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  "body": "\n## Proof\n\n1. Let $U$, $u$, $c$ and $r$ be as in the statement, and let $K = \\set{z : \\abs{z - c} \\le r} \\subseteq U$.\n2. Claim: there is $R' \u003e r$ with $B = \\set{z : \\abs{z - c} \u003c R'} \\subseteq U$. Suppose not. Then for each $n \\ge 1$ there is $w_n \\notin U$ with $\\abs{w_n - c} \u003c r + 1/n$.\n3. The points $w_n$ lie in the closed disc of radius $r + 1$ around $c$, which is compact. By [[sequential compactness|0K8SGQSV7G]], a subsequence converges to some $w$ with $\\abs{w - c} \\le r$.\n4. Then $w \\in K \\subseteq U$. Since $U$ is open, the subsequence eventually lies in $U$. This contradicts $w_n \\notin U$, which proves the claim.\n5. On $B$, let $P = u_x$ and $Q = -u_y$, and let $g = P + iQ$. Since $u$ is twice continuously differentiable, $P$ and $Q$ have continuous partial derivatives.\n6. So $P$ and $Q$ are differentiable on $B$, by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]].\n7. We have $P_x = u_{xx}$ and $Q_y = -u_{yy}$. These are equal because $u_{xx} + u_{yy} = 0$.\n8. We have $P_y = u_{xy}$ and $-Q_x = u_{yx}$. These are equal by [[Schwarz's theorem|0NNQY8WCW6]].\n9. By steps 6 to 8 and the [[Cauchy–Riemann equations|0T71SSYZ39]], $g$ is complex differentiable at each point of $B$. So $g$ is [[holomorphic|0DG0Z72CQD]] on $B$.\n10. By [[holomorphic functions on a disc have primitives|02AC7JR6G6]], there is a holomorphic $F$ on $B$ with $F' = g$. Write $F = U_1 + i V_1$ with $U_1, V_1$ real.\n11. By the Cauchy–Riemann equations for $F$, $F' = \\partial_x U_1 + i \\partial_x V_1$ and $\\partial_x V_1 = -\\partial_y U_1$. So $F' = \\partial_x U_1 - i \\partial_y U_1$.\n12. Comparing with $g = u_x - i u_y$ gives $\\partial_x U_1 = u_x$ and $\\partial_y U_1 = u_y$ on $B$.\n13. Let $w = U_1 - u$ on $B$. It is differentiable, since $U_1$ is (step 11) and $u$ is (step 6 applied to $u$). Its total derivative is $0$ by step 12.\n14. The disc $B$ is path-connected along segments, so it is connected by [[path-connected spaces are connected|0ZMT1GP4KM]]. By [[zero derivative implies constant|0BYZTFM0DJ]], $w$ equals a real constant $k$.\n15. Let $H = F - k$. Then $H$ is holomorphic on $B$, and $\\operatorname{Re} H = U_1 - k = u$ on $B$.\n16. The closed disc $K$ lies in $B$. By [[Cauchy's integral formula|0DG10TVYPN]] with $a = c$ and the parametrization of the [[circle integral|0T7ABX4CWE]], $$H(c) = \\frac{1}{2\\pi i} \\int_0^{2\\pi} \\frac{H(c + r e^{i\\theta})}{r e^{i\\theta}} \\, i r e^{i\\theta} \\, d\\theta = \\frac{1}{2\\pi} \\int_0^{2\\pi} H(c + r e^{i\\theta}) \\, d\\theta.$$\n17. The integral of a complex function is defined through its real and imaginary parts, as for the [[contour integral|00FWNP9TXY]]. Taking real parts in step 16 and using step 15 gives $u(c) = \\frac{1}{2\\pi} \\int_0^{2\\pi} u(c + r e^{i\\theta}) \\, d\\theta$.\n\n## Idea\n\nOn a disc slightly larger than the given one, the harmonic function $u$ is the real part of a holomorphic function $H$. The function $u_x - i u_y$ is holomorphic by the Cauchy–Riemann equations, and $H$ is its primitive up to a constant. Cauchy's integral formula at the center says that $H(c)$ is the average of $H$ on the circle. Taking real parts gives the same for $u$.\n",
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