# conformal but not real differentiable, still holomorphic?

can one prove that a conformal map (one which preserves angles (in sense as well as in size) between arcs with non-zero derivatives at a point z and for which the limit of the absolute value of the difference quotient exists at z) has a complex derivative at z without assuming that it is differentiable at z as a map between IR^2? If so, how? If not can you give me a counterexample?

Thanks

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Since holomorphic functions are always real differentiable (just use the fact that holomorphic functions are analytic), I doubt you will get far. – Johannes Kloos Jul 9 '12 at 20:44
How exactly do you define the angle between the images of two arcs under a map which is not necessarily differentiable along curves? – Sam Jul 9 '12 at 21:10
I don't understand the downvote, and I understand the remark by @JohannesKloos even less. The question is quite reasonable. – user31373 Jul 9 '12 at 21:26

As @Sam pointed out, it is not immediately clear how to define the angle between images of curves. One way is to simply assume

(V) for every differentiable curve with nonzero velocity vector its image under $f$ is also differentiable and has nonzero velocity vector

One may suspect that assumption (V) is equivalent to $\mathbb R$-differentiability of $f$, but it is not. For example, the function $f(re^{i\theta})=r\exp (\theta+i\sin 2\theta)$ satisfies (V) but is not $\mathbb R$-differentiable at $0$. (Aside: there was a similar question A sufficient condition for differentiability of a function of several variables in terms of differentiability along paths. but it had the extra assumption that the directional derivative is linear.) So the questions makes sense with the addition of (V), but is somehow less appealing in this form since (V) does not look a very natural thing.

So I'm inclined to forget the angles and to consider only the second condition in the question: $$(*)\qquad \lim_{z\to a}\frac{|f(z)-f(a)|}{|z-a|}$$ exists at every point $a$. If such $f$ is a homeomorphism (at least locally), then it satisfies the definition of quasiconformal map with the coefficient of quasiconformality $K=1$. It is a well known, but nontrivial fact, that $1$-quasiconformal maps are conformal. The linked article by Heinonen attributes it to Menshov (1937) but does not give a precise reference and I can't track it down right now. Nowadays one simply appeals to the equivalence of the metric and analytic definitions of quasiconformality, which is established, e.g., in Lectures on Quasiconformal Mappings by Väisälä.

By the way, 1-quasiconformality is substantially weaker than (*): it only requires that $|f(z)-f(a)|$ is close to $|f(\zeta)-f(a)|$ whenever $|z-a|=|\zeta-a|$. There is no requirement for $|f(z)-f(a)|$ to be close to $|z-a|$.

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Thank you for your answer and link. They have been very useful. (V) is what I had in mind, indeed. Also, I think the Looman-Menchoff theorem might be closely related to the result by Menshov you mentioned. A proof can be found in Complex Analysis in One Variable by Raghavan Narasimhan on pp. 43-50. The only other reference in a related context to a result by Menshov which I found is D.E. Menshov, Les Conditions de Monogeneité, Actualités scientifiques et industrielles, No. 329 – Jorge Jul 9 '12 at 23:07
@Jorge I think this paper by Menshov answers your original question. In fact, his assumption are much weaker and involve only limits along finitely many directions at every point (directions may be different for different points). Full text in French is available, with an abstract in Russian. – user31373 Jul 9 '12 at 23:21