If $|f(z)|\leq |f(z^2)|$ on the unit disk, show that $f$ is a constant function. 
Suppose $f$ is holomorphic in the unit disk $D=\{z: |z|<1\}$ and $|f(z)|\leq |f(z^2)|$ on $D$. Show that $f$ is a constant function.

My Try:
I wanted to use Strong Maximum Principle, but failed to prove that $f$ is bounded on $D$. I want to try this problem in my own. Can anybody please give me just a hint?
 A: Hint: Note that for every $z \in \mathbb{D}$, $|f(z)| \leq |f(z^{2^n})|$ for every $n$ by repeated application of the inequality $|f(z)| \leq |f(z^2)|$. As $|z|< 1$, what does $z^{2^n}$ converge to?
A: Amusingly, we can also say something about holomorphic functions $f: \mathbb{D} \to \mathbb{C}$ satisfying the opposite inequality $$|f(z^2)| \le |f(z)|$$ If $f$ satisfies this inequality, then $$|f(z^{2^n})| \le |f(z)| \to |f(0)| \le |f(z)|$$ by sending $n\to\infty$. If $|f(0)|>0$, then this contradicts the minimum modulus principle; so $f(0) = 0$. Note that $f$ can have no other zeroes in $\mathbb{D}$, since if it had some zero $w \neq 0$, then $w^{2^n}$ is a sequence of zeroes accumulating at $0$.
Let $k$ be the order of the zero at $0$. Then we may write $f(z) = z^k g(z)$ where $g$ is holomorphic and $g$ is nonvanishing in $\mathbb{D}$. The original inequality yields $$|z^k g(z^2)| \le |g(z)| \implies \left \vert \frac{z^k g(z^2)}{g(z)} \right \vert \le 1$$ Since $g$ is nonvanishing, $z^k g(z^2)/g(z)$ is holomorphic from $\mathbb{D} \to \mathbb{D}$ and vanishes at $0$. So we can apply the Schwarz lemma to get $$\left\vert \frac{z^k g(z^2)}{g(z)} \right \vert \le |z| \implies\left \vert \frac{z^{k-1} g(z^2)}{g(z)} \right \vert \le 1 \implies \left \vert \frac{z^{k-1} g(z^2)}{g(z)} \right \vert \le |z| \implies \cdots \implies \left \vert \frac{ g(z^2)}{g(z)} \right \vert \le 1$$ Now, as before, the inequality implies $|g(z)| \ge |g(0)|>0$, so by minimum modulus, we have $g$ is constant. Hence $f(z) = cz^k$ for some $c\in \mathbb{C}$.
