I am trying to follow an example in my literature and I am pretty lost. It says that if $u$ and $v$ are complex-valued functions that are $C^1$-smooth in some open set, and $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \hspace{5mm}\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}, $$ then we can show that $u$ and $v$ are $C^{\infty}$-smooth. Then they go on to show it, by stating that it follows from the assumptions that $f=u+iv$ and $g = \bar{u} + i\bar{v}$ are analytic, and hence that $$u = \frac{f + \bar{g}}{2} $$ and $$v=\frac{f-\bar{g}}{2i} $$ are analytic, and so $u$ and $v$ must be $C^{\infty}$-smooth, since analytic functions always are.

The issues I have with this is:

1) Why would it even follow that $f=u+iv$ is analytic? I mean, if $u$ and $v$ were real-valued, then I know that it does follow. But $u$ and $v$ aren't said to be real-valued in this case.

2) I also don't see how we can draw from the assumption the conclusion that $g =\bar{u} + i\bar{v}$ would be analytic. I suppose that if I understood (1), then maybe I would understand (2) as well.

3) And lastly, I also don't get how we can assume that for example $\frac{f-\bar{g}}{2}$ is analytic. I mean, $g$ being analytic doesn't imply that $\bar{g}$ is, does it?

I wish I had any of my own work to show, but at this point I am simply looking through the literature, trying to find something that explains these conclusions that are drawn in the example and I can't find any... So I'm hoping for help here.


I think I have figured out why $f=u+iv$ and $g=\bar{u} + i\bar{v}$ are analytic. So I suppose that what I need help with right now is to understand why that implies that for example $$u = \frac{f + \bar{g}}{2} $$ is $C^{\infty}$-smooth.

  • $\begingroup$ It's not really clear what you have available, but I'll try to add some thoughts. First, there is a difference between real- and complex-analytic that's important here, and neither are equivalent to (nor implied by) $C^{\infty}$. Since you're talking about the real and imaginary parts, real-analyticity is probably the right thing to look at. The proof that solutions to the Cauchy-Riemann equations are analytic is non-trivial and typically goes through the Cauchy integral formula or a variant on Weyl's lemma. $\endgroup$ – user296602 Dec 7 '18 at 0:59
  • $\begingroup$ (I've never heard of Weyl's lemma before.) In my literature, there hasn't really been a distinction made between real-analytic and complex-analytic. Which now makes me feel really frustrated. :p But there is a theorem stating that if a function $f$ is analytic on some domain, then it is $C^{\infty}$-smooth there... And also a theorem stating that if $u,v$ are real-valued and satisfy the Cauchy-Riemann equations, then $f=u+iv$ is analytic. Are you saying that this is true even if $u$ and $v$ are not necessarily real? @T.Bongers $\endgroup$ – j.eee Dec 7 '18 at 1:10
  • $\begingroup$ The real part of a complex-analytic function is generally not complex-analytic, so there is an important distinction to make. Perhaps this question will help. $\endgroup$ – user296602 Dec 7 '18 at 1:12
  • $\begingroup$ One should be careful. If $f$ is analytic, then $\overline{f(\bar z)} $ is analytic. The real and imaginary parts of a complex analytic function aren't in general complex analytic, so you can't expect them to be a sum of complex analytic functions. We do have that the real and imaginary parts are Harmonic though, which minimally implies $C^\infty$ $\endgroup$ – Brevan Ellefsen Dec 7 '18 at 16:51

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