The derivative of a holomorphic function on the boundary of the unit circle Let $f$ be holomorphic on $D(0,1)\cup \{1\}$, and $f(0)=0$, $f(1)=1$, $f(D(0,1))\subset D(0,1)$. Prove that $|f'(1)|\geq 1$.
I have no idea. Maybe $f'(\xi)=[f(1)-f(0)]/(1-0)$ for some $\xi \in (0,1)$, and the maximal modules principle applies? This is not exact.
 A: It is easier to argue by contradiction. We are given that $f'(1)$ exists. Suppose $|f'(1)|<1$. Pick $c$ so that $|f'(1)|<c<1$. There  is a neighborhood of $1$ in which $$|f(z)-f(1)|\le c|z-1|$$ By the reverse triangle $$|f(z)|\ge |f(1)|-c|z-1| = 1-c|z-1| \tag{1}$$
On the other hand, by the Schwarz lemma 
$$|f(z)|\le |z|  \tag{2}$$
From (1) and (2) we get 
$$1-|z|\le c|z-1|\tag{3}$$
which is impossible when $z\in (0,1)$.
A: Very strong hint:  You can actually prove that $|f'(1)| \geq \dfrac{2}{1+|f'(0)|}$, and the result then follows from Schwarz's lemma.  Establishing the above inequality will use some ideas from the proof of Schwarz lemma.  Seems very very tricky to me.  I would not have been able to solve when I was a first year graduate student... maybe there is a simpler way than what I found however.  Perhaps expanding in power series about 0 and 1 and solving systems of equations will give it to you.  In any case the Schwarz lemma must enter into the picture somehow.
A: I have got it using Schwarz lemma. It follows that $|f(z)|\leq |z|$. And hence 
$$f'(1)=\lim_{R\ni r\to 1^-}\frac{f(r)-f(1)}{r-1}$$ that $|f'(1)|\geq 1$. It is not easy as this. Since $f(r)$ may be complex-valued.
A: The result follows from a simple calculus observation:
Lemma: Suppose $u:[0,1]\to \mathbb R$ is differentiable, with $u(1)=1$ and $u(x)\le x$ for all $x\in [0,1].$ Then $u'(1)\ge 1.$
Proof: Since $u(x)\le x,$ the line through $(x,u(x))$ and $(1,1)$ has slope $\ge 1.$ I.e.,
$$\frac{u(x)-u(1)}{x-1}\ge 1,\,0\le x <1.$$
The limit of this as $x\to 1^-,$ which is $u'(1),$ is thus $\ge 1.$
Now to our problem: By the Schwarz Lemma, $|f(z) |\le |z|$ for all $z\in D(0,1).$ Let $u=\text {Re }f.$ Since for $x\in [0,1]$ we have $|f(x)|\le |x|=x,$ we have $u(x)\le x.$ Now $u(1)=1,$ so by the lemma, $u'(1)\ge 1.$ This implies $|f'(1)|\ge 1$ as desired.
