# [SOLVED]Analytic on D

#### dwsmith

##### Well-known member
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
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
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
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
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
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
Look up the identity theorem. For the rest, surely you can argue that if $f^{(m)}\equiv 0$ then $f$ is a polynomial.