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I want to make sure that my proof to the following problem is right. I came up with this proof before I checked any stuff online. However, none of those solutions use my approach, which in my opinion, is simpler. So I wonder there may be some mistakes or gaps in my proof.

Show that if $f$ is holomorphic in the unit disc, is bounded, and converges uniformly to zero in the sector $\theta < \text{arg} z < \varphi$ as $|z| \rightarrow 1$, then $f = 0$.

I will not include every detail in the following proof, but the skeleton is there:

Proof: WLOG we may assume $\theta = 0$; otherwise a rotation will do. $f$ now converges uniformly to $0$ on $0 < \text{arg} z < \varphi$. Also, this means we can extend $f$ to $\partial \mathbb{D} \cap \text{sector}$ continuously by setting $f = 0$ on it.

Now there must exist $N \in \mathbb{N}$ s.t. $N \varphi > 2\pi$. Consider $f^N$ on $\mathbb{D}$. We know that $f^N$ on $\partial \mathbb{D}$ is zero by the nature of $N$ and $\varphi$. Hence $f^N \equiv 0$ on $\mathbb{D}$ by the max. mod. priciple, and the result follows.

Thank you very much!

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    $\begingroup$ How do we know that $f^N$ is zero on the boundary? $\endgroup$
    – Zach Boyd
    Oct 6, 2017 at 22:06
  • $\begingroup$ Oh I confused myself with the values and the domain. $\endgroup$
    – S. D. Z
    Oct 6, 2017 at 22:08
  • $\begingroup$ If you can find a way to repair the argument I would be interested to see it. I suspect the argument may be broken though. $\endgroup$
    – Zach Boyd
    Oct 6, 2017 at 22:09
  • $\begingroup$ Could be useful: math.stackexchange.com/questions/893114/… $\endgroup$
    – sranthrop
    Oct 6, 2017 at 22:39

3 Answers 3

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OK. I tried sth different to rebuild my proof, with all the necessary details and explanation (I believe) contained. Here is the sketch and the complete version:

Sketch of the proof: first show that $f$ can be continuously extended to $0$ on the arc of the sector. Define a new function depending on $f$, possibly by rotating, such that it equals to $0$ on the boundary. Then apply the maximum modulus principle.

Proof: WLOG we assume that $\theta = 0$. $S:= \left\{ z\in\mathbb{D}: 0< \text{arg} z <\varphi\right\}$ is the sector. $f \rightrightarrows 0$ on $S$, i.e. $\forall \epsilon >0$, $\exists \delta >0$, s.t. $\forall z\in S$, and $\forall w\in \partial \mathbb{D}$, whenever $|z-w|<\delta$, we have $\left |f(z) \right |<\epsilon$. Here we explains the uniformity. Basically we can imagine it dies to 0 very well on the boundary.

Now we may extend $f$ to $\partial \mathbb{D}_{(0, \varphi)}$ (the arc of the sector) by letting $f = 0$.

For $\varphi/2$, there is a nature number $N\in \mathbb{N}$ s.t. $N \varphi /2 > 2\pi$. We want to somehow cover the whole boundary of the disc so we may apply the max. mod. principle. However, the interval of argument is open. It is necessary and convenient to choose a number smaller than$\varphi$so that we can rotate it many times to cover all of the boundary with no holes at all. Consider $$F(z):= f(z)f(e^{-i\varphi/2}z)f(e^{-i2\varphi/2}z)\cdots f(e^{-iN\varphi/2}z)$$

on $\mathbb{D}$. First, it is bounded on $\mathbb{D}$. Next, we show that it can be continuously extended to $\bar{\mathbb{D}}$: $\forall w\in \partial \mathbb{D}$, $\exists k\in \left\{ 0, 1, 2, \cdots, N\right\}$ s.t. $k\varphi/2 < \text{arg} w \leq (k+1)\varphi/2$. Now given $\epsilon >0$, $\exists \delta >0$ s.t. $|z-e^{-ik\varphi/2}w|<\delta$ (Rotate it back to where we start) implies $\left |f(z) \right |<\epsilon$; for the other numbers $j \neq k$ in $\left\{ 0, 1, 2, \cdots, N\right\}$, $\left |f(e^{-ij\varphi/2}z) \right |<B$ as $f$ is bounded. Hence $$\left |F(z) \right |= \prod_{j=0}^{N}\left |f(e^{-ij\varphi/2}z) \right |<B^N\epsilon.$$

So we may define $F$ to be $0$ on $\partial \mathbb{D}$. Apply the maximum modulus principle. We conclude $F \equiv 0$ on $\bar{\mathbb{D}}$.

If $f \not \equiv 0$, then $f$ has at most countably many roots since it is analytic. We use this fact to conclude a contradiction. Now fix an arbitrary $\alpha$, $\forall re^{i\alpha}\in \mathbb{D}$, at least one of the term in $F(\alpha) = 0$ is zero. This gives a root for each $r \in (0, 1)$. But $(0, 1)$ is uncountable. Contradiction.

Therefore $f\equiv 0$. $\square $

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  • $\begingroup$ Try not copy this to your 503 homework XD $\endgroup$
    – S. D. Z
    Oct 7, 2017 at 3:59
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Your solution if $f$ is holomorphic and bounded on $\mathbb{D}$, applying the maximum modulus principle to $\prod_{l=1}^n f(z e^{2i \pi l/n})$ is good. But in general you'll need Schwarz's reflection principle.

If $f(z)$ is holomorphic on $U = \mathbb{D} \cap \{ \text{arg}(z) \in (a,b)\}$, continuous on $\overline{U}$ and real on $B = \overline{U} \cap \partial \mathbb{D}$ then let $g(z) = f(z) 1_{|z| \le 1} + \overline{f(1/\overline{z})} 1_{|z| > 1}$ which is holomorphic on $U \cup 1/U$ and continuous on $V = U \cup 1/U \cup B$. Therefore by Morera's theorem $g$ is holomorphic on $V$.

If $f$ vanishes on $B$ then $g = 0, f=0$

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Consider the function $f\colon\mathbb C\longrightarrow\mathbb C$ defined by$$f(z)=\begin{cases}0&\text{ if }\operatorname{Re}z\geqslant 0\\1&\text{ otherwise.}\end{cases}$$Then, within the boundary of the unit disk, $f\equiv0$ on the right half of the boundary and only there. The same thing occurs with any power of $f$.

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  • $\begingroup$ Yes. Thanks. That is what I know as an alternative way to do. However, I want to repair my proof. I will think about it later. $\endgroup$
    – S. D. Z
    Oct 6, 2017 at 22:15

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