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I was stuck on the following problem:

Let $f$ be analytic on $D = \{z \in \Bbb{C} : |z| < 1\}$ and $f(0) = 0$.

Define $g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$

Then which of the following option(s) is/are correct?

  1. $g$ is discontinuous at $z = 0$ for all $f$.

  2. $g$ is continuous, but not analytic, at $z = 0$ for all $f$.

  3. $g$ is analytic at $z = 0$ for all $f$.

  4. $g$ is analytic at $z = 0$ only if $f'(0) = 0$.

Can someone point me in the right direction with some explanation? Thanks in advance for your time.

EDIT: I have posted an answer .Feel free to comment if I missed anything in my answer.

  • 5
    $f$ is analytic in some neighbourhood of $0$, so you can write it as a power series in $z$. This should make it pretty obvious which of those statements are true, and which not.2012-11-19

1 Answers 1

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We have, $g(z) = \begin{cases} \displaystyle \frac{f(z)}{z} & z \neq 0, \\ f'(0) & z = 0. \end{cases}$.

Since $f$ is analytic in some neighbourhood of $0$, so I can write it as a power series in $z$. So,$f(z)=a_0+a_1z+ \cdots \cdots +a_nz^n $ for $|z|<1$. Now,$f(0)=0$ gives $a_0=0.$ Hence $ \begin{align}f(z) &=a_1z+ \cdots \cdots +a_nz^n \\ \implies {f(z) \over z} &=g(z) =a_1+a_2z+ \cdots +a_nz^{n-1} \\ \lim_{z \to 0} g(z) &= a_1=f'(0)=g(0)\\ \text{Since},f'(z) &=a_1+ \cdots +na_nz^{n-1} \implies f'(0)=a_1\end {align}$ Hence $g$ is continuous at $z=0.$ This eliminates option 1.

Also,we notice $\lim_{z \to 0}{g(z)-g(0) \over z-0}=a_2$, which exists for all $f$. This eliminates options 2 and 4.

So, option 3 is the right choice.

In this context Riemann's Removable Singularity Theorem is useful.