z < 0$, so $\operatorname{Log}(z - 1) = \ln|z-1| + i\pi$ and $\operatorname{Log}(z) = \ln|z| + i\pi$ (approaching from above) or $\ln|z| - i\pi$ (from below). The imaginary parts cancel when approached from the same side, giving $\frac{1}{\pi}\ln\frac{|z-1|}{|z|}$ (real) for $z < 0$. **Step 2: Boundary behaviour from above and below.** For $x_0 \in (0,1)$ and $\epsilon > 0$ small: \begin{align*} \mathcal{C}\mu(x_0 + i\epsilon) &= \frac{1}{\pi}\left(\operatorname{Log}(x_0 - 1 + i\epsilon) - \operatorname{Log}(x_0 + i\epsilon)\right). \end{align*} As $\epsilon \to 0^+$: $x_0 + i\epsilon \to x_0 > 0$, so $\operatorname{Log}(x_0 + i\epsilon) \to \ln x_0$. And $x_0 - 1 + i\epsilon \to x_0 - 1 < 0$ with positive imaginary part, so $\operatorname{Log}(x_0 - 1 + i\epsilon) \to \ln(1 - x_0) + i\pi$. Therefore: \begin{align*} \lim_{\epsilon \to 0^+} \mathcal{C}\mu(x_0 + i\epsilon) = \frac{1}{\pi}\left(\ln(1 - x_0) + i\pi - \ln x_0\right) = \frac{1}{\pi}\ln\frac{1 - x_0}{x_0} + i. \end{align*} From below ($\epsilon \to 0^-$, i.e., $z = x_0 - i|\epsilon|$): $\operatorname{Log}(x_0 - i|\epsilon|) \to \ln x_0$ and $\operatorname{Log}(x_0 - 1 - i|\epsilon|) \to \ln(1 - x_0) - i\pi$. Therefore: \begin{align*} \lim_{\epsilon \to 0^+} \mathcal{C}\mu(x_0 - i\epsilon) = \frac{1}{\pi}\left(\ln(1 - x_0) - i\pi - \ln x_0\right) = \frac{1}{\pi}\ln\frac{1 - x_0}{x_0} - i. \end{align*} **Step 3: The jump.** \begin{align*} \mathcal{C}\mu(x_0 + i0^+) - \mathcal{C}\mu(x_0 - i0^+) = 2i. \end{align*} This is consistent with the Plemelj–Sokhotski formulas: the density of $\mu$ with respect to arc length on $[0,1]$ is $\varphi = 1$, and the jump in the $1/(2\pi i)$ convention is $\varphi(x_0) = 1$. In our $1/\pi$ convention, the Cauchy kernel has an extra factor of $-1/(2i)$ relative to the contour integral, giving a jump of $2i \cdot 1 = 2i$. (The precise relationship is: $\frac{1}{\pi} \cdot \frac{1}{z - t} = \frac{-1}{2\pi i} \cdot \frac{2i}{t - z}$, so the jump in the $1/\pi$ convention is $-2i \cdot (-\varphi) = 2i\varphi$ — but one must be careful with orientation conventions.) **Step 4: The principal-value integral.** The average of the boundary values is \begin{align*} \frac{1}{2}\left(\mathcal{C}\mu(x_0 + i0^+) + \mathcal{C}\mu(x_0 - i0^+)\right) = \frac{1}{\pi}\ln\frac{1 - x_0}{x_0}, \end{align*} which is the principal-value integral $\frac{1}{\pi}\operatorname{p.v.}\int_0^1 \frac{dt}{x_0 - t}$. This can be verified directly: $\operatorname{p.v.}\int_0^1 \frac{dt}{x_0 - t} = \ln\frac{x_0}{1 - x_0}$ (by the same computation as the Hilbert transform of $\mathbb{1}_{[0,1]}$, with the opposite sign convention), giving $\frac{1}{\pi}\ln\frac{x_0}{1-x_0}$. The discrepancy in sign is due to the kernel being $1/(z - t)$ vs. $1/(t - z)$; with our convention $1/(z - t)$, the principal value is $\frac{1}{\pi}\operatorname{p.v.}\int_0^1 \frac{dt}{x_0 - t} = \frac{1}{\pi}\ln\frac{1 - x_0}{x_0}$ (negative for $x_0 > 1/2$, positive for $x_0 < 1/2$), which matches. [/solution] [problem] Let $f(w) = \bar{w}^2$ for $w \in B(0,1)$ and $f(w) = 0$ for $w \notin B(0,1)$. Compute $\mathcal{C}f(z)$ for $|z| < 1$ and verify that $\bar{\partial}(\mathcal{C}f) = f$ there. [/problem] [solution] **Step 1: Set up the integral.