How to show that a continuous function analytic in a deleted neighbourhood is analytic in the entire neighbourhood?

Question

Suppose $$ F(z)= \begin{cases} \dfrac{f(z)-f(a)}{z-a}, & z\ne a\\[2ex] \quad f'(a), & z=a \end{cases} $$ Here $f$ is analytic on a simply connected domain $D$ containing $a$. Clearly, $F$ is analytic on $D\setminus\{a\}$ since it is a composition of analytic functions. I want to show that $F$ is analytic on $D$.

I tried to prove it using the definition. To show analyticity on $D$, we can show differentiability at $a$. So I tried evaluating the limit: $$ \lim_{z\to a} \dfrac{F(z)-F(a)}{z-a} $$

Substituting $F(z)$, $$ \lim_{z\to a} \dfrac{\dfrac{f(z)-f(a)}{z-a}-f'(a)}{z-a} $$

Since $f$ is analytic, $\lim_{z\to a}\dfrac{f(z)-f(a)}{z-a}=f'(a)$. It's been a long time since I evaluated limits but I know that I can't just substitute this in the numerator since I need to account for the denominator as well. Any help would be appreciated.

Answer 1

If the Taylor series of $f$ centered at $a$ is $\sum_{n=0}^\infty a_n(z-a)^n$, then $a_0=f(a)$ and therefore near $a$ you have$$F(z)=\frac{f(z)-f(a)}{z-a}=\sum_{n=1}^\infty a_n(z-a)^{n-1}=\sum_{n=0}^\infty a_{n+1}(z-a)^n.$$

Answer 2

Since $f$ is analytic on $D$, we have for $z\in D$ and $a\in D$

$$\begin{align} f(z)-f(a)-f'(a)(z-a)&=\int_a^z\int_a^{z'} f''(w)\,dw\,dz'\\\\ &=\int_a^z f''(w)(z-w)\,dw\tag1 \end{align}$$

where the integrals in $(1)$ are taken along rectifiable curves that lie in $D$.

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Next, dividing $(1)$ by $(z-a)^2$ and letting $z\to a$ reveals

$$\begin{align} \lim_{z\to a}\frac{f(z)-f(a)-f'(a)(z-a)}{(z-a)^2}&=\lim_{z\to a}\frac1{(z-a)^2}\int_a^z f''(w)(z-w)\,dw\\\\ &=f''(a)\lim_{z\to a}\frac1{(z-a)^2}\int_a^z (z-w)\,dw\\\\ &+\lim_{z\to a}\frac1{(z-a)^2}\int_a^z (f''(w)-f''(a))(z-w)\,dw\\\\ &=\frac12 f''(a)\\\\ &+\lim_{z\to a}\frac1{(z-a)^2}\int_a^z (f''(w)-f''(a))(z-w)\,dw\tag2 \end{align}$$

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We will now show that the limit on the right-hand side of $(2)$ is $0$.

First, since $f''$ is continuous, for any given $\varepsilon>0$, there exists a number $0<|z-a|<\delta$ such that $|f''(z)-f''(a)|<\varepsilon$ whenever $|z-a|<\delta$.

We also take $\delta$ so small that we can parameterize $w\in D$ as $w=a+(z-a)t$, $t\in[0,1]$. Then, we have the estimates for $|z-a|<\delta$

$$\begin{align} \left|\int_a^z (f''(w)-f''(a))(z-w)\,dw\right|&\le \int_0^1 |f''(a+(z-a)t)-f''(a)| (1-t)|z-a|^2\,dt\\\\ &=\frac{\varepsilon}2|z-a|^2 \end{align}$$

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Putting it all together yields the coveted result

$$\lim_{z\to a}\frac{F(z)-F(a)}{z-a}=\frac12 f''(a)$$

from which we find that $F$ is analytic on $D$. And we are done!

Answer 3

1. Since $F$ is analytic in $D \setminus \{a\}$, the function $F$ has an isolated singularity at $a$.

2. From $ \lim_{z \to a}F(z)=f'(a)$ we see that the singularity at $a$ is a removabvle singularity (Riemann !).