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This question is a cross-post from MSE. it is also a special case of this question.

Let $B=\{z\in \mathbb C \,|\,|z|\le 1\}$, and let $f:B \to \mathbb{C}$ be holomorphic on the interior $B^o$, and smooth on $B$.

Suppose that $f'(z)$ is not identically zero. 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 $h_x,h_y,g_x,g_y \in L^1(B)$.

The assumption implies that $f'(z) \neq 0$ a.e. on $B$.

For any $\epsilon >0$, the derivatives are integrable on $\{z\in \mathbb C:|z|\le 1-\epsilon\}$. The question is to determine what happens near the boundary.

Around any isolated zero of $f'$ in $B^o$, everything is fine*; $f'$ may vanish at infinitely many points (with an accumulation point on $\partial B$).

*Proof:

Suppose that $f'(0)=0$; write $f'(z)=z^ng(z)$ for some holomorphic $g$ 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 around $z=0$, 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 $\theta$ (the case of $\frac{z}{\|z\|}$). What is left is a straightforward calculation.

Edit:

It seems that there might be explosions near the boundary, even when $f'$ is never zero. So, I guess that the answer is negative?

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    $\begingroup$ You have enough trouble even without zeroes: just consider the function $f$ with the derivative $f'(z)=e^{-(1-z)^{-1/2}}e^{-\frac{1+z}{1-z}}$ (the first factor is there just to ensure $C^\infty$ up to the boundary). $\endgroup$
    – fedja
    Commented May 12, 2019 at 9:49
  • $\begingroup$ I am probably missing something stupid, but: The normalized derivative is measurable and bounded, so it is surely in $L^1(B)$, assuming that you use the Lebesgue measure on $B$. $\endgroup$
    – PhoemueX
    Commented May 12, 2019 at 11:37
  • $\begingroup$ @Asaf: Right, I am stupid. Thanks for confirming :P $\endgroup$
    – PhoemueX
    Commented May 12, 2019 at 13:39
  • $\begingroup$ Around $1$, of course. $\endgroup$
    – fedja
    Commented May 12, 2019 at 14:05
  • $\begingroup$ Yes, they diverge. The essential part iz $1/(1-z)^2$ near $1$. The first factor doesn't contribute much. If you still have trouble seeing why, let me know and I'll post a detailed explanation when I have time. $\endgroup$
    – fedja
    Commented May 23, 2019 at 20:19

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