2 added 775 characters in body; deleted 2 characters in body

Let $\psi=f+ig=\rho e^{i\theta}$ be a complex function on some open subset of $\mathbb{R}^n$, where $f,g,\rho$ and $\theta$ are real-valued. I happened to find that the identity of differentiation for the polar coordinate expression $$(1)\qquad\nabla\psi=e^{i\theta}\nabla\rho + i\rho e^{i\theta}\nabla\theta$$ is rather formal. In general, $f$ and $g$ has the same regularity as $\psi$: $\psi\in C^1$ iff $f,g\in C^1$; $\psi\in H^1$ iff $f,g\in H^1$, etc. But this is not the case for $\rho$ and $\theta$. For example, even $f,g\in C^{\infty}$, $\rho=\sqrt{f^2+g^2}$ may not be differentiable at points where $f=g=0$. One can see this by considering the one dimensional example $f(x)=x$, $g(x)=0$, and $\rho=|x|$ is not differentiable at $0$. Assuming $f,g\geq 0$ is a way to remedy this problem. In fact, $f,g\geq 0$ and $C^1$ implies $\rho\in C^1$. A more natural way is to simply consider weak differentiability. For example if $f,g\in H^1$, I can show $\rho\in H^1$ (of course with weak derivatives in the same form as the usual ordinary ones. But although it seems to be very natural, I found this is a not so easy exercise in real analysis). I proved it by using approximation of $f,g$ by smooth functions. And I failed when using only the definition of weak derivatives. So, if only weak differentiability of $f,g$ is known and there is no approximation theorems as in Sobolev spaces, I don't know whether $\rho$ is also weakly differentiable. But anyway the result in the Sobolev setting may be sufficiently satisfactory.

The big problem comes when considering $\theta$. I have no idea about the regularity of $\theta$ at all. See the second term in the right hand side of identity (1), if $\psi\in H^1$, we should have $\rho\nabla\theta\in L^2$. But we know $\rho\in L^2$ (in fact $H^1$), so $\nabla\theta\in L^{\infty}$? $\theta\in W^{1,\infty}$? There is also a disturbing problem that on where $f$ and $g$ are both zero, $\theta$ can in fact be defined arbitrarily. So, a more correct question may be is there a natural choice of $\theta$ so that (1) is true in some sense?

Thanks to Denis Serre's references, I think my question may better be divided into two parts.

First, let $f+ig=\rho u$ with $\rho=\sqrt{f^2+g^2}$ and $u$ some complex function with $|u|=1$. Then, what is the best possible regularity of $u$ expected, in terms of $f$ and $g$?

And second, write $u=e^{iv}$, what is the best regularity of $v$ in terms of $u$?

The references deal with the second question. From their introductions it looks like the natural framework for this question are such spaces as VMO, BMO, etc. But does the first question lead to such answer as $u\in$VMO or $u\in$BMO, under for example the assumption $f,g\in H^1$? I have little knowledge about the spaces BMO and VMO. Is the answer to my last question naturally "yes"?

1

# What is the regularity of the argument of a complex function?

Let $\psi=f+ig=\rho e^{i\theta}$ be a complex function on some open subset of $\mathbb{R}^n$, where $f,g,\rho$ and $\theta$ are real-valued. I happened to find that the identity of differentiation for the polar coordinate expression $$(1)\qquad\nabla\psi=e^{i\theta}\nabla\rho + i\rho e^{i\theta}\nabla\theta$$ is rather formal. In general, $f$ and $g$ has the same regularity as $\psi$: $\psi\in C^1$ iff $f,g\in C^1$; $\psi\in H^1$ iff $f,g\in H^1$, etc. But this is not the case for $\rho$ and $\theta$. For example, even $f,g\in C^{\infty}$, $\rho=\sqrt{f^2+g^2}$ may not be differentiable at points where $f=g=0$. One can see this by considering the one dimensional example $f(x)=x$, $g(x)=0$, and $\rho=|x|$ is not differentiable at $0$. Assuming $f,g\geq 0$ is a way to remedy this problem. In fact, $f,g\geq 0$ and $C^1$ implies $\rho\in C^1$. A more natural way is to simply consider weak differentiability. For example if $f,g\in H^1$, I can show $\rho\in H^1$ (of course with weak derivatives in the same form as the usual ordinary ones. But although it seems to be very natural, I found this is a not so easy exercise in real analysis). I proved it by using approximation of $f,g$ by smooth functions. And I failed when using only the definition of weak derivatives. So, if only weak differentiability of $f,g$ is known and there is no approximation theorems as in Sobolev spaces, I don't know whether $\rho$ is also weakly differentiable. But anyway the result in the Sobolev setting may be sufficiently satisfactory.

The big problem comes when considering $\theta$. I have no idea about the regularity of $\theta$ at all. See the second term in the right hand side of identity (1), if $\psi\in H^1$, we should have $\rho\nabla\theta\in L^2$. But we know $\rho\in L^2$ (in fact $H^1$), so $\nabla\theta\in L^{\infty}$? $\theta\in W^{1,\infty}$? There is also a disturbing problem that on where $f$ and $g$ are both zero, $\theta$ can in fact be defined arbitrarily. So, a more correct question may be is there a natural choice of $\theta$ so that (1) is true in some sense?