[proofplan] Existence follows by verifying that the Cauchy transform satisfies analyticity, the jump condition, and the decay at infinity, using the [Analyticity Of The Cauchy Transform](/theorems/555) and the [Plemelj Formulae](/theorems/556). Uniqueness is proved by showing that any two solutions differ by a bounded entire function, which must be constant by Liouville's theorem, and the decay condition forces that constant to zero. [/proofplan] [step:Verify that the Cauchy transform satisfies all three conditions] Define $G(z) := \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\,d\xi$. We verify the three conditions in turn. By the [Analyticity Of The Cauchy Transform](/theorems/555), $G$ is analytic on $\mathbb{C} \setminus \gamma$. This is condition (1). The jump formula from the [Plemelj Formulae](/theorems/556) gives $G^+(z_0) - G^-(z_0) = f(z_0)$ for every interior point $z_0 \in \gamma$. This is condition (2). Again by the [Analyticity Of The Cauchy Transform](/theorems/555), $G(z) = O(1/z)$ as $z \to \infty$, so $G(z) \to 0$. This is condition (3). [/step] [step:Prove uniqueness by reducing to Liouville's theorem] Suppose $G_1$ and $G_2$ both satisfy conditions (1)--(3). Define $\Phi := G_1 - G_2$. Then $\Phi$ is analytic on $\mathbb{C} \setminus \gamma$ by condition (1). For every interior point $z_0 \in \gamma$, the jump of $\Phi$ across $\gamma$ is: \begin{align*} \Phi^+(z_0) - \Phi^-(z_0) = (G_1^+(z_0) - G_1^-(z_0)) - (G_2^+(z_0) - G_2^-(z_0)) = f(z_0) - f(z_0) = 0. \end{align*} Since $\Phi$ is analytic on both sides of $\gamma$ and continuous across it (zero jump), Morera's theorem implies that $\Phi$ extends to an entire function $\Phi: \mathbb{C} \to \mathbb{C}$. By condition (3), $\Phi(z) = G_1(z) - G_2(z) \to 0$ as $z \to \infty$. In particular, $\Phi$ is bounded on $\mathbb{C}$. Liouville's theorem states that a bounded entire function is constant. Therefore $\Phi$ is a constant $c \in \mathbb{C}$. The vanishing at infinity forces $c = 0$, so $G_1 = G_2$. [guided] The uniqueness argument has two ingredients: removal of the discontinuity and Liouville's theorem. Why can we extend $\Phi$ across $\gamma$? The function $\Phi$ is analytic on each connected component of $\mathbb{C} \setminus \gamma$, and the boundary values from either side agree: $\Phi^+ = \Phi^-$ on $\gamma$. Morera's theorem provides the extension: if $\Phi$ is continuous on a region $\Omega$ and the integral of $\Phi$ over every triangle in $\Omega$ vanishes, then $\Phi$ is analytic on $\Omega$. For any triangle $T$ that crosses $\gamma$, we can split $T$ into pieces on each side of $\gamma$ and use the analyticity of $\Phi$ on each side plus the continuity across $\gamma$ to conclude that $\int_T \Phi\,dz = 0$. Why does boundedness hold? On any compact subset of $\mathbb{C}$, $\Phi$ is continuous (hence bounded). Near infinity, $\Phi(z) \to 0$, so $|\Phi(z)| < 1$ for $|z|$ sufficiently large. Combining gives a global bound. Liouville's theorem then forces $\Phi$ to be constant, and the decay at infinity pins this constant to zero. [/guided] [/step]