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Does it exist a holomorphic function on $D(0,1)$ such that for all $z\in D(0,1/2)$ $f(2z)=f(z)$

Iterating we have $f(1/2)=f(2\cdot\frac1{4})=f(\frac1{4})=\cdots=f(\frac1{2^n}).$

By continuity we must have $f(\frac1{2})=f(0)$ so that when I write $f(z)=\sum_{k=0}^\infty a_kz^k,$ we have:

$$a_0+\frac{a_1}{2}+\frac{a_2}{4}+\cdots+\frac{a_n}{2^n}+\cdots=a_0$$ so that $a_k=0$ for all $k\ge 1$, so f must be constant egual to $a_0.$

Is that correct ?

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    A series can converge to $0$ without the terms all being zero. – 2017-02-15
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    A more elegant way would be to show that the function is equal to $a_0$ on a set with a limit point, therefore it must be equal to $a_0$ everywhere. You still haven't shown that the series $\frac{a_1}{2}+\frac{a_2}{4}+\dots$ being zero implies that all $a_i$ are also zero. – 2017-02-15
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    @JonasMeyer I don't understand your comment, I must have $a_0=a_0+a_1/2+\cdots$ by unicity it implies that all $a_k=0$ for $k\ge 2$, isn't it ? – 2017-02-15
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    @JonasMeyer ah I see your poi,t! – 2017-02-15
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    @5xum thanks! How can I show that the function is equal to a_0 on a set with a limit point ? – 2017-02-15
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    @Alex You already did. – 2017-02-15
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    @Alex Also, https://en.wikipedia.org/wiki/Identity_theorem – 2017-02-15
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    You can use the identity theorem just from the sequence you gave, but you can also show just using continuity that $f$ is constant on $D(0,\frac12)$. But you'd still need to use the identity theorem (with a weaker hypothesis) or the power series at $0$ to extend the result to $D(0,1)$. See http://math.stackexchange.com/questions/666857/let-f-mathbbc-rightarrow-mathbbc-be-a-continuous-function-and-assume – 2017-02-15
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    @5xum hum so $f(z)=a_0$ on the set where the series $\frac{a_1}{2}+\frac{a_2}{4}+\cdots$ and such a set contains a limit point (I must prove that I know) and so by identity theorem, $f(z)=a_0$ everywhere? – 2017-02-15
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    @JonasMeyer Ok, I see! Thank you. – 2017-02-15
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    @Alex Just... forget the series. – 2017-02-15

3 Answers 3

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Directly from the power series: $$\sum_{n=0}^\infty a_nz^n = \sum_{n=0}^\infty a_n(2z)^n = \sum_{n=0}^\infty 2^na_nz^n.$$ By uniqueness of coefficients, $$\forall n\in\Bbb N:\ a_n = 2^n a_n,$$ and this implies $a_n = 0$ for $n\ge 1$.

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From $f(2z)=f(z)$ for $|z|<1/2$, we see , by induction:

$2^n f^{(n)}(2z)=f^{(n)}(z)$ for $|z|<1/2$ and for $ n \ge 1$. It follows that $f^{(n)}(0)=0$ for $ n \ge 1$.

The power series expansion of $f$ around $0$ shows now that $f$ is constant in $D(0,1/2)$.

By the identity theorem, $f$ is constant 0n $D(0,1)$.

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As you observed: $f(1/2^n) = f(0), n = 1,2, \dots$ Thus $f,$ which is holomorphic on $D(0,1),$ equals $f(0)$ on a set in $D(0,1)$ with limit point in $D(0,1).$ By the identity principle, $f\equiv f(0)$ in $D(0,1).$