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Claim: The real and imaginary parts of an analytic function are harmonic

Proof:
Let $f=u+iv $ be analytic in some open set of the complex plane.
$$ \frac {\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} $$ $$ =\frac{\partial}{\partial x}\frac{\partial u}{\partial x} + \frac{\partial}{\partial y}\frac{\partial u}{\partial y} $$ $$ =\frac{\partial}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial}{\partial y}\frac{\partial v}{\partial x}(using \ Cauchy-Riemann) $$ $$ =\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial x\partial y}=0 $$

What I am confused about:
In line 4, how do we know the mixed partials are equal. Doesn't v have to be $ C_2 $ for the mixed partials to be equal. Does analytic imply $C_2?$

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    You can talk about harmonicity only once you know that $u$ and $v$ are $C^2$. Once they are $C^2$, it follows that the mixed partials are equal. But do you know a proof of why they are $C^2$ in the first place? (By that, I mean that analyticity does imply $C^2$ property of the component functions, but do you know the reason for that?) If you do not, then look up power series expansions of holomorphic functions. – 2017-01-06
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    Every analytic function is $C^\infty$. – 2017-01-06
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    May I know why it is $C^\infty$ – 2017-01-06
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    @matt see Ponnusamy S., Silverman H., Complex Variables with Applications, Birkhauser (2006), page 243. – 2017-01-06

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