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I was reading about this problem yesterday:

Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is zero. Show that $f$ is a polynomial.

I was wondering, if instead of

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function. Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is zero. Show that $f$ is a polynomial.

we ask

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a holomorphic function. Suppose that for each $a\in \mathbb{C}$ at least one coefficient of the Taylor's series $f$ about $a$ is real. Then, is $f$ still a polynomial?

It seems that I have fall into a logic mistake in approaching it:

For $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$. Still, the coefficient for the series at $n$ is $c_n=\frac{f^{(n)}(a)}{n!}=r$, for $n$ is the index and $r$ is the real coefficient. Does this means the coefficient $c_{n+1}$ should be $c_{n+1}$ because $f^{(n)}(a)=n! r$ seems like a constant? And then everything back to the "zero" statement.

Am I right or wrong? If wrong, is there any other approaches to the new question?

Zizheng Yang
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1 Answers1

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Your approach does not work. For each $a \in \Bbb C$ there is an index $n$ and a real number $r$ such that $\frac{f^{(n)}(a)}{n!} = r$, but that real number depends on $a$.

But one can mimic this proof and define the sets $$ A_{n}=\left\{z\in\mathbb{C}:f^{(n)}(z) \in \Bbb R\right\} \, . $$ All $A_n$ are closed, and their union is equal to $\Bbb C$. By the Baire category theorem, one of the sets must have a non-empty interior, i.e. there is an open disk $U$ and an index $k$ such that $U \subset A_k$.

Then use the open mapping theorem to conclude that $f^{(k)}$ is constant on $U$ and therefore constant in $\Bbb C$. It follows that $f$ is a polynomial.

Alternatively one can proceed as suggested by Conrad: With the

Lemma: Let $g: \Bbb C \to \Bbb C$ be a non-constant holomorphic function. Then the set $A = \{ z \in \Bbb C : g(z) \in \Bbb R\}$ has Lebesgue measure zero.

we can argue as follows: If $f$ is not a polynomial then all derivatives $f^{(k)}$ are non-constant, so that all sets $A_n$ have the Lebesgue measure zero. Then the countable union $\bigcup_n A_n$ has Lebesgue measure zero as well. That is a contradiction to the assumption that $\bigcup_n A_n = \Bbb C$.

Remark: The Baire category theorem is not used in this argument.

Proof of the Lemma: First assume that $g(z_0) = a \in \Bbb R$. Then $$ g(z) = a + (h(z))^n $$ in a neighborhood $V$ of $z=z_0$, where $n$ is the multiplicity of $g$ at $z=z_0$, and $h$ is an injective mapping from $V$ to a disk $B_r(a)$. It follows that $$ \{ z \in V : g(z) \in \Bbb R \} $$ is the union of $n$ analytic curves, and therefore has Lebesgue measure zero.

We have thus shown: Every $z\in A$ has a neighborhood $V_z$ such that $V_z \cap A$ has Lebesgue measure zero.

Now one can conclude that all compact sets $A \cap \overline{B_R(0)}$ have Lebesgue measure zero (because they can be covered with finitely many sets $V_z$).

Finally conclude that $A$ has Lebesgue measure zero.

Martin R
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    without Baire theorem, one notices that if $g(z_0)=b_0, b_0 \in \mathbb R$ analytic (non constant); for small enough $r, g$ behaves like $b_0+c(z-z_0)^n, n \ge 1$ on $D(z_0, r)$ where $n$ is the order of $g-b_0$ at $z_0$ so $g^{-1}([b_0-b_1, b_0+b_1]) \cap D(z_0, r)$ is a union of at most $n$ analytic segments so has measure zero in the plane; but now there is an $m$ st $A_m$ as above has non zero measure and then there is $b_0 \in A_m$ st all disks centered at it intersected with $A_m$ have non zero measure; the above argument applied to $g=f^{(m)}$ gives the required contradiction – Conrad Sep 09 '21 at 15:45
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    @Conrad: Yes, that is also a good argument. – Martin R Sep 11 '21 at 19:33
  • @Conrad, I am not very clear, what is the required contradiction by the way? I didn't see any assumptions. Sorry... – Zizheng Yang Sep 11 '21 at 21:08
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    We know that $g$ sends a set of non zero measure to the real axis in any neighborhood of $z_0$ (where $g$ is a derivative of $f$ as above), but the preimage of the real axis there is a finite union of analytic arcs so has zero measure – Conrad Sep 11 '21 at 21:12
  • @MartinR, just want to make sure, I understand open mapping theorem tells us that $f^{(k)}$ maps to open set in $\mathbb{C}$, but why it is constant mapping? – Zizheng Yang Sep 11 '21 at 21:12
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    @ZizhengYang: Some set $A_k$ contains an open disk $U$, i.e. $f^{(k)}(U) \subset \Bbb R$. But $\Bbb R$ has no interior points, so $f^{(k)}(U)$ cannot be an open set, which means that $f^{(k)}$ must be constant. – Martin R Sep 11 '21 at 21:16
  • @MartinR, thank you very much for the explanation, I think I fully understand your statement. It's just, can you explain me about the "without BCT" statement, I still don't understand that part very well. – Zizheng Yang Sep 11 '21 at 21:43
  • @MartinR, thank you very much it is really helpful. I am thinking about this problem today, also. I recall that Liouville's Theorem stated that any bounded entire function are constant. Is there way to show there exists some $N$ such that for all $k>N$, $f^{(k)}(z)$ are bounded? I think I will open anther question. – Zizheng Yang Sep 12 '21 at 22:06