# Prove that if $f : D(0,1) \to D(0,1)$ is analytic with $f(0) = 0$, then $\frac{f(z)}{z}$ has a removable singularity at 0

Prove that if $f : D(0,1) \to D(0,1)$ is analytic with $f(0) = 0$, then $g(z) = \frac{f(z)}{z}$ has a removable singularity at 0.

My thoughts so far:

Is this even a question? If $f$ is analytic, then we can write $f(z) = \sum_{n=0}^\infty a_n z^n$ valid for all $z \in D(0,1)$. Then $f(0) = 0$ gives that $a_0 = 0$. Then we can write a Laurent series $g(z) = \frac{a_0}{z} + \sum_{n = 0}^\infty a_{n+1} z^n$, and as $a_0 = 0$, the point 0 is necessarily a removable singularity (and it's already been removed by setting $f(0) = 0$!) Am I missing something?

Thanks

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I don't think your missing anything. By the way, I can't see any point in requiring $f$ to take values in $D(0,1)$. Also, a related problem is to show that if $f : \mathbb{R} \to \mathbb{C}$ is $\mathcal{C}^{\infty}$ with $f(0) = 0$, then $x \mapsto f(x)/x$ is also $\mathcal{C}^{\infty}$ . – Joel Cohen May 8 '11 at 0:29

Yes, that's one way to see it. In general, if $f$ has a zero of order $m$ at $a$, then $f(z)=(z-a)^mg(z)$ for some analytic function $g$ with $g(a)\neq 0$, and $g$ is the analytic extension of $\frac{f(z)}{(z-a)^m}$ whose domain includes $a$. The statement that $f$ maps into the unit disk is irrelevant.

In this case, you could also use the definition of the derivative to observe that $\displaystyle{\lim_{z\to 0}g(z)=f'(0)}$.

You write, "it's already been removed." That is not quite accurate, but this is just a technicality due to the fact that $\frac{f(z)}{z}$ can't be evaluated directly at $0$; you would get $\frac{0}{0}$. It is removable because there is a limit at $0$, and defining $g(0)$ to be that limit gives the unique analytic extension to all of $D(0,1)$.

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Thanks. I'm looking at a solution to part of a follow up question: is it obvious that $|f(z)| < 1$? – user938272 May 8 '11 at 0:40
@user938272: Do you mean $|g(z)|<1$? If so, it is true (unless $f(z)=z$) but requires proof. (That would explain why the codomain of $f$ was stated to be $D(0,1)$.) It's a good idea to know how to prove it before calling it "obvious". If you really mean $f$, then yes it is obvious because it is part of the hypothesis. – Jonas Meyer May 8 '11 at 0:44
I did mean $f$, and so am an idiot. Thank you! – user938272 May 8 '11 at 0:46

Let $g(z)= \frac{f(z)}{z}$. By a known result we can say $0$ is a removable singularity for $g(z)$ if and only if $\lim_{z\rightarrow0} g(z)$ finite. Here we have $\lim_{z\rightarrow0} g(z)=f'(0)$ and is finite as f is analytic at zero, so $0$ is a removable singularity of $\,g$ .

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I think what you are missing is that, basically, $f(z) =1$ and $g(z)=z/z$ are not the same functions ("though" they agree everywhere except at $0$). $g$ is not defined at $0$ because its value must be evaluated as $0/0$ at $0$ which is not defined.

However, clearly, $g$ has a removable singularity at $0$ and its analytic extension is $f$.

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The function $f(z)/z$ is holomorphic on the punctured unit ball and so the singularity at $0$ is either removable, a pole of essential. It can't be essential by Big Picard, and it can't be a pole because the Schwarz lemma gives that $|f(z)| \leq |z|$ for all $z \in D$. Thus the singularity is removable.

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