Cauchy estimate on angular domain

Question

This question is a (perhaps naive) 'simplification' of a result in a paper, so the answer could be negative.

Define the cone $\Sigma(\theta)$ for $\theta\in(0,\pi/2]$, $$\Sigma(\theta) = \left\{ z = x+iy : x>0, |y|<(\tan\theta)x\right\}, $$ and define the norm $\|f\|_\theta$ for functions $f$ analytic on $\Sigma(\theta)$ by

$$\|f\|_\theta = \sup_{z\in \Sigma(\theta)}|f|$$ Let $Y_\theta$ be the space of functions analytic on $\Sigma(\theta)$ with $\|f\|_\theta < \infty$. Also let $\chi$ be a distinguished analytic function in $Y_{\pi/2}$ with $\chi(0) = 0$. It would seem then, that the following inequality is true: Let $f\in Y_{\theta}$. Then for any $\theta'<\theta$, $$ \|\chi f'\|_{\theta'} \leq C \frac{\|f\|_{\theta}}{\theta-\theta'} $$ where the constant $C$ can depend on $\chi$. How is this proven, and how does $\chi$ help?

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Remarks

• (basic Cauchy Estimate) Let $B(r)$ denote the ball around $0$ of radius $r$, and let $|f|_r$ denote $\sup_{z\in B(r)} |f(z)|$. Lets say $f\in X_r$ if $f:B(r)\to \mathbb C$ is analytic on its domain, with $|f|_r<\infty$. From the usual Cauchy formula $f'(z) = \frac1{2π i}\int_{\partial D} \frac{f(w) dw}{(w-z)^2}$ it is not hard to prove that for any $f\in X_r$, with any $r'<r$, $$ |f'|_{r'} \leq C \frac{|f|_{r}}{r-r'}.$$

Answer 1

If you didn't include the factor $\chi,$ the result would imply $f'$ is bounded in $\Sigma (\theta').$ There are certainly counterexamples to that. But if you take, say, $\chi(z) = z,$ then $\chi$ tamps down the possible growth of $f'$ near $0$ and we have a chance.

Here's a sketch with $\chi(z) = z:$ Show that if $re^{it} \in \Sigma (\theta'),$ then

$$\tag 1 d(re^{it}, \partial \Sigma (\theta)) \ge cr(\theta - \theta').$$

Here $c$ is a constant that depends only on $\theta, \theta'.$ Now we can apply Cauchy's estimates directly:

$$|\chi(re^{it})f'(re^{it})| = r|f'(re^{it})| \le r\frac{\|f\|_\theta}{d(re^{it}, \partial \Sigma (\theta))}.$$

Apply $(1)$ to finish.

Answer 2

Expanding on the accepted answer.

If you didn't include the factor $χ$, the result would imply $f′$ is bounded in $Σ(θ′)$.There are certainly counterexamples to that.

One family of counterexamples is $z^s e^{-z}$, $s∈(0,1)$, where the branch cut of $z^s$ is chosen to lie outside the sector. The derivatives near the origin are like $1/|z|^{1-s}$ (c.f. the bound $\|zf'(z)\|_{\theta'}<\|f\|_\theta$)

Show that if $re^{it}∈Σ(\theta′)$, then $\tag 1 d(re^{it}, \partial \Sigma (\theta)) \ge cr(\theta - \theta').$

The distance $d(re^{it}, \partial \Sigma (\theta))$ is obtained by dropping a perpendicular to the closest ray of $\partial \Sigma (\theta)$. Therefore, it is half a chord on a circle of radius $r$, across a sector of angle $2(\theta-t)$, hence, $d(re^{it}, \partial \Sigma (\theta)) = \frac{r}{2}\sin(\theta-t)$. For the range $\alpha\in(\theta-\theta',\theta] \subset (0,\pi/2)$, $$\sin \alpha > \frac{\sin(\theta) }{\theta}\alpha > \frac{2}{\pi} \alpha,$$ so setting $t = \theta - \alpha$, we see that $$d(re^{it}, \partial \Sigma (\theta)) \ge \frac{r}{\pi} (\theta-t) \ge \frac{r}{\pi} (\theta-\theta') .$$