Suppose that $f$ has an isolated singularity at $0$ and satisfies $|f(z)|\leq \frac{1}{\sqrt{|z|}}$ in some punctured neighborhood of $0$. Show that $f$ has a removable singularity at $0$.
I tried defining a function $g =f(z)\sqrt{|z|}$. From here we can see that $$0 \leq |g(z)|=|f(z)\sqrt{|z|}|\leq |f(z)|z|| \leq |\sqrt{|z|}| $$ and by squeeze theorem it is clear that $g(z)$ would go to $0$ and by Riemann $g(z)$ has a removable singularity at $0$. Since $g(z)$ has a removable singularity at $0$ so does $f$ and we are done, or am I missing something here?
Best Answer
You cannot apply Riemann's theorem on removable singularities to $f(z)\sqrt{|z|}$ because that function is not holomorphic in a punctured neighbourhood of $0$.
But you can apply the theorem to $f$ directly: $\lim_{z \to 0} z f(z) = 0$ implies that $f$ has a removable singularity at $z=0$.