Maximum of an analytic function on the unit disk. This question is a old question but in that question one condition was not explained well.
Let $f$ be analytic on the unit disk $D$. Assume that $f(r)=\max\limits_{|z|=r} |f(z)|$. (Note that here we are not defining a new function. It just means  that $f(z)$ attains its maximum at a point $z=r$.)
Why $f′(r)>0$, if $f$ is not a constant?
And why if $f(0)=0$, then $rf'(r)\geq f(r)$ and the equality holds if and only if $f(z)=cz$ for some nonnegative constant $c$ ?
Why $f'(r)$ is real number and even positive? Why not negative or some complex number? It is pretty strange for me!
 A: Define $g(z) = f(z)/z$ if $z \neq 0$ and $g(z) = f'(0)$ if $z = 0$.
$g$ is analytic in the unit disk. Also, $|g(z)| \leq g(r)/r$ on the circle $|z| = r$.
By the maximum modulus principle, $|g(z)| \leq g(r)/r$ for all $|z| \leq r$.
Converting back to $f$, we have $|f(z)/z| \leq f(r)/r^2$, or $|f(z)| \leq |z|f(r)/r^2$ for all $|z| \leq r$.
I'm kind of stuck at this point. I'm guessing you've probably gotten at least to here.
A: If the derivative were ever negative, we would contradict the maximum principle: increase r slightly and you get a boundary with smaller modulus than in the interior. If it were ever zero, then by Rolle's theorem (the one from calculus) there are distinct $r_1,r_2$ with $f(r_1)=f(r_2)$. This implies $f$ is constant by the maximum modulus principle.
Ask yourself exactly which of the assumptions are actually required here. The other part seems to be something you are making progress on. It's definitely Schwarz. 
A: We assume that $|f(r)|>0$ otherwise $f=0$ and the rest is trivial.
Define $g(z)=f(z)/f(r)$. Therefore $g:D\to D$ is analytic, so we can apply Schwartz's lemma. This tells us that
$|g(z)|\leq |z|$ on $D$ and we have $|g(r)|=1$. This can only hold if $|r|=1$.
Let us write $r=\mathrm e^{\mathrm i\rho}$. 
As we have the equality $|g(z)|=|z|$ for $z=r$, the lemma says that $g(z)=az$ with $|a|=1$. Let us write $a=\mathrm e^{\mathrm i\alpha}$.
We have obtained $f(z)=f(\mathrm e^{\mathrm i\rho})\mathrm e^{\mathrm i\alpha}z$, which solves the problem. (Note in particular that $f'(r)=f(r)$)
