Let $f : A(a, 0, 1) → \Bbb C$ be an analytic function, where $A(a, 0, 1) = \{z : 0 < |z − a| < 1\}$. Prove that $f$ has an essential singularity at a if and only if $\lim_{z→a} f(z)$ does not exist as an element of $\hat{\Bbb C} = \Bbb C ∪ \{∞\}$.

I am trying to solve the above. I attempted the problem below but I believe it is a little shaky at parts, please let me know if it is sufficient or how to fix it:


Suppose $f$ has an essential singularity at $a$. We know by Casorati -Weierstrass that given a punctured disk $\Delta := \{z : 0< |z-a| < R\}$, there is a $z \in \Delta$ such that $\forall w \in \Bbb C, \forall \epsilon_1 >0$, $|w-f(z)| < \epsilon_1$. Now suppose $\lim_{z→a} f(z) = c \in {\Bbb C}$. Then $\forall \epsilon_2,\ \exists \delta$ such that $|z-a|< \delta \implies |f(z) - c| < \epsilon_2$. But by choosing $R< \delta$, we have a contradiction as we can choose a $w$ for which we require $|w - f(z)|< \epsilon_1,\ z \in \Delta$, and $|w-c|>\epsilon_2+\epsilon_1$. Thus, $\lim_{z→a} f(z)$ does not converge to an element of $\Bbb C$ nor $\{\infty \}$, because then $a$ would be a pole.

We can prove the converse by observing first that the limit's nonexistence in the extended complex plane implies that $f$ has a singularity as $a$. The limit not approaching $\infty$ means $a$ cannot be a pole. Now assume the singularity is removable, we can then define a function $g(z)$ such that it is analytic on an open ball $B_{\epsilon}(a)$ and $f=g$ on $B_{\epsilon}(a) \backslash \{a\} $, but analyticity of $g$ means it must be continuous at $a$. This is impossible as $g$ must be equal to $f$ in the neighborhood of $a$.


You don't need Casorati-Weierstrass - the very definition is enough: If the limit exists and is finite, then we have a removable singularity. If the limit exists and is $\infty$, then $\frac1{f(z)}$ has a removable singularity, which happens to be a zero of some order. Then $f$ has a pole of the same order.

  • $\begingroup$ Is that "by definition" true for removable singularities? And other than that is it sufficient? $\endgroup$ – Meecolm Nov 3 '15 at 22:18

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