Show that a limit of the derivative of a complex function is $0$

Question

Question: For some $\alpha>0$, define $S=\{re^{i\theta}, r>0, 0<\theta<\alpha\}$. $f$ is bounded and holomorphic in $S$. Show that $\lim_{r\to\infty}f'(re^{i\theta})=0$ for each $0<\theta<\alpha$.

Below $\Gamma$ is the counter-clockwise contour in S with only one loop around $z$ (here I choose $\Gamma$ to be a circle).

My attempt is to represent $f'$ by Cauchy’s integral formula, i.e., I write $$ f'(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(w)}{(w-z)^2}dw.$$ Then $$ f'(re^{i\theta})=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(w)}{(w-re^{i\theta})^2}dw.$$ Since $r\to\infty$, we can get $\frac{f(w)}{(w-re^{i\theta})^2}\to 0$, and then the integral goes to $0$ so that $f'(re^{i\theta})$ goes to $0$.

But I begin to doubt myself since my attempt is too easy. So, is the method above correct? Is there any other ways to approch the problem.

Answer 1

Assume $f$ is holomorphic on $S,$ with $|f|\le M$ on $S.$ Let $\theta \in (0,\alpha).$ Then there exists $\rho >0$ such that $D(e^{i\theta},\rho) \subset S.$ By Cauchy's estimates,

$$|f'(e^{i\theta})| \le \frac{M}{\rho}.$$

We can apply this to all of the functions $f_r(z) = f(rz), r>0,$ since they are all holomorphic in $S$ with same bound. Thus for all $r>0,$

$$\tag 1 |(f_r)'(e^{i\theta})| \le \frac{M}{\rho}.$$

But notice that $(f_r)'(z) = rf'(rz).$ Thus $(1)$ becomes

$$|rf'(re^{i\theta})| \le \frac{M}{\rho}$$

for all $r>0.$ The result follows immediately.

Answer 2

The idea is good, though you may want to be more specific how you choose $\Gamma$. To begin with, a $\Gamma_r$ is chosen for each $r$ (with $\theta$ fixed). Suppose we want to use the estimate ¹ $$|f'(re^{i\theta})|\leq\frac1{2\pi}\left|\int_{\Gamma_r}\frac{f(w)}{(w-re^{i\theta})^2}dw\right|\ll\frac{\ell(\Gamma_r)}{\inf_{w\in\Gamma_r}|w-re^{i\theta}|^2}.$$

It makes sense to choose $\Gamma_r$ to be a circle of certain radius $R_r$ around $re^{i\theta}$, because this optimizes our estimate.² This gives

$$|f'(re^{i\theta})|\ll\frac1{R_r}$$

The question is now whether we can choose radiuses $R_r$ such that $R_r\to\infty$. This is the whole point: $\theta$ is fixed and between $0$ and $\alpha$, so the distance from $re^{i\theta}$ to the exterior of $S$ increases linearly with $r$. We can thus choose $R_r\asymp r$.³

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¹ "$\ll$" is a notation due to Vinogradov, which means "$\leq$" up to a positive factor, here: $\|f\|_\infty/(2\pi)$.
² Think about it: we want $\Gamma_r$ to have small length $\ell(\Gamma_r)$, while keeping the minimal distance to a fixed interior point ($re^{i\theta}$) large. Given any contour $\Gamma_r$, the circle with radius $\inf_{w\in\Gamma_r}|w-re^{i\theta}|$ will always do better.
³ To be precise: the distance is $r\sin\beta$ where $\beta=\min(\theta,\alpha-\theta)$, so $R_r=0.99r\sin\beta$ works.
Note that when $\theta$ is close to $0$ or $\alpha$, this $R_r$ is very small, so the convergence is typically slower, and at the boundary values $0$ and $\alpha$ we have no control at all. This is often the case in analysis: if convergence (more generally, an estimate) is guaranteed for certain values of a parameter, it is typically slower (resp. worse) near boundary values. I use it as a quick check whether estimates make sense: I would have worried if the convergence speed did not depend on $\theta$.

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Self-test:

1. With the same set-up, what is the weakest condition on the growth of $f$ if we want to conclude that the derivatives of $f$ up to the $n$th vanish at infinity, for fixed $\theta$?
2. Can we guarantee $\lim_{r\to\infty} f(re^{i\theta})=0$ uniformly for $0<\theta<\alpha$? Can you think of a counterexample?

Answers:

1. $f(re^{i\theta})=o_\theta(r^n)$ (and the implicit constant may depend on $\theta$)
2. No. Examples: $f(z)=e^{iz}$, any $\alpha\leq\pi$: for any large $r$, $f\approx1$ close the the real axis. Also: rational functions $p/q$ with $\deg p=\deg q$ and no poles in $S\cup\{0\}$ but poles on the boundary of $S$. Weirder things like $\Gamma'/\Gamma$, which grows at most logarithmically on fixed angular regions $|\theta|<\pi-\delta$, and thus has vanishing derivative at infinity by the same argument.