When does a bounded continuous function extend continuously to its closure Let $\Omega$ be a domain in $\mathbb{C}^n$. Let $f:\Omega\longrightarrow\mathbb{C}$ be a bounded continuous function. I wanted know if there are any necessary and sufficient conditions for $f$ to be extended continuously to $\bar{\Omega}$?
 A: One necessary and sufficient condition is that the restriction of $f$ to any bounded subset $A \subset \Omega$ be uniformly continuous. In fact this is a necessary and sufficient condition even under weaker hypotheses, namely that $f : \Omega \to \mathbb C$ be a continuous function whose restriction to every bounded subset of $\Omega$ is bounded.
To see that this is necessary, suppose that $f$ has a continuous extension to $\overline\Omega$. If $A \subset \Omega$ is bounded then $\overline A \subset \overline \Omega$ is bounded, and since $\overline A$ is also closed it follows that $\overline A$ is compact. Thus $f$ is bounded and uniformly continuous on $\overline A$ (these are theorems of topology), and so it is bounded and uniformly continuous on $A$.
To see that this is sufficient, suppose that $f$ is uniformly continuous on each bounded subset of $\Omega$. Consider the nested set of closed balls
$$B(O,1) \subset B(O,2) \subset \cdots \subset B(O,n) \subset \cdots
$$
where $O$ is the origin. Let $A_n = \Omega \cap B(O,n)$ and so
$$A_1 \subset A_2 \subset \cdots \subset A_n \subset \cdots \qquad\text{and} \quad \Omega = \bigcup_{n=1}^{\infty} A_n
$$
$$\overline A_1 \subset \overline A_2 \subset \cdots \subset \overline A_n \subset \cdots \qquad\text{and} \quad \overline\Omega = \bigcup_{n=1}^\infty \overline A_n
$$
(To prove the inclusion $\overline\Omega \subset \bigcup_{n=1}^\infty \overline A_n$, given $x \in \overline\Omega$ choose $n \in \mathbb N$ so that $n > 1 + d(O,x)$, and so $x \in B(O,n-1)$. If $0<r<1$ then $\Omega \cap B(x,r) \ne \emptyset$ and $B(x,r) \subset B(O,n)$ so $A_n \cap B(x,r) = (\Omega \cap B(O,n)) \cap B(x,r) \ne \emptyset$. Since this holds for all $0<r<1$ it follows that $x \in \overline A_n$.)
Since $f$ is bounded and uniformly continuous on the bounded set $A_n$ it follows that $f$ has a unique continuous extension to $\overline A_n$ (this is another theorem of topology). From uniqueness it follows that if $m<n$ then the continuous extension to $\overline A_m$ is the restriction of the continuous extension to $\overline A_n$. Thus, the continuous extensions of $f$ to $\overline A_n$ all piece together to define a continuous extension to $\overline\Omega$.
