# Proof that “holomorphic implies continuously differentiable” using real analysis

I am interested in how one could prove that $H(\Omega)\subset \mathscr C^1(\Omega)$ without using the Cauchy integral formula, or any other complex analytic techniques, only real analysis.

Thank you in advance

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Why would you want to do that? That said, it's probably possible to show that $\bar\partial$ is elliptic, or at least hypoelliptic using purely real methods. –  mrf Jun 14 '13 at 15:20
But I'm afraid this will not be enough : it does not seem obvious that if a function $f$ is holomorphic (i.e. just differentiable in the complex sense) then $\overline\partial f=0$ in the distribution sense. –  Etienne Jun 14 '13 at 18:59

If $f$ is holomorphic then you can write;
$f(z+h) - f(z) + hf'(z) = h \phi(h)$
where $\phi$ has the property that $\phi(h) \rightarrow 0$ as $h \rightarrow 0$, then; $f(z+h) \rightarrow f(z)$, as $h \rightarrow 0$
Hence $f$ is continuous
The OP wants to prove that $f'$ is continuous. Not $f$. –  1015 Jun 14 '13 at 14:35