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I guess this could be a very elementary question. Anyway I can not find an answer in literature.

Let $f:U\rightarrow\mathbb{C}$ be an holomorphic function on an upen subset $U\subseteq\mathbb{C}$. Let us write $f(z) = u(x,y)+iv(x,y)$, where $z = x+iy$.

Is there a way of proving that $u,v\in C^{1}(U)$ without using complex integrals, and in particular without using the Cauchy integral formula?

Here holomorphic in $U$ means holomorphic in any $z_0\in U$ that is for any point $z_0\in U$ there exists the limit $$f^{'}(z) = \lim_{z\mapsto z_0}\frac{f(z)-f(z_0)}{z-z_0}.$$

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    $\begingroup$ This has probably been addressed before on MathOverflow, but can you get what you want from the theory of "elliptic regularity"? $\endgroup$
    – Yemon Choi
    Commented Apr 21, 2015 at 1:17
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    $\begingroup$ "Without using integrals" is a very strong requirement. $\endgroup$ Commented Apr 21, 2015 at 2:00
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    $\begingroup$ What is your definition of "holomorphic function"? $\endgroup$ Commented Apr 21, 2015 at 2:24
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    $\begingroup$ @YemonChoi Then it boils down to how to prove elliptic regularity without integrals. $\endgroup$
    – Fan Zheng
    Commented Apr 21, 2015 at 3:18
  • $\begingroup$ Can you define precisely what you mean by holomorphic function? Depending on the definition you use, the question may have an immediate answer. $\endgroup$ Commented Apr 21, 2015 at 11:01

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You can solve the Dirichlet problem for Laplace's equation on a square by separation of variables and Fourier series. You can check explicitly from the Fourier series that the solution is $C^\infty$ inside the square. Now use the maximum principle to infer uniqueness.

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    $\begingroup$ But how can we make sense of the Laplace equation without assuming $C^1$ regularity? In the weak sense? But that is integration, again! $\endgroup$
    – Fan Zheng
    Commented Apr 21, 2015 at 5:28

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