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I have some homework problems from Greene and Krantz' Function Theory of One Complex Variable. They come from Chapter 5. I definitely do not want answers, just light prodding in the right direction.

Let $f_j: D(0, 1)\to\mathbb C$ be holomorphic and suppose that each $f_j$ has at least $k$ roots in $D(0,1)$, counting multiplicities. Suppose that $f_j\to f$ uniformly on compact sets. Show by example that it does not follow that $f$ has at least $k$ roots counting multiplicities. In particular, construct examples, for each fixed $k$ and each $\ell$, $0\le\ell\le k$, where $f$ has exactly $\ell$ roots. What simple hypothesis can you add that will guarantee that $f$ does have at least $k$ roots?

I know we require continuity of the $f_j$ on the boundary for the number of zeroes to be the same in the limit, but I'm not clear on why this is. Presumably, that's the purpose of the question. In another question, the goal is to prove this when the disk and its boundary are in the region of holomorticity of the $f_j$.

I'm tempted to think that the problem we run in to is when the zeroes move to the boundary in the limit, but maybe I'm just not familiar enough to construct an actual example of this. I'm also confused by their use of at least $k$ roots. It seems simplest to start with a sequence of functions that all have exactly $k$ roots, but I'm afraid I'm missing something, and that it will only work if the functions have different numbers of zeroes, for some reason.

Basically, I would like some intuition about how things can go wrong when we only have holomorticity on the interior of the disk, and maybe a (small) clue about the form the sequence will take.

Prove: If $f$ is a polynomial on $\mathbb C$, then the zeroes of $f^\prime$ are contained in the closed convex hull of the zeroes of $f$. (Here the closed convex hull of a set $S$ is the intersection of all closed convex sets that contain $S$.) [Hint: If the zeroes of $f$ are contained in a halfplane $V$, then so are the zeroes of $f^\prime$.

I would really like to use the maximum modulus theorem to say that the zeroes of $f^\prime$ occur at maxima (minima) of $f$, and therefore that these only happen on the boundary of some set $U$ (where $f$ is continuous on $\overline U$ and holomorphic on $U$), but I can't see a way to relate this statement about general bounded domains and convex hulls.

I could be looking at these entirely wrong. If I need to clarify my thoughts or say more, please let me know, and again, I definitely don't want more than little hints. Thanks all.

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up vote 3 down vote accepted

It seems to me that your intuition of zeroes wandering off towards the boundary is the correct one and you don't have to look for nastiness of the behavior on the boundary. In fact, Hurwitz's theorem can be stated as follows:

Let $G$ be open and connected and let $f_{n},f: G \to \mathbb{C}$ be holomorphic such that $f_{n} \to f$ uniformly on compact sets. Assume that $U \subset G$ is bounded and open, that $\overline{U} \subset G$ and that $f$ has no zeroes on $\partial U$ then there exists an index $n_{0} = n_{0}(U)$ such that $f_{n}$ and $f$ have the same number of zeroes (counted with multiplicity) in $\overline{U}$ for all $n \geq n_{0}$.

If you want to produce a sequence of functions with a zero wandering towards $\zeta \in \partial D$, simply put $f_{n} = (z - (1 - \frac{1}{n})\zeta)g$ where $g$ is an arbitrary non-constant holomorphic function on $D$.

Concerning the second question(*), I don't quite understand why one would want to use the maximum principle. It can be done completely explicitly:

The logarithmic derivative of a polynomial $f$ of degree $n$ is \[ \frac{f'(z)}{f(z)} = \frac{1}{z-z_{1}} + \cdots + \frac{1}{z-z_{n}} = \overline{\sum_{k=1}^{n} \frac{z - z_{k}}{\vert z-z_{k}\vert^2}} \] where the $z_{1},\ldots,z_{n}$ are the (not necessarily distinct) zeroes of $f$. This is easily proved by induction on the degree of $f$. Note that for $g(z) = (z-z_{0})f(z)$ we have $\frac{g'}{g} = \frac{1}{z-z_{0}} + \frac{f'}{f}$.

If $f'(c) = 0$ we want to show that $c$ is in the convex hull of the $z_{k}$. Now if $c \in \{z_{1},\ldots,z_{n}\}$ this is trivial, so we may assume that this is not the case. But $0 = \overline{\frac{f'(c)}{f(c)}}$ then gives \[ \left(\sum_{k=1}^{n} \frac{1}{|c-z_{k}|^2} \right) \cdot c = \sum_{k=1}^{n} \frac{1}{|c-z_{k}|^2} z_{k} \] and this easily gives $c$ as a convex combination of the $z_{k}$. Recall that a convex combination is a sum of the form $\sum_{k=1}^{n} \lambda_{k} z_{k}$ with $\lambda_{k} \geq 0$ and $\sum \lambda_{k}= 1$.

(*) Much Later: This is called the Gauß–Lucas theorem.

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Whoa I completely missed the word "polynomial" in the second one. These are good hints, thank you, I think I can proceed now. – leif Mar 15 '11 at 2:18
@leif: You're welcome. I'm slightly puzzled about the title of your question: Three basic complex... I must have missed the third one. – t.b. Mar 15 '11 at 2:27
oh I decided I could figure out one of them while I was typing it up, I should fix that :) – leif Mar 15 '11 at 3:53
Thanks again! All problems resolved. – leif Mar 15 '11 at 5:07

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