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My book says that for any holomorphic function $f(z)=u(x,y)+iv(x,y)$, $u$ and $v$ satisfy Laplace's equation.

$f$ is holomorphic $\implies$ * $u_x=v_y$ and $u_y=-v_x$, so $u_{xx}=v_{yx}=v_{xy}=(-u_y)_y=-u_{yy}$. My question is in regards to symmetry of the mixed partials $v_{yx}=v_{xy}$. From what I understand, if $f$ is holomorphic, all we know (from the Cauchy-Riemann eqs) is that the first partials for $u$ and $v$ satisfy *, but nothing about the existence or continuity of the mixed partials. Is this just an extra implicit assumption that $u$ and $v$ satisfy Clairaut's theorem, or does this actually follow from $f$ being holomorphic?

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    $\begingroup$ A holomorphic function is not only smooth, it is analytic (has a power series expansion). So the mixed partials are equal. $\endgroup$ – Daniel Fischer Jul 16 '14 at 19:16
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    $\begingroup$ It may be worth noting that a harmonic distribution - and for distributions, the partial derivatives commute as for smooth functions - is actually induced by a real-analytic function (and therefore a harmonic function in the classical sense). $\endgroup$ – Daniel Fischer Jul 16 '14 at 19:20
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Summary of comments by Daniel Fischer:

A holomorphic function is not only smooth, it is analytic (has a power series expansion). So the mixed partials are equal.

It may be worth noting that a harmonic distribution - and for distributions, the partial derivatives commute as for smooth functions - is actually induced by a real-analytic function (and therefore a harmonic function in the classical sense). Weyl's lemma

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