For which $K$ is $\varphi: H(\Omega)\xrightarrow{f\, \mapsto \, f\vert_K} C(K)$ injective/surjective?

Question

Let $\Omega$ be an open connected subset of $\Bbb C$, and let $K$ be a nonempty closed subset of $\Omega$. $H(\Omega)$ denotes the vector space of all holomorphic functions $f:\Omega\to \mathbb C$. Define $\varphi: H(\Omega)\to C(K)$ by $\varphi(f) = f\vert_K$, i.e. $\varphi$ is the restriction map. With proof, deduce which of the following statements are true:

1. If $K$ is uncountable, then $\varphi$ is injective.
2. If $K$ is uncountable, then $\varphi$ is surjective.
3. If $K$ is finite, then $\varphi$ is injective.
4. If $K$ is finite, then $\varphi$ is surjective.

What can be said about $\varphi$ if $K$ is countably infinite?

My try:

1. Finite: Let $K = \{x_1, \ldots, x_n\}$. Every function from a finite set into $\mathbb C$ is continuous. (A function on $K$ is just an element of $\mathbb C^n$.) We can extend this function to $\Omega$ in infinitely many ways (so $\varphi$ is not injective), but we can surely find a holomorphic extension of a function on $K$, to a function on $\Omega$ (so $\varphi$ is surjective.)

2. Uncountable: $\varphi$ is injective, but not surjective. Consider $f, g \in H(\Omega)$ with $\varphi(f) = \varphi(g)$, i.e. $f\vert_K = g\vert_K$. Using the identity theorem, $f-g\equiv 0$ on $\Omega$ ($f-g$ either has isolated, hence countably many zeros, or is identically zero on $\Omega$) giving $f=g$, hence injectivity of $\varphi$. Now, I must find $h\in C(K)$ with no holomorphic extension to $\Omega$.

It'd be great if I could get help with filling in the gaps in the above "proof", and the case where $|K| = |\mathbb N|$. Thanks!

Answer 1

The results are as follow;

$K$ finite, map surjective (even polynomial restrictions work with the usual Lagrange formula) but not injective (use Lagrange to the set plus another point where you choose the value in various ways)

$K$ countable - there are two cases; $K$ has an accumulation point in it (as $K$ closed) or $K$ discrete in $\Omega$ ($a_n$ accumulates to $\partial \Omega$ for a good ordering of $K$ ={$a_n$}, where for $\Omega=\mathbb C$ this means $|a_n| \to \infty$ etc)

In the first case clearly $\phi$ injective by the identity theorem, but $\phi$ not surjective as pick an accumulation point $a$ and a non-accumulation point $b$ which exists since $K$ countable so not perfect, $K-b$ is also closed hence $f=0$ there and $f(b)=1$ is continuous on $K$ but clearly cannot be extended to a holomorphic function on $\Omega$

In the second case, $\phi$ is surjective as a straightforward application of Weierstrass and Mittag Leffler for general regions shows (those theorems are generally more difficult to prove for a general domain different from the plane but the proof is similar) and like in the finite case adding an extra point in $\Omega$ to $K$ preserves its properties and we can specify continuous functions on the new set arbitrarily at the extra point and extend them to holomorphic functions so $\phi$ is not injective

$K$ uncountable - now we are essentially in the case one above, so $\phi$ is injective because now $K$ has an accumulation point and $\phi$ is not surjective since picking an accumulation point $a$ and a small enough closed disc centered at it st $D \cap a$ is not all $K$ so there is a $b$ in $K$ but outside it, one can extend the continuous function that is zero on $D$ and $1$ at $b$ continuously to $\Omega$ hence to $K$, so the argument above for $K$ countable case $1$ applies again

(the difference from the countable case is that $K$ may be perfect here like say a small closed disc or a closed segment, so we need to use some nontrivial topological extension theorem as opposed to the countable case)