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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.

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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 $0$?.

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  • $\begingroup$ 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 $|g(\zeta_k)|\to \infty$? $\endgroup$ – Sham Jan 5 at 1:11
  • $\begingroup$ I have edited my answer. @sham $\endgroup$ – Kavi Rama Murthy Jan 5 at 1:46
  • $\begingroup$ 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 $m_k$ $\endgroup$ – Sham Jan 5 at 2:07
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    $\begingroup$ @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 $0$. $\endgroup$ – Kavi Rama Murthy Jan 5 at 4:44

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