# Show that $g$ can be extended to a holomorphic function on $\mathbb{D}$.

Let $$g$$ be a holomorphic function on $$\mathbb{D}\setminus\{0\}$$, and denote $$g_m(z)=g(z/m)$$ for each positive integer $$m$$. Suppose that $$\{g_m\}_{m=1}^{\infty}$$ has a subsequence $$\{g_{m_k}\}_{k=1}^{\infty}$$ which is uniformly bounded by 1 on the circle $$\{z;|z|=1/2\}$$, i.e., $$\max_{|z|=1/2}|g_{m_k}(z)|\leq 1\quad \text{for all k\geq 1}.$$ Show that $$g$$ can be extended to a holmorphic function on $$\mathbb{D}$$.

It follows by Montel's Theorem that $$\mathcal{F}:=\{g_{m_k}\}_{k=1}^{\infty}$$ is a normal family of holomorphic functions. So there exists a subsequence $$\mathcal{F}_j:=\{g_{m_{k_j}}\}_{j=1}^{\infty}$$ that converges uniformly to a holomorphic $$h$$ on all compact subsets of the disc $$\{z;|z|\leq 1/2\}$$.

I'm not sure if this is the right direction.

Apply MMP to the annulus between the circles of radii $$\frac 1 {2m_k}$$ and $$\frac 1 {2m_{k+1}}$$. Do you see now that $$g$$ is bounded near $$0$$?.

• So here you are supposing that $g$ is unbounded in a neighborhood of 0. Why is it sufficient to suppose there is a subsequence $\zeta_k\to 0$ such that $|g(\zeta_k)|\to \infty$? – Sham Jan 5 at 1:11
• I have edited my answer. @sham – Kavi Rama Murthy Jan 5 at 1:46
• By the MMP we have $$|g(z)|\leq 1$$ for all $\frac{1}{2m_{k+1}}<|z|<\frac{1}{2m_k}$. This must be true for all positive integers $m_k$ – Sham Jan 5 at 2:07
• @Sham Since $m_k \to \infty$ the union of these annuli is the set $\{z: 0<|z|<\frac 1 {2m_1}\}$. So $g$ is bounded on this set which implies that it has a removable singularity at $0$. – Kavi Rama Murthy Jan 5 at 4:44