# If $u$ and $v$ are complex-valued and satisfy the Cauchy-Riemann equations, does it follow that $u$ and $v$ are $C^{\infty}$-smooth?

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.

EDIT

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.

• 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. – user296602 Dec 7 '18 at 0:59
• (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 – j.eee Dec 7 '18 at 1:10
• 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. – user296602 Dec 7 '18 at 1:12
• 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 $C^\infty$ – Brevan Ellefsen Dec 7 '18 at 16:51