# Is the normalized derivative of a holomorphic function Sobolev?

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.

• 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. – user647486 Apr 28 '19 at 5:58
• Is $\mathbb D^2=\{z\in \mathbb C:|z|\le 1\}?$ If so, that is strange notation in this context. – zhw. May 6 '19 at 16:27
• It seems to me the problem is on the boundary. At points in the open disc there is no problem. – zhw. May 6 '19 at 18:26
• 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. – Asaf Shachar May 7 '19 at 9:56
• 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. – zhw. May 7 '19 at 17:02