# “Close-to-analytic”, but “not quite”, functions on the complex numbers.

I heard of a way by which one can say that a given complex function is "close to analytic", namely if its Wirtinger partial $\frac{\partial f}{\partial \bar{z}}$ is small, meaning it "depends only a little bit on $\bar{z}$".

So one may consider a kind of complex function which is more general than analytic (here, I mean "entire") functions: namely where $\frac{\partial f}{\partial \bar{z}}$ is bounded, and the function is real-smooth everywhere. So my question is: can we get "close-to-analytic" "bump function"-like objects this way? Namely, can we make a function with compact support in the complex plane, real-smooth, and whose $\frac{\partial f}{\partial \bar{z}}$ is close to, but not quite, zero, across the domain -- i.e. $\left| \frac{\partial f}{\partial \bar{z}} \right| < \epsilon$ for all $z$ in the domain and some real $\epsilon > 0$ which is sufficiently small? If no, what's the reason and proof? If yes, what happens as we reduce $\epsilon$ toward 0? What that last part means is, what is the behavior of a parameterized function (or function family) $f_\epsilon(z)$ with $\epsilon > 0$ where as a function of $z$ it meets the given criteria and its Wirtinger $\bar{z}$-partial is bounded by $\epsilon$, as $\epsilon \rightarrow 0$? My guess is the bump gets flatter (i.e. $\max |f_\epsilon(z)|$ gets smaller). Is this right, or always right? If the bump gets flatter, then does that imply the existence of a limiting amplitude for a given $\epsilon$? If so, what is it? What happens if one considers such a family for a fixed support set?

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Given a $C^\infty$ function $g$ with compact support on $\mathbb C$, there always exists a $C^\infty$ function $f$ on $\mathbb C$ with $$\frac{\partial f}{\partial \bar{z}}=g .$$ This is the special case $n=1$ of a theorem in complex analysis of several variables due to Dolbeault and Grothendieck.
You can find a proof for the one-variable case in Forster's admirable book Lectures on Riemann Surfaces, page 104.
In exercise 13.3 page 108 Forster states that moreover you can find an $f$ which also has compact support iff for all $n\in \mathbb N$ you have $$\iint _\mathbb C z^n g(z)dz\wedge d\bar z =0$$

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For the converse, how can we show? I have tried several methods but I cannot solve it. Any hints? –  enoughsaid05 Dec 18 '13 at 22:50

Cauchy's generalized formula states that for a $C^1$ map states : $$f(z)= \frac{1}{2i\pi} \int_{|z-\zeta|=1} \frac{d\zeta}{z-\zeta} + \frac{1}{2i\pi} \int_{|z-\zeta| \leq 1} \frac{\overline{\partial}f}{z-\zeta} |d\zeta|^2$$

Therefore, for a smooth bump function like you mentionned, you'll get $\max |f_\epsilon| =O(\| \overline{\partial}f \|_{L^1} ) = O(\epsilon).$

On a related note (though not quite what you asked, but you might find this of interest ): the so-called Alhfors-Bers theorem says that if you choose any mesurable bounded map $\mu : \mathbb{P}^1 \rightarrow \mathbb{C}$ with $\|\mu\|_\infty \leq 1$, there exists a unique (up to composition with Möbius transformation) homeomorphism $f_\mu$ such that $f_\mu$ satisfies the so-called Beltrami equation (in the sense of distributions) :
$$\overline{\partial} f_\mu = \mu \partial f_\mu$$

These maps are called quasiconformal, and they retain some of the rigidity of the conformal maps. See Gardiner and Lakic, "Quasiconformal Teichmüller theory".

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