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Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function. And assume that at each point, one of it's derivatives vanishes.

What can you say about $f$?

A hint suggests that $f$ must be a polynomial.

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  • $\begingroup$ What are your thoughts on the problem? $\endgroup$ – Omnomnomnom Oct 28 '14 at 15:32
  • $\begingroup$ It sounds like Baire's catgeory theorem. If a derivative vanished on an open ball it's clear. $\endgroup$ – Curious Droid Oct 28 '14 at 15:35
  • $\begingroup$ Thinking again, I have the solution:) $\endgroup$ – Curious Droid Oct 28 '14 at 15:35
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Let $A_n$ be the set of points where $f^{(n)}(z)$ vanishes. Each $A_n$ is closed. By the Baire category theorem one of those sets must have a non empty interior which in turn implies that some derivative vanishes on an open ball. Therefore it is identically zero and $f$ is a polynomial.

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  • $\begingroup$ Sir, may be i am wrong. But i think what you had proved is already given. It is already given that, for some integer $n$ , $f^{n}(z)$ is identically zero. So Sir, what you proved? Kindly please Explain me if I am wrong..... $\endgroup$ – Akash Patalwanshi May 30 '18 at 13:14
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Let $n: [0,1] \to \{0,1,... \}$ be defined such that $n(x)$ is the smallest $k \ge 0$ such that $f^{(k)}(x) = 0$.

Since $[0,1]$ is uncountable, there must be some $k$ such that $Z=n^{-1} \{k\}$ has an infinite number of points. Since $[0,1]$ is compact, $Z$ has a limit point $p \in [0,1]$, and hence we have $f^{(k)}(z) = 0$ everywhere.

Elaboration: Let $p_n \in Z$ such that $p_n \to p$. Since $f^{(k)}(p_n) = 0$, the identify theorem (https://en.wikipedia.org/wiki/Identity_theorem) shows that $f^{(k)}(z) = 0$ everywhere from which it follows that $f$ is a polynomial (take the power series expansion around any point).

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  • $\begingroup$ Sir. Please explain, isn't it is already given that "at each point" some derivative of $f$ vanishes, that is for some integer $k$ , $f^k(z) =0$ everywhere. So why you proved it again? Sir please, Explain me if I am wrong $\endgroup$ – Akash Patalwanshi May 30 '18 at 13:21
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    $\begingroup$ @AkashPatalwanshi: The statement is that for any $z$ there is some $k$ such that $f^{(k)}(z) = 0$. What the above shows is that there is some $k$ such that $f^{(k)}(z) = 0$ for all $z$. The latter is a much stronger statement. $\endgroup$ – copper.hat May 30 '18 at 15:12
  • $\begingroup$ Sir, isn't in Question it was given that, "At all point" that means, for every $z$? Is am i wrong? $\endgroup$ – Akash Patalwanshi May 30 '18 at 15:25
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    $\begingroup$ @AkashPatalwanshi: The question just states that at each point, one of the derivatives vanishes. The $k$ may depend on the $z$. $\endgroup$ – copper.hat May 30 '18 at 15:29
  • $\begingroup$ Thank you so much sir, for giving your valuable time. $\endgroup$ – Akash Patalwanshi May 30 '18 at 15:37
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we can also solve this without using baire category theorem..

we can use the fact that zeros of non zero analytic function is countable. Note that union of $A_{n}$ is $\mathbb{C}$ thus we must have some fix derivative identically 0. Now use power series and analytic continuation.

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