Let $V$ be an open subset of $\mathbb{C}$ and $B:=B(a,r)$ a ball whose closure is in $V$. Denote by $E$ the space of all complex-valued functions holomorphic on $B$ and continuous on its boundary/closure. Note that $E$ is a Banach space under the supremum norm. Let $\mathscr{H}$ be the space of entire functions and $F$ the closure of $\mathscr{H}$ in $E$.
Question: Show that if $f$ is analytic on $V$, then $f \in F.$
My Attempt: Clearly, $f$ is in $E$. So we only need to show that $f$ is a limit of entire functions under the supremum norm of $E$. Since $f$ is analytic on $V$, it is analytic on some open ball $B'$ that contains $\bar{B}$. So $f$ is a power series on $B'$ and this power series converges uniformly to $f$ on $\bar{B}$. As such, $f$ is the uniform limit of entire functions. Therefore, $f \in F$.
Is this correct?
Bingo! That's exactly the way I'd do it.