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[SOLVED] Analytic on D

dwsmith

Well-known member
Feb 1, 2012
1,673
If $f$ is analytic on the disc D and for each $a\in D$, the power series of $f$ expanded at
a has at least one coefficient equal to zero, then f is a polynomial on D.

I am at a loss here.
 

Jose27

New member
Feb 1, 2012
15
Take \(D'\subset D\) any closed subdisk, and consider the sets \(A_n=\{ x\in D' : f^{(n)}(x)=0 \}\). Prove that one of these, say \(A_k\), has an accumulation point in \(D\), what can you say about \(f^{(k)}\)?
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
Take \(D'\subset D\) any closed subdisk, and consider the sets \(A_n=\{ x\in D' : f^{(n)}(x)=0 \}\). Prove that one of these, say \(A_k\), has an accumulation point in \(D\), what can you say about \(f^{(k)}\)?
Since D' is bounded and closed, by the Heine-Borel Theorem, D' is compact. In D', $A_k$ would have an accumulation point. How can I extended that into D? How does that help with showing $f$ is a polynomial?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So $\displaystyle\bigcup A_n = D'$ and at least one $A_n$ is infinite. Let $A_m$ be infinite.
We have a sequence $a_m\in A_m$ in $D'$. By the Heine-Borel Theorem, $D'$ is compact and has a convergent subsequence of $a_m$. Therefore, $f^{(m)} = 0$ and $f$ is a polynomial.

Is this good?
 

Jose27

New member
Feb 1, 2012
15
So $\displaystyle\bigcup A_n = D'$ and at least one $A_n$ is infinite. Let $A_m$ be infinite.
We have a sequence $a_m\in A_m$ in $D'$. By the Heine-Borel Theorem, $D'$ is compact and has a convergent subsequence of $a_m$. Therefore, $f^{(m)} = 0$ and $f$ is a polynomial.

Is this good?
As long as you know why each of your claims is valid then yes, everything's fine.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
As long as you know why each of your claims is valid then yes, everything's fine.
I think I am unsure of is $f^{m} = 0$ and $f$ is a polynomial. Can you explain why that is the case?
 

Jose27

New member
Feb 1, 2012
15
Look up the identity theorem. For the rest, surely you can argue that if $f^{(m)}\equiv 0$ then $f$ is a polynomial.