Let $f$ be a non-constant analytic function on $\mathbb{D}$ satisfying $|f(z^2)| ≤ |f(z)|$ for $|z| < 1.$ Let $f$ be a non-constant analytic function on $\mathbb{D}$ satisfying $|f(z^2)| ≤ |f(z)|$ for $|z| < 1.$
$(a)$ Show that $f(0) = 0.$
$(b)$ Show that $f(z)\neq 0$ for $0<|z|<1.$
$(c)$ Show that $f(z) = az^n,$ where $a ∈ \mathbb{C},n ∈ N.$
I know how to do $(b)$ but I'm not sure how to do the other two parts.  Some help would be great thanks.
 A: (a) Suppose that $f(0) =a\neq 0 .$ If for all $z$ such that $0<|z|<1$ we have $|f(z)|\geq |a|$ then $f$ must be constant. Hence there exists $z_0 $ such that $0<|z_ 0|<1$ and $|f(z_0 ) |<|a| .$ The sequence $z_n =z_0^{2^n} \to 0$ and therefore $|f(0)| =\lim_{n\to \infty } |f(z_n )|\leq |f(z_0 )|<|a|=|f(0)|.$ Contradiction.
A: Less about economy, and more about clarity: 
Part (c)
From part (a), $f(z)$ has a zero of some order $n$ at $z = 0$, so you can factor $f(z) = z^n g(z)$, where $g(z)$ is analytic in the disk and $g(0) \ne 0$.
From part (b), $g(z)$ is never $0$ in $\mathbb{D}$.
Note that $g(z) = z^{-n}f(z)$.  I'm
Let $h(z)$ be the analytic function $\frac{g(z^2)}{g(z)}$ and notice that
$$
|h(z)|
  =\Bigg| \frac{g(z^2)}{g(z)} \Bigg|
  = \Bigg| \frac{z^{-2n} f(z^2)}{z^{-n} f(z)} \Bigg|
  = \frac{1}{|z^n|} \Bigg| \frac{f(z^2)}{f(z)} \Bigg|
  \le \frac{1}{|z^n|}
$$
On a disk $|z| \le r < 1$, $|h(z)|$ attains it's maximum somewhere on $|z| = r$.
So $h(z) \le 1/r^n$.
As in the proof of the Schwarz Lemma, let $r \to 1$ to see that $|h(z)| \le 1$ if $|z| < 1$.
Now we know $|g(z^2)| \le |g(z)|$ on $\mathbb{D}$.
Since $g$ is never $0$ on $\mathbb{D}$, you can apply the Minimum Modulus principle to conclude that $g(z)$ is constant.
Details: $|g|$ attains a minimum value on $|z| \le r$ at some $z_0$ with $|z_0| = r$, but now $z_0 ^2$ is an interior point of $|z| \le r$ where the modulus is no more than the modulus at $z_0$.  
So $f(z) = a z^n$.
An easier way to do part a) 
If $f(0) \ne 0$ then by continuity there's a smaller closed disk of radius $r \le 1$ on which $f$ is not $0$.
By the Minimum Modulus principle $|f(z)|$ attains its minimum on the smaller disk at some $z_0$ with $|z_0| = r$. But now $z_0^2$ is an interior point of this disk with $|f(z_0^2)| \le |f(z_0)|$ and so $f$ is constant, a contradiction.
