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Let $\{f_n\}$ be a family of functions in $H(\Omega)$ the space of holomorphic functions, and suppose that $f_n$ converges to $f$ uniformly in compacts.

(1) Let $\overline{D_r(a)}\subset \Omega$ be a disck of radius $r$ centered in $a$ and $f(z)\neq 0 \forall z \in \partial D_r(a)$. Then there is $n_0$ such that for every $n>n_0$ $|f_n(z)-f(z)|<|f(z)|$ $\forall z\in \partial D_r(a)$

Proof: I think it should follow trivially from $f_n\to f$. I think of it this way: $f$ is defined in a compact disk, therefore it reaches a minumum $M$. Take $\epsilon>0$ such that $\epsilon < M$. Because $f_n\to f$ there is a $n_0$ such that $|f_n(z)-f(z)|<\epsilon \leq |f(z)|$.

(2) There is $n_0$ such that for every $n>n_0$ $f_n$ has as many roots as $f$ in the closure of $D_r(a)$.

Proof: Follows from (1) like this: If $z_0$ is a zero then $|f_n(z_0)-f(z_0)|<|f(z_0)|=0$ then $f_n(z_0)=0$

(3) If $f_n$ has no roots, then either $f$ has no roots or $f=0$.

Proof: Seems the contrapositive of (2), if $f$ had a root then $f_n$ would have roots.

(4) If $f_n$ is injective, then either $f$ is injective or constant.

Proof: I don't know how to prove this one... But I want to prove it by means of an example, this is, taking a non-constant nor ind injective $f$ and showing that $f_n$ cannot be injective. I haven't been able to figure out a function for this.

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  • $\begingroup$ Is $D_r(a)$ supposed to be the closed or the open disk with radius $r$? $\endgroup$ – Daniel Fischer Nov 19 '17 at 21:59
  • $\begingroup$ @DanielFischer Open, but it doesn't matter - there were some typos in (1). We consider $f $ in the boundary. $\endgroup$ – Cure Nov 19 '17 at 22:19
  • $\begingroup$ Right. That makes more sense, since otherwise part (2) would be silly. But for $f$ to be defined on $\partial D_r(a)$ we need $\overline{D_r(a)} \subset \Omega$. Only $D_r(a) \subset \Omega$ is not enough. (Aside: your muscle memory made you type $H(\Omega)$ instead of $\Omega$ there.) $\endgroup$ – Daniel Fischer Nov 19 '17 at 22:23
  • $\begingroup$ And the condition ought to be $f(z) \neq 0$ for $z \in \partial D_r(a)$. $\endgroup$ – Daniel Fischer Nov 19 '17 at 22:24
  • $\begingroup$ @DanielFischer The condition $\overline{D_r(a)}$ is right. I mistyped that too :( $\endgroup$ – Cure Nov 19 '17 at 22:27
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You have the right argument for (1), but you haven't written it up correctly. Since $f$ is complex-valued, it doesn't make sense to say that $f$ reaches a minimum (on $\partial D_r(a)$). You need to take the modulus, $\lvert f\rvert$ is a real-valued continuous function on $\Omega$, hence it attains a minimum on the compact set $\partial D_r(a)$. By the hypothesis, this minimum $M$ is strictly positive. Then the uniform convergence of $f_n$ to $f$ on compact subsets of $\Omega$ - in particular on $\partial D_r(a)$ - gives you that there is an $n_0$ such that $\lvert f_n(z) - f(z)\rvert < \epsilon$ for $n \geqslant n_0$. Which by your choice of $\epsilon$ implies the desired conclusion.

Your argument for (2) doesn't work. We only know $\lvert f_n(z) - f(z)\rvert < \lvert f(z)\rvert$ on $\partial D_r(a)$, not in the interior. Also, $\lvert f_n(z_0) - f(z_0)\rvert < 0$ is impossible. Part (1) gave us the premises of Rouché's theorem, which yields the conclusion.

Part (3) is not completely the contrapositive of part (2), but mostly. What is missing is that if $f$ is not identically $0$ [assuming that $\Omega$ is connected, otherwise insert "on any connected component of $\Omega$"], then it has only isolated zeros. For an isolated zero $a$, we can choose a positive $r$ small enough that the premise of part (1) holds, and hence part (2). Thus: if $f$ has zeros, but does not identically vanish, by parts (1) and (2) we know that $f_n$ has zeros for all large enough $n$. That the condition $f \not\equiv 0$ is necessary is easily seen by considering $f_n \equiv \frac{1}{n}$.

For part (4), suppose that $f$ is neither constant nor injective. Then there are $z_1, z_2 \in \Omega$ with $f(z_1) = w = f(z_2)$ and $z_1 \neq z_2$. By considering $f - w$ and $f_n - w$ in place of $f$ and $f_n$, we can assume that $w = 0$. Since $f$ is not constant, for all small enough positive $r$, we have $\overline{D_r(z_k)} \subset \Omega$ and $f(z) \neq 0$ on $\partial D_r(z_k)$ for $k = 1$ and $k = 2$. If we choose $r < \frac{1}{2} \lvert z_1 - z_2\rvert$, these two disks are disjoint. By parts (1) and (2), each $f_n$ has a zero in $D_r(z_1)$ and a zero in $D_r(z_2)$ for large enough $n$, so is not injective.

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