$f$ constant on the open unit disk $\implies$ $f$ is constant on the closed unit disk Suppose $f$ continuous on $\{z\in \mathbb{C}||z|\leq1\}$ and analytic on $\{z\in \mathbb{C}||z|<1\}$ and that $f$ is constant on $\{z\in \mathbb{C}||z|<1\}$. Is $f$ being continuous enough to say that $f$ has to be constant on all of $\{z\in \mathbb{C}||z|\leq1\}$ ? 
 A: Yes. If $f$ is constant on the open disk $\Bbb D$, then $f(\Bbb D)=c$ for some $c$. Since the set $\{c\}$ is closed and $f$ is continuous, $f^{-1}(c)$ is a closed set containing $\Bbb D$. As such, $f^{-1}(c)$ must also contain the closure of $\Bbb D$, which is the closed disk. Hence, $f$ is constant on $\overline{\Bbb D}$.
A: Yes, and continuity is indeed sufficient. To see this fix $z\in\mathbb C,|z|=1$. Since $f$ is constant in $D=\{w\in\mathbb C\ |\ |w|<1\}$, we have $f(tz)=f(0)$ for all $t\in[0,1)$. Since $f$ is continuous in $\overline D$, we conclude
\begin{align*}
f(z)=\lim_{t\to 1-}f(tz)=\lim_{t\to 1-}f(0)=f(0).
\end{align*}
Since $z$ was arbitrary, $f(z)=0$ for all $z\in\overline D$.
A: By sequential continuity, which follows from continuity ( and the function $c$ being uniformly continuous) and $|z|=1$ being the closure of $z:|z|<1 $}, take a sequence {$x_n$}converging to a point on $|z|=1$, then we must have $(x_n \rightarrow x ) \rightarrow (f(x_n) \rightarrow f(x) )$  then , since $f(x_n)=c $ , we have $f(x_n)=c \rightarrow f(x)=c $. 
