Characterizations of removable singularities

If $a$ is removable, let $F$ be a holomorphic extension to $\mathrm{\mid }z-a\mathrm{\mid }<\epsilon$; by [L3], $F$ is continuous at $a$, so $F$ is bounded on some smaller disc, and hence $f$ is bounded on the corresponding punctured disc.

If $f$ has a finite limit at $a$, then $f$ is bounded on some punctured neighbourhood of $a$.

Suppose $\mathrm{\mid }f(z)\mathrm{\mid }\le M$ whenever $0<\mathrm{\mid }z-a\mathrm{\mid }<\epsilon$. For $m\ge 1$ and $0<\rho <\epsilon$, the coefficient formula gives $\mathrm{\mid }{c}_{-m}\mathrm{\mid }\le \frac{1}{2\pi }{\int }_{\mathrm{\mid }\zeta -a\mathrm{\mid }=\rho }\mathrm{\mid }f(\zeta )\mathrm{\mid }\mathrm{\mid }\zeta -a{\mathrm{\mid }}^{m-1}\mathrm{\mid }d\zeta \mathrm{\mid }\le M{\rho }^m\mathrm{.}$

If the principal part is $0$, then $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ on the punctured disc, and [L1] makes this regular part holomorphic on $\mathrm{\mid }z-a\mathrm{\mid }<R$; defining $F(a)=c_0$ therefore extends $f$ holomorphically across $a$, so the singularity is removable.

Since step 1.3 holds for every sufficiently small $\rho >0$, letting $\rho \to 0$ gives $c_{-m}=0$ for every $m\ge 1$; so the principal part is $0$.

The extension from step 1.4 is continuous at $a$ by [L3], so $f(z)\to F(a)=c_0$ and, multiplying by $z-a$, one gets $(z-a)f(z)\to 0$.

Suppose $(z-a)f(z)\to 0$, and put $g(z):=(z-a)f(z)$ on the punctured disc. Then $g$ is holomorphic there and bounded near $a$, so the argument of steps 1.3 and 2.1 applied to the Laurent expansion $g(z)={\sum }_{n\in \mathbb{Z}}{c}_n(z-a{)}^{n+1}$ gives $c_{-m}=0$ for every $m\ge 2$.

With the coefficients from step 3.1 gone, $g(z)=c_{-1}+{\sum }_{n\ge 0}{c}_n(z-a{)}^{n+1}$, and [L1] makes the tail a holomorphic function vanishing at $a$; the hypothesis $g(z)\to 0$ therefore forces $c_{-1}=0$. So the whole principal part of $f$ is $0$.

Step 1.1 proves $1\Rightarrow 3$, step 1.2 proves $4\Rightarrow 3$, steps 1.3 and 2.1 prove $3\Rightarrow 2$, step 1.4 proves $2\Rightarrow 1$, step 2.2 proves $2\Rightarrow 4$ and $2\Rightarrow 5$, and steps 3.1 and 4.1 prove $5\Rightarrow 2$; therefore all five conditions are equivalent, and the extension value is the finite limit from step 2.2.