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Let $f \in H(\Omega)$, where $\Omega \subset \mathbb{C}$ open. Suppose that for all $z_0 \in \Omega$ the Taylor series associated to $f$ $$\sum_{n=0}^\infty a_n(z-z_0)^n$$ has one null coefficient $a_N$. How do I prove that $f$ is then a polynomial (without applying Baire's category theorem)?

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    This question is similar: http://math.stackexchange.com/questions/1959597/show-that-fn0-0-for-infinitely-many-n-ge-0 – 2017-01-05
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    If $X_i$ is the set of points for which the $i$'th coefficient is $0$, then they are all closed and their union is $\Omega$. By the Baire Category Theorem, one must have a nonempty interior, so that the $i$'th derivative is $0$ in this interior, and hence everywhere in the domain. – 2017-01-05
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    @BobJones I'm looking for a solution that does not involve Baire's theorem. – 2017-01-05
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    @MartinR The difference is that I need a more hands-on solution that does not apply Baire's theorem. – 2017-01-06
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    @Axel: The question that I linked to has an answer http://math.stackexchange.com/a/995129/42969 which does not use Baire. – 2017-01-06

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There is a map $n:\Omega \to \{0,1,...\}$ such that $f^{(n(z))}(z) = 0$ for each $z \in \Omega$.

Note that $\Omega = \cup_k n^{-1}(k)$. Pick some $z_0$, $r>0$ such that $\overline{B}(z_0,r) \subset \Omega$. Note that $\overline{B}(z_0,r)$ is compact.

Since $\overline{B}(z_0,r)= \cup_k (\overline{B}(z_0,r) \cap n^{-1}(k))$ and $\overline{B}(z_0,r)$ is uncountable, then there is some $k$ such that $\overline{B}(z_0,r) \cap n^{-1}(k)$ is at least countable.

Hence there is a sequence $z_n \in \overline{B}(z_0,r)$ such that $f^{(k)}(z_n) = 0$, and since $\overline{B}(z_0,r)$ is compact, we can assume, by taking a subsequence if necessary, that $z_n \to z^*$ for some $z^* \in \Omega$.

A standard result then shows that $f^{(k)}(z) = 0$ for all $z \in B(z_0,r)$. Hence the Taylor series expansion of $f$ shows that $f(z) = p(z)$ for some (entire) polynomial $p$ of degree $\partial p

Since $\Omega $ is open and connected, the same standard result shows that $f(z) = p(z)$ for all $z \in \Omega$.

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    Do you mean uncountable number of points? – 2017-01-06
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    @Cat: No, one only needs a countable number of points to ensure that an accumulation point exists. I should have written "at least a...". – 2017-01-06
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    I understand that you only need countably infinite to get an accumulation point in the bounded set, but since we can cover $\Omega$ with countably many balls, if all such intersections contained only countably many points, then it would imply that $\Omega$ is countable. So indeed we can conclude that there are uncountably many points in the intersection, right? – 2017-01-06
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    @Cat: You can, but all we need for the conclusion is a countable number of points. – 2017-01-06
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    The second part of your answer isn't quite clear to me. Could you add some details? "Since $\Omega = \cup_k n^{-1}(k)$, we see that for some $k$, $z_0$ and $r>0$, we have $\overline{B}(z_0,r) \subset \Omega$ and $\overline{B}(z_0,r) \cap n^{-1}(k)$ contains at least a countable number of points, which must have an accumulation point." Why is that? "Hence $f^{(k)}(z) = 0$ and so $f$ is a polynomial of degree $$k$-th derivative imply that $f$ is a polynomial of degree $k$? – 2017-01-06
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    @Axel: I rewrote the latter part of my answer, hopefully it is less cryptic. – 2017-01-06
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  1. Show that there exists $n$ such that the set of all $z_0$ with $a_n = 0$ is uncountable.
  2. Show that this uncountable set has an accumulation point in $\Omega$.
  3. Use the identity theorem for holomorphic functions to conclude the result.

Tips:

  1. Look at the sets $\Omega_n := \{z_0 \in \Omega \mid a_n = 0\}$ and observe that $\bigcup_{n \in \mathbb{N}} \Omega_n = \Omega$. What do you know about countable unions of countable sets? Can all $\Omega_n$ be countable?

  2. Take an open ball $B$ in $\Omega$ whose boundary is in $\Omega$. What do you know about $\Omega_n \cap B$? What can you conclude about their cardinality? And what does this say about an accumulation point in $\Omega$?

  3. Let $g = f^{(n)}$. Then $g = 0$ by the identity theorem. What can you conclude for $f$?

I just saw, that 1 is not necessary to prove. You can skip it.

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    Actually it's not clear to me how to prove 1 and 2 and what's your aim with 3. – 2017-01-06
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    Added tips, but I saw, that 1 isn't even necessary to show. – 2017-01-06