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$f(z_1 z_2) = f(z_1) f(z_2)$ for $z_1,z_2\in \mathbb{C}$ then $f(z) = z^k$ for some $k$

How can we characterize the analytic functions defined in the open unit disc $D\subset\mathbb{C}$ that satisfy $f(ab)=f(a)f(b)\text{ }$ for all $a,b\in D$.

What happens if we consider larger domains?

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marked as duplicate by Eric Naslund, t.b., Qiaochu Yuan Jul 18 '11 at 22:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is the standard procedure here? Should I delete this question, or leave it as a reference to the other? – RHP Jul 18 '11 at 22:13
You can't delete a question as soon as there are answers. You can choose an answer to accept, if you wish to do so or leave the question as it is. That it is closed simply means that no more answers can be added. – t.b. Jul 18 '11 at 22:16
up vote 4 down vote accepted

Differentiating with respect to $a$ yields $f^{(n)}(ab)b^n=f^{(n)}(a)f(b)$ for all $a$ and $b$ in $D$ and for each positive integer $n$. In particular, $f^{(n)}(0)b^n=f^{(n)}(0)f(b)$ for all $b$ in $D$ and $n\in\mathbb N$. If $f^{(n)}(0)=0$ for all $n>0$, then $f$ is constant, $0$ or $1$. If $f^{(n)}(0)\neq 0$ for some $n>0$, then $f(b)=b^n$ for all $b\in D$.

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Do you think that this is the "right" proof? It seems to me that analyticity is way too much to ask for a map satisfying $f(a)f(b) = f(ab)$ to be automatically of the required form. Shouldn't measurability already be enough? – t.b. Jul 18 '11 at 22:31
Theo, it's just a simple proof for a simple special case, I guess, and not suited for generalizing outside of analyticity; so no, it isn't "right". Perhaps the only measurable nonzero examples have the form $f(z)=z^k\overline{z}^j$ for nonnegative integers $k$ and $j$, but no proof comes to mind. – Jonas Meyer Jul 18 '11 at 23:55
Thanks! It reminds me a bit of standard automatic continuity situations, e.g. this, but I can't turn that idea into a proof (and it's way too late once more). – t.b. Jul 19 '11 at 0:02
Theo, oops, that guess is definitely false. Perhaps the only measurable nonzero examples have the form $f(z)=|z|^cz^k$ or $f(z)=|z|^c\overline z^k$ where c∈[0,∞) and k is a nonnegative integer, but I wouldn't call that an educated guess. – Jonas Meyer Jul 19 '11 at 1:45
@Theo: I asked a question about this. – Jonas Meyer Jul 19 '11 at 3:23

We have $f(0) = f(0)f(b)$ hence, if $f$ is not the constant function equal to $1$, we have $f(0) =0$. If $f$ is not the constant function equal to $0$, then we can find $k\in\mathbb{N}^*$ such that $f(z) = z^kg(z)$ with $g(0)\neq 0$ and $g$ analytic. We get for $a,b\neq 0$ that $g(a)g(b) =g(ab)$ and by continuity for all $a$ and $b$. We have $g(0)=g(0)g(b)$ hence $g(z)=1$ for all $z$. Finally the only solutions are $f(z)=0$, $f(z)=1$ and $f(z)=z^k, k\in\mathbb{N}$.

We only used the fact that the unit disc is connected; the result can be extended to each open connected subset of $\mathbb C$ which contains $0$.

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Suppose $f(0)\neq 0$. Then since $f(0)=f(0)f(b)$ for any $b$, we see that $f(b)=1$ for every $b$, so $f$ is identically the constant function. Suppose $f(0)=0$. Then $f(z)=z^m g(z)$ for some $g$ with $g(0)\neq 0$. Since $f$ and $z^m$ are multiplicative, so is $g(z)$. But by the first part, $g(z)=1$ for all $z$ and is identically the constant function. Hence $f(z)=z^m$.

Conclusion: $f(z)=z^m$ for some nonegative integer.

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