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It is well known by a Theorem of Rademacher that any locally Lipschitz-continuous function $f:\mathbb R^m\to\mathbb R^n$ is (Frechét-)differentiable almost everywhere (with respect to the Lebesgue-measure), i.e. for almost all $x_0\in\mathbb R^m$ there is a linear map $Df(x_0):\mathbb R^m\to\mathbb R^n$ such that \begin{align*} \lim_{x\to x_0}\frac{||f(x)-f(x_0)-Df(x_0)(x-x_0)||}{||x-x_0||}=0. \end{align*} My question now is:

Is there a similar result for complex functions, i.e. is every locally Lipschitz-continuous function $f:\mathbb C\to\mathbb C$ almost everywhere COMPLEX differentiable?

I don't think that this is true. We certainly get the real differentiability in the above mentioned sense, but to make $f$ complex differentiable at a point, it needs to satisfy the Cauchy-Riemann-Equations there.

Any help is highly appreciated. Thanks in advance!

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Indeed it is not true; if $f(z)=\overline{z}$ then $f$ is even globally Lipschitz but is nowhere $\mathbb C$-differentiable.

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    $\begingroup$ Ohh sometimes I overlook the most trivial things... Thank you! $\endgroup$ – sranthrop Oct 10 '16 at 10:06
  • $\begingroup$ no problem. it was indeed the first thing to try ;) $\endgroup$ – Glougloubarbaki Oct 10 '16 at 10:07

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