Proving $ f (z)=5$ for all $ z \in \mathbb {C} $ Suppose $ f: B \to \mathbb {C} $ is analytic on the disk $ B= \{  z \in \mathbb {C} : |z|<2 \}$, that $f (1)=5$, and that $ f (w)=f (w/2)$ for every $ w \in \mathbb{B} $. Show that $ f (z)=5$ for all $ z $ in $ \mathbb {B} $
My general idea here is to prove that the function f (z) is bounded in $\mathbb {C} $ then invoke Liovilles theorem. I basically saying that $f (z)=5 $implies that the real part of $ f $ is equal to 5 which implies |e^f|=e^5. Therefore by Liovilles theorem, $e^f$ is constant,  taking log of both sides,$ f (z)=5$. Is this correct?
 A: in fact all you need is continuity at $0$. take any $z$ in $B$, then $\frac{z}{2^n} \rightarrow 0$. iterating the identity from the question you get $f(\frac{z}{2^n} = f(z)$. if $f$ is continuous at $0$ then the left hand side converges to $f(0)$ hence $f(0) = f(z)$ for all $z$
A: The way I would do this is to consider the function $g(z) = f(z)-5$. Then $g$ has an accumulation point of zeroes in the ball. (What is the set of zeroes and what is the accumulation point?) What would you know then about $g$ and hence about $f$ from this observation?
A: $f$ is bounded and analytic on compact $D= \{z: |z| \leq 1\}$. By the maximum modulus principle, $|f|$ attains a maximum on the boundary, say $M = |f(e^{i\theta})| \geq |f(z)|$ for all $z \in D$.  But then $|f(e^{i\theta}/2)| = |f(e^{i\theta})|=M$. So $f$ attains it's maximum at an interior point and thus by the maximum modulus principle must be constant.
A: "So I take it my way is not the correct approach"
Your approach CAN work. $f$ is bounded and analytic on compact $\bar{D} = {z: |z| \leq 2 - \epsilon}$. You can define a new function $g$ such that $g(z) = f(z)$ when $z \in  \bar{D}$, and $g(z) = f(z/2^n)$ when $z \notin \bar{D}$. 
Then prove that this $g$ is entire and bounded, and apply Liouville's Theorem.