** For $|z| < 1$: \begin{align*} \mathcal{C}f(z) = \frac{1}{\pi}\int_{B(0,1)} \frac{\bar{w}^2}{z - w} \, d\mathcal{L}^2(w). \end{align*} Use polar coordinates $w = re^{i\theta}$, $\bar{w} = re^{-i\theta}$, $d\mathcal{L}^2(w) = r \, dr \, d\theta$: \begin{align*} \mathcal{C}f(z) = \frac{1}{\pi}\int_0^1 \int_0^{2\pi} \frac{r^2 e^{-2i\theta}}{z - re^{i\theta}} \cdot r \, d\theta \, dr. \end{align*} **Step 2: Evaluate the angular integral.** For fixed $r \in (0,1)$, define $I(r, z) := \int_0^{2\pi} \frac{r^3 e^{-2i\theta}}{z - re^{i\theta}} \, d\theta$. Set $w = re^{i\theta}$, $d\theta = dw/(iw)$, $e^{-2i\theta} = r^2/w^2$: \begin{align*} I(r, z) = \int_0^{2\pi} \frac{r^3 \cdot r^2/w^2}{z - w} \cdot \frac{dw}{iw} \cdot \frac{1}{1} = \frac{r^5}{i} \oint_{|w|=r} \frac{dw}{w^3(z - w)}. \end{align*} Wait, let us be more careful. With $w = re^{i\theta}$, $e^{-2i\theta} = (re^{i\theta})^{-2} \cdot r^2 = r^2/w^2$, and $d\theta = dw/(iw)$: \begin{align*} I(r, z) = \oint_{|w|=r} \frac{r^2 \cdot (r^2/w^2)}{z - w} \cdot r \cdot \frac{dw}{iw} = \frac{r^5}{i} \oint_{|w|=r} \frac{dw}{w^3(z - w)}. \end{align*} Using the partial fraction expansion: $\frac{1}{w^3(z - w)} = \frac{1}{z}\left(\frac{1}{w^3} + \frac{1}{w^2 z} + \frac{1}{wz^2} + \frac{1}{z^2(z - w)}\right)$. Actually, let us use the geometric series: $\frac{1}{z - w} = \frac{1}{z} \cdot \frac{1}{1 - w/z} = \frac{1}{z}\sum_{n=0}^\infty (w/z)^n$ for $|w| < |z|$ (i.e., $r < |z|$), and $\frac{1}{z - w} = -\frac{1}{w}\cdot\frac{1}{1 - z/w} = -\frac{1}{w}\sum_{n=0}^\infty (z/w)^n$ for $|w| > |z|$ (i.e., $r > |z|$). **Case $r < |z|$:** \begin{align*} \frac{1}{w^3(z - w)} = \frac{1}{w^3 z} \sum_{n=0}^\infty \left(\frac{w}{z}\right)^n = \sum_{n=0}^\infty \frac{1}{z^{n+1} w^{3-n}}. \end{align*} By the residue theorem, $\oint_{|w|=r} w^{k} \, dw = 0$ unless $k = -1$, in which case it equals $2\pi i$. We need $3 - n = 1$, i.e., $n = 2$: \begin{align*} \oint_{|w|=r} \frac{dw}{w^3(z-w)} = \frac{2\pi i}{z^3}. \end{align*} So $I(r,z) = \frac{r^5}{i} \cdot \frac{2\pi i}{z^3} = \frac{2\pi r^5}{z^3}$. **Case $r > |z|$:** \begin{align*} \frac{1}{w^3(z - w)} = -\frac{1}{w^4} \sum_{n=0}^\infty \left(\frac{z}{w}\right)^n = -\sum_{n=0}^\infty \frac{z^n}{w^{n+4}}. \end{align*} We need $n + 4 = 1$, i.e., $n = -3$, which is impossible for $n \ge 0$. So $\oint_{|w|=r} \frac{dw}{w^3(z-w)} = 0$ when $r > |z|$. **Step 3: Integrate over $r$.** \begin{align*} \mathcal{C}f(z) = \frac{1}{\pi} \int_0^{|z|} \frac{2\pi r^5}{z^3} \, dr + \frac{1}{\pi}\int_{|z|}^1 0 \, dr = \frac{2}{z^3} \int_0^{|z|} r^5 \, dr = \frac{2}{z^3} \cdot \frac{|z|^6}{6} = \frac{|z|^6}{3z^3}. \end{align*} Since $|z|^6 = (z\bar{z})^3 = z^3\bar{z}^3$: \begin{align*} \mathcal{C}f(z) = \frac{z^3\bar{z}^3}{3z^3} = \frac{\bar{z}^3}{3} \qquad \text{for } |z| < 1. \end{align*} **Step 4: Verify $\bar{\partial}(\mathcal{C}f) = f$.** Using $\bar{\partial}(\bar{z}^n) = n\bar{z}^{n-1} \cdot \bar{\partial}(\bar{z}) = n\bar{z}^{n-1}$ (since $\bar{\partial}(\bar{z}) = 1$): \begin{align*} \bar{\partial}\left(\frac{\bar{z}^3}{3}\right) = \frac{1}{3} \cdot 3\bar{z}^2 = \bar{z}^2 = f(z). \end{align*} This confirms $\bar{\partial}(\mathcal{C}f) = f$ on the interior of the disc. [/solution] ## References - Ahlfors, L. V., *Bounded Analytic Functions* (lecture notes, 1947); see also *Conformal Invariants* (1973). - Garnett, J. B., *Analytic Capacity and Measure* (1972). - Muscalu, C. and Schlag, W., *Classical and Multilinear Harmonic Analysis*, Vol. I (2013). - Tolsa, X., *Analytic Capacity, the Cauchy Transform, and Non-homogeneous Calderón–Zygmund Theory* (2014). - Vekua, I. N., *Generalized Analytic Functions* (1962).