Let $D$ be the open unit disc in the complex plane and $U=D/\{\frac{-1}{2},\frac{1}{2}\}.$ Also let $H_{1}=\{f:D\rightarrow\mathbb{C} \mid f\text{ is holomorphic and bounded}\}$ and $H_{2}=\{f:U\rightarrow\mathbb{C}\mid f\text{ is holomorphic and bounded}\}.$ Then the map $r:H_1\rightarrow H_2$ given by $r(f)=f|_U$, the restriction of $f$ to $U,$ is

$A.$ Injective but surjective.

$B.$ Surjective but not injective.

$C.$ Both injctive and surjective.

$D.$ Neither injective nor surjective.

According to me it is both injective and surjectve . As for injective $r(f)=r(g)\Rightarrow f|_U=g|_U\Rightarrow f=g$ by identity theorem . For surjective take preimage of $f$ as $f$ itself as $f$ can have atmost removable singularities at $\frac{\pm 1}{2}.$ Please suggest me. Thanks in advance.

  • $\begingroup$ yes u r correct,,neelkant?..exam kaisa rha? $\endgroup$ – user476275 Dec 17 '17 at 20:48

It is of course surjective: Let $f \in H_2$. Then we can construct $f_0 \in H_1$ by $f_0(z)=\frac{z+1/2}{z+1/2} \frac{z-1/2}{z-1/2}f(z)$. Of course, $r(f_0)=f$, so we have surjectivity.

To prove injectivity, we only need Riemann's Theorem, which says that if $a$ is a singularity of a function $f$ that is bounded in some neighborhood of $a$, then $f$ can be uniquely extended by defining $\bar{f}(a)=lim\lim_{z\to a}f(z)$. Since any $f \in H_1$ is bounded by assumption, we see that $r$ must map the functions uniquely [i.e. $f \neq g \implies r(f) \neq r(g)$].


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