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Suppose $f$ is holomorphic on $B(0,1) \cup \{1\}$, $f(0)=0$, $f(1)=1$, $f(B)\subset B$, I want to prove $f'(1)\geq1$.

I tried to use first-order Taylor expansion to get an estimate of $f'(1)\geq 0$ using $f(1)=1$, but I don't know how to use $f(0)=0$ to get a deeper insight.

Any help will be appreciated.

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We have $f$ analytic in $B \cup D(1,r)$ for some $r>0.$ Because $|f|<1$ in $B,$ we have $|f|\le 1$ on $\partial B \cap D(1,r).$ Writing $f = u + iv,$ we see $u\le 1$ on $\partial B \cap D(1,r).$

For small $t \in \mathbb R,$ $u(e^{it})$ is well defined and differentiable. Because $u(1) = 1, u(e^{it})$ has a maximum at $t=0.$ Thus $d\, u(e^{it})/dt|_{t=0} = 0.$ This implies $u_y(1) = 0.$ By the CR equations, $v_x(1) = 0.$ Since $f'(1) = u_x(1) + iv_x(1),$ we see see $f'(1) = u_x(1).$

Now use $f(0)=0$ and the Schwarz Lemma to see $u(x,0)\le x$ for $x\in (0,1).$ Hence

$$\frac{u(1,0)- u(x,0)}{1-x} \ge \frac{1- x}{1-x} = 1$$

for $x\in (0,1).$ Thus $u_x(1) \ge 1$ and we're done.

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  • $\begingroup$ I edited my answer. I saw later that first proving $f'(0)\ne 0$ is not needed. $\endgroup$
    – zhw.
    Oct 5, 2016 at 0:40

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