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I often read that the rigidity and smoothness properties of holomorphic functions can be explained by the fact that the Cauchy-Riemann equations are elliptic. In which sense is that true?

Obviously they are not elliptic in the sense of the Wikipedia definition or that of Lecture notes of Ambrosio, both of which are restricted to second order PDE. There is also a definition of hypoellipticity out there that is simply based on smoothness of solutions. That one would hardly be useful to make sense of how ellipticity explains smoothness properties of holomorphic functions.

It remains the definition of elliptic equations as those with definite principal symbol. However, I don't believe the CR equations can be reasonably be classified as elliptic under this definition. Of course we can say that the symbol of the Cauchy-Riemann operator $\partial_x+i\partial_y$ is $x+iy$ for $x,y\in \mathbb{R}$. If that's it, I already have my answer, because the last expression is definite. However, I feel like this would be cheating. Since the differentials $\partial_x f$ and $\partial_y f$ occuring in the Cauchy-Riemann equations are elements of $\mathbb{C}$ the variables in the symbol should probably be too, and in that case $x+iy$ can be zero even for nonzero $x$ and $y$. Is there any insight to gain in using the other form of the symbol and deriving properties of holomorphic functions from there?

Bananach
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  • i always thought that's not really related to the CR itself but rather to the fact for each component function - as a direct consequence of CR - $\Delta u =0$ and $\Delta v = 0$. That is an elliptic operator. – JayTuma Aug 18 '18 at 21:54
  • It's elliptic in the sense that the principal symbol is always nonzero. – Angina Seng Aug 18 '18 at 22:03
  • @LordSharkTheUnknown ... if you admit only real values for the monomials in the principal symbol. Do you know of a justification to do that? The other answer does not go along this direction; it'd be great if you added one – Bananach Aug 19 '18 at 06:07

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Suppose $f(z)$ is holomorphic and switch from thinking of $f$ as $f:\mathbb{C} \rightarrow \mathbb{C}$ to $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ via: \begin{align*} z &= x + \mathrm{i} y \\ f(x,y) &= f(x+\mathrm{i}y) = f(z) \\ u(x,y) &= \Re(f(x,y)) \text{, and} \\ v(x,y) &= \Im(f(x,y)) \text{,} \\ \text{so } f(x,y) &= u(x,y) + \mathrm{i}v(x,y) \text{.} \end{align*} Then the C-R equations are $u_x = v_y, u_y = -v_x$, where subscripts denote partial differentiation. These lead to the system of decoupled elliptic equations \begin{align*} u_{xx} = -u_{yy} \\ v_{xx} = -v_{yy} \text{.} \end{align*} ... and it is fair to say that this "everywhere saddle in each component" property does lead to rigidity.

Eric Towers
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