8

Let $f$ be an entire function and let for each $a\in \mathbb R$, there exists at least one coefficient $c_n$ in $f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n$, which is zero. Then

  1. $f^{(n)}(0)=0$ for infinitely many $n\geq 0$

  2. $f^{(n)}(0)=0$ for every $n\geq 0$

  3. $f^{(2n+1)}(0)=0$ for every $n\geq 0$
  4. There exists $k\geq 0$ such that $f^{(n)}(0)=0$ for all $n\geq k$

We know that $c_n=\frac{f^{(n)}(a)}{n!}$ for all $n\in \{0,1,2\ldots\}$. Thus for $a=0$, $c_n=\frac{f^{(n)}(0)}{n!}$. By hypothersis, atleast one $c_n=0$. After that I could not do anything. Please help!

robjohn
  • 345,667
Anupam
  • 4,908
  • 1
    Something more I can add. Since $\mathbb R$ is uncountable and $\mathbb N \cup {0}$ is countable, therefore, there exists $n$ such that $c_n=\frac{f^{n}(a)}{n!}=0$ for uncountably many $a$. But a holomorphic function can not have uncountably many zeros. Is it OK? – Anupam Jun 10 '15 at 05:38
  • You should be able to work that into a solution. That is what I was in the process of using. Remember that any uncountable set in $\mathbb{R}$ contains an accumulation point. – robjohn Jun 10 '15 at 05:41
  • 1
    I think you can be a little more explicit. without much effort. You're idea is good. You know that $f^{(n)}(a) = 0$ for uncountably many $a$, but the $n$ could vary. So you have to use infinite pigeonhole to say that some $n_0$ appears unaccountably many times. Then $f^{(n_0)}$ is identically $0$. So we're happy. – Zach Stone Jun 10 '15 at 05:59
  • @ZachStone, what if say $c_0 = 0$, $c_i \ne 0$ for $i>0$? – Michael Galuza Jun 10 '15 at 06:46
  • I'm not entirely sure what you mean. Keep in mind $c_0$ really depends on the base point of the expansion. Let $c_0(a)$ be the $0$ coefficient in the $a$ expansion. Then if $c_0(a) $ is 0 for unaccountably a, then $f$ is identically $0$. So $c_i(a) =0$ for all $i$ and $a$. – Zach Stone Jun 10 '15 at 06:51
  • @ZachStone, oh, you're right, $a$ is not a constant. +1 to you) – Michael Galuza Jun 10 '15 at 06:56

1 Answers1

5

Let $A_n =\{z\in \mathbb{C} : f^{(n)} (z) =0 \}.$ By assumption $$\mathbb{R} =\bigcup_{n\in\mathbb{N}\cup \{0\}} A_n $$ and since $\mathbb{R} $ is uncountable at least one of the sets $A_n $ is uncoutable. Let $$k=\min\{ n\in\mathbb{N}\cup \{0\} :\mbox{card}(A_n ) >\aleph_0 \}$$ then the set $A_k$ is uncountable and therefore it has an accumulation point. Hence the $f^{(k)} (x) $ is identically zero on $\mathbb{C}$ and therefore $f$ is an polynomial with degree less or equal to $k.$

So 1. and 4. are true.