Let $B=\{z\in \mathbb C \,|\,|z|\le 1\}$ be the closed unit disk, and let $f:B \to \mathbb{C}$ be holomorphic.

More precisely, I assume that $f$ is holomorphic on the interior $\text{int}(B)$, and smooth on the closed disk $B$.

Assume also that $f'(z) \neq 0$ almost everywhere on $B$. Are the derivatives of $\frac{f'(z)}{\|f'(z)\|}$ in $L^1(B)$?

More precisely, if $f=u+iv$, then $f'(z)=u_x+iv_x$, and I consider $$h(x,y)=\frac{u_x}{\sqrt{u_x^2+v_x^2}},g(x,y)=\frac{v_x}{\sqrt{u_x^2+v_x^2}}.$$

The question is whether or not $h_x,h_y,g_x,g_y \in L^1(B)$.

I can show that for any $\epsilon >0$, the derivatives are integrable on $\{z\in \mathbb C:|z|\le 1-\epsilon\}$. The "remaining part" of the question is to determine what happens near the boundary.

I thought that if $f'(z) \neq 0$ a.e., then $f'(z)$ may vanish only at finitely many points, but this is not true in general, since we have a non-empty boundary (where the usual identity theorem doesn't work). So, the case where $f'(z)$ has infinitely many zeros (with an accumulation point on the boundary) is still open.

Here is a solution for the case where $f'$ has only finitely many zeroes: (this implies that we are OK on every disk of radius $1-\epsilon$).

This problem reduces to analyzing what happens around a single zero, say at $z=0$. We can write $f'(z)=z^ng(z)$ for some holomorphic function $g$ satisfying $g(0) \neq 0$. Then $\frac{f'(z)}{\|f'(z)\|}=\frac{z^n}{\|z^n\|}\frac{g(z)}{\|g(z)\|}$. $g$ is smooth and non-zero in a neighbourhood of $z=0$, so the factor $\frac{g(z)}{\|g(z)\|}$ causes no problems. We are left with $\frac{z^n}{\|z^n\|}=(\frac{z}{\|z\|})^n=e^{in\theta}$, which reduces the problem to the analysis of the "argument function" $\theta$ (i.e. the case of $\frac{z}{\|z\|}$). A complete treatment of this can be found here in this previous question of mine.

  • $\begingroup$ Near a point $a$ where $f'(a)=0$ you can write $f'(z)=(z-a)^ng(z)$, where $g(a)\neq0$. Then $f'(z)/|f'(z)|=\frac{(z-a)^n}{|z-a|^n}\frac{g(z)}{|g(z)|}$, which is a product of two bounded functions. By the way, $f'$ might vanish at infinitely many points, although finitely many in a compact. $\endgroup$ – user647486 Apr 28 '19 at 5:58
  • $\begingroup$ Is $\mathbb D^2=\{z\in \mathbb C:|z|\le 1\}?$ If so, that is strange notation in this context. $\endgroup$ – zhw. May 6 '19 at 16:27
  • $\begingroup$ It seems to me the problem is on the boundary. At points in the open disc there is no problem. $\endgroup$ – zhw. May 6 '19 at 18:26
  • $\begingroup$ 1) Yes, this is what I meant by $\mathbb D^2$. What notation do you suggest instead? 2) I agree with you that there is no problem at interior points, since they are isolated. The "remaining part" of the question is to determine what happens near the boundary. $\endgroup$ – Asaf Shachar May 7 '19 at 9:56
  • $\begingroup$ It's just that very often $\mathbb D$ denotes the open unit disc in $\mathbb C.$ No need to change the notation, just thought I'd mention it. $\endgroup$ – zhw. May 7 '19 at 17:02

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