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Let $\mathbb{D}= \{ z\in \mathbb{C}: |z|<1\}$. For $t\in \mathbb{R}$, let $f_t$ denote the holomorphic function on $\mathbb{D}$ defined by $f_t(z)= (\frac{1+z}{1-z})^{it}$, $z\in \mathbb{D}$ with respect to the principal branch of the logarithm.

Show that for every infinite bounded subset $X \subset \mathbb{R}$ there is a sequence $(t_n)_{n \in \mathbb{N}}$ of distinct points in $X$ such that the sequence $(f_{t_n})_{n\in \mathbb{N}}$ converges uniformly on compact subsets of $\mathbb{D}$ to a holomorphic function on $\mathbb{D}$.

From Show that a complex function is bounded, we know that $\sup|f_t(z)|\le C^t$ for all $t\in X$.

I'm having trouble finding such sequence $(t_n)$ so that $(f_{t_n})_{n\in \mathbb{N}}$ converges. I think once I find such sequence, then I can do something similar to Existence of holomorphic function on the unit disk. Also, I'm not sure where the compactness takes place in the proof.

Thanks in advance!

Toasted_Brain
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This is an immediate consequence of Montel's Theorem on normal families. [Theorem 14.6 of Rudin's RCA]. If $X$ is a bounded subset of $\mathbb R$ then $(f_t)_{t \in X}$ is uniformly bounded (since $C^{t}$ is bounded. Hence any sequence in this family has subsequence which converges uniformly on compact sets.

geetha290krm
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