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Let $ f(z) $ be an entire function satisfying $f(2z)=f(z)^{2} \text{ for all } z \in \Bbb C \ .$

I have the following question: Suppose there exist a $z_0 \in \Bbb C\ $ such that $f(z_0)=0.$ Prove that $f$ is identically zero.

I got the hint: Show that the origin 0 must be one of the zeros of $f$. Check the mutiplicilty of $f$ at z=0. See if $f$ has finite mutiplicilty at z=0.

It is obvious that $f(0)(f(0)-1)=0$ and it has finite mutiplicilty but how does it imply that $f$ is identically zero?

I saw others using Identity theorem to prove it in All entire functions which satisfying : $f(2z)=f(z)^{2}$. However, it doesn't solve my problem.

Felix
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    I'm not understanding how the identity theorem doesn't solve your question. Are you interested in the particular case the zero happens at $z=0$? – hellofriends Dec 20 '23 at 21:30
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    If $f(0)=0$ and $f$ is non-constant then its zeros are isolated so apply the Argument Principle to a very small closed curve around $0$ to generate a contradiction. – user8675309 Dec 20 '23 at 21:33
  • @hellofriends I am interested in how multiplicity of $f$ helps in this question. – Felix Dec 20 '23 at 22:22

2 Answers2

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If $f(z_0) = 0$ then $0 = f(2 \cdot \frac{z_0}{2}) = f(\frac{z_0}{2})^2$ implying $f(\frac{z_0}{2}) = 0$. This repeats so that $f(\frac{z_0}{2^n}) = 0$ for any $n \in \mathbb{N}$. By continuity, this implies $f(0) = 0$ which is perhaps how you have shown that 0 is also a root. Regardless of this, its clear that the zero set has an accumulation point, namely $\frac{z_0}{2^n}$ tend to 0, so the Identity theorem implies $f$ must be identically 0.

Of course, this doesn't work if we start with $z_0 = 0$. In this case, we know automatically that $f(0) = 0$. Since $f$ is analytic, $f(z) = 0 + a_1 z + ... = z h(z)$ for some $h$. We want to show that $h(0) = 0$ so then $f(z) = z^2 h_1(z)$. We can write $2zh(2z) = f(2z) = f(z)^2 = z^2 h(z)^2$ and dividing by $z$ see that indeed 0 is a zero of $h$. So then $f(z) = z^2 h_1(z)$, etc... By some induction argument, you can show that we keep getting $f(z) = z^k h(z)$, i.e., $f^{(n)}(0) = 0$, which then implies all the coefficients $a_i$ of $f$ must be 0. So actually, this last paragraph shows you don't need the Identity theorem at all since you know $f(0) = 0$ in either case.

JZweifler
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  • May I ask which part is related to the multiplicity of $f$ at z=0? I just can’t understand how multiplicity help. – Felix Dec 20 '23 at 22:17
  • @Felix $f$ has a zero at $0$ of infinite multiplicity. – Gary Dec 21 '23 at 01:14
  • @Gary $f$ has a zero at $0$ of infinite multiplicity means the zeros of $f$ acccumulate at $z=0$? If yes, then how can we conclude that $f$ is identically zero from it? – Felix Dec 21 '23 at 08:32
  • @Felix No, that is not what it means. Read the complex analysis section of https://en.m.wikipedia.org/wiki/Multiplicity_(mathematics) – Gary Dec 21 '23 at 08:48
  • @Gary I have read but I still don't understand how it helps in this question. – Felix Dec 21 '23 at 13:04
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    @Felix Showing that all the coefficients $a_i$ are 0, is just another way of saying that $0$ has infinite multiplicity as a root. If you assume $0$ has finite multiplicity then $f(z) = z^k h(z)$ where $h(0) \neq 0$ (pull out all the finitely many factors of $z$, ensuring $h(0) \neq 0$). But by above, we can keep pulling out more factors so 0 must have infinite multiplicity. – JZweifler Dec 21 '23 at 22:54
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To begin with, as pointed out in your link, $f(z_0)=0$ implies $f(z_0/2)=0$, and by induction and continuity, $f(0)=0$.

On the other hand, the functional equation provides $$ \left|f\left(\frac {z} {2}\right)\right| = \sqrt{|f(z)|},$$ and by induction on $n\in\mathbb{N}$, $$ \left|f\left(\frac z {2^n}\right)\right| = |f(z)|^{1/2^n}.$$ If $f(z)\neq 0$ for some $z$, this yields $|f(0)|=1$ by letting $n$ going to $+\infty$, hence a contradiction.

The only regularity assumption you need is the continuity of $f$ at $0$.

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