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  "body": "\n## Proof\n\nWe differentiate the Cauchy–Riemann equations once more and use the symmetry of mixed partial derivatives.\n\n1. Since $f$ is [[holomorphic|0DG0Z72CQD]] on $U$, [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]]. So, as a function of $(x, y)$, $f$ has continuous partial derivatives of all orders on $U$.\n2. The partial derivatives of $f$ are those of $u$ plus $i$ times those of $v$, by the definition of [[real and imaginary parts|02ZE6JHJ7G]]. So all partial derivatives of $u$ and $v$ exist and are continuous on $U$.\n3. In particular, $u$ and $v$ are of [[class $C^2$|0TCBZMEYGN]] on $U$. This is the first claim.\n4. Since $f$ is complex differentiable at each point of $U$, the [[Cauchy–Riemann equations|0T71SSYZ39]] hold at each point of $U$: $u_x = v_y$ and $u_y = -v_x$ on $U$.\n5. These are equalities of functions on the open set $U$. So we may differentiate both sides. Differentiating $u_x = v_y$ in $x$ gives $u_{xx} = (v_y)_x$.\n6. Differentiating $u_y = -v_x$ in $y$ gives $u_{yy} = -(v_x)_y$.\n7. Since $v$ is of class $C^2$, [[Schwarz's theorem|0NNQY8WCW6]] gives $(v_y)_x = (v_x)_y$ on $U$.\n8. Adding steps 5 and 6 and using step 7 gives $u_{xx} + u_{yy} = (v_y)_x - (v_x)_y = 0$ on $U$.\n9. Differentiating $u_y = -v_x$ in $x$ gives $(u_y)_x = -v_{xx}$. Differentiating $u_x = v_y$ in $y$ gives $(u_x)_y = v_{yy}$.\n10. Schwarz's theorem for $u$ gives $(u_y)_x = (u_x)_y$. So $v_{xx} + v_{yy} = -(u_y)_x + (u_x)_y = 0$ on $U$.\n11. By steps 3, 8 and 10, $u$ and $v$ are of class $C^2$ with zero Laplacian. So they are [[harmonic|0W5NJQ8MKY]] on $U$.\n\n## Idea\n\nThe Cauchy–Riemann equations tie the first derivatives of $u$ to those of $v$. Differentiating once more, the two second derivatives $u_{xx}$ and $u_{yy}$ become the same mixed derivative of $v$ with opposite signs. The real work is hidden in step 1: we need to know that $f$ has second derivatives at all.\n",
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