Let $α ∈ (0, 1)$ be a fixed exponent, and let $C > 0$
Denote as usual by $D$ the unit disc, and suppose that $f$ is holomorphic on $D$ \ {$0$} , and $|f(z)| ≤ C/|z|^α$
Show that $f$ extends holomorphically to $D$.
My attempt:
I already have tgat $f$ is holomorphic on $D$ \ {$0$} . So I still have to prove that $f$ is bounded on $D$ \ {$0$}, because if that's the case, I can apply Riemann's Extension Theorem and finally $f$ extends holomorphically to $D$
So, let $z=x+iy$ $\in$ $D$ \ {$0$} then $|z| = \sqrt{(x^2+y^2)} = 1$ because $D$ is the unit disc and thus $D$ = {$(x, y) \in \mathbb R^2 : x^2+y^2 = 1$}.
Hence $|z|^α = 1$ and finally $|f(z)| ≤ C/|z|^α = C$ and thus $f$ is bounded on $D$ \ {$0$}. Therefore, $f$ extends holomorphically to $D$
Is my attempt correct?
The function $f$ is not directly bounded, as $\lim_{z\to 0}\frac{C}{|z|^a}=+\infty$. However, we can mimic the proof of Riemann's Extension theorem (in some versions of the theorem this case is already considered, as @MartinR ntoted in the comments below):
We consider the function $g(z):=zf(z)$, it is bounded and so by Riemann's theorem is analytic in the complex unit disc. In particular, since $|g(z)|\le C|z|^{1-a}$, $g(0)=0$. Let $n$ be the order of this zero of $g$ (note that $n\ge 1$); we can write $g(z)=z^nh(z)$, with $h$ a holomorphic function which is nonzero near $0$. It is now trivial to verify that $z^{n-1}h(z)$ is an analytic extension to the unit disc of $f$.