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Let $D$ be the open unit disc in the complex plane and $U=D\setminus \{-1/2 ,1/2\}$.

Also let $$H_{1}=\{f:D\rightarrow\mathbb{C}\mid \text{$f$ is holomorphic and bounded} \}$$ $$H_{2}=\{f:U\rightarrow\mathbb{C}\mid \text{$f$ is holomorphic and bounded} \}$$

Then the map $r:H_{1} \rightarrow H_{2}$ given by $r(f)=f|_{U}$, the restriction of $f $ to $U$, is injective or bijective or both...

My Efforts: I tried to solve this , using the concept of restriction map but didn't get any hint.Please give some idea...

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  • $\begingroup$ Use \{ and \} for braces, and \text{ some text } for text inside math. $\endgroup$ – metamorphy Nov 1 '19 at 9:00
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If $r(f)=r(g)$ then $f(z)=g(z)$ except when $z \in \{-\frac 1 2 , \frac 1 2\}$. But then continuity implies $f(z)=g(z)$ for all $z$ so $r$ is injective. It is surjective because any bounded holomorphic function on $U$ can be extended to a holomorphic function on $D$ since it has removable singularities at $\pm \frac 1 2$.

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