CONTEXT
The question is taken from Exercise VII.4.5 in Conway's "Functions on One Complex Variable I".
Let $f$ be analytic on $G = \{z: \mbox{Re}\ z > 0\}$, one-one, with $\mbox{Re}\ f(z) > 0$ for all $z$ in $G$, and $f(a) = a$ for some real number $a$. Show that $|f'(a)| \leq 1$.
MY ATTEMPT
Since $G$ is simply connected, so is $f(G)$, since it is conformally equivalent to $G$ by means of $f$. Using Riemann Mapping Theorem, let $h$ be the unique one-one analytic function that maps $G$ onto the unit disk, such that $h(a) = 0$ and $h'(a)>0$. Similarly, let $g$ be the unique one-one analytic function that maps $f(G)$ onto the unit disk, such that $g(a) = 0$ and $g'(a)>0$.
Now, the function
$$\psi (z) = g(f(h^{-1}(z))) $$
maps the unit disk onto itself and $\psi (0) = 0$, hence it must be a constant function.
Thus we have
$$\psi'(z) = g'(f(h^{-1}(z)))\cdot f'(h^{-1}(z))(h^{-1})'(z)=0.$$
In particular
$$\psi'(0) = g'(a)\cdot f'(a) \cdot \frac1{h'(a)} = 0.$$
But this leads to the conclusion that $f'(a) = 0$. Of course something is wrong here. I am not even using all the hypotheses.
QUESTION
Can you
- Tell me what is wrong in my attempt to a solution and
- Give me some hints on how to proceed?
I am self studying the subject, and relatively new to it. So no explaination is too simple for me.
Best Answer
For $a>0$ let $$\varphi(z)={z-a\over z+a}$$ Then $\varphi $ is a bijection from the open right half-plane onto the open unit circle so that $\varphi(a)=0.$ Denote $$g(z)=(\varphi\circ f\circ \varphi^{-1})(z)$$ Then $g$ maps the open unit disc into itself, $g(0)=0$ and $$g'(0)={1\over 2\pi i}\int\limits_{|z|=r}{g(z)\over z^2}\,dz$$ Hence $$|g'(0)|\le {1\over r},\quad 0<r<1$$ Therefore $|g'(0)|\le 1.$ We have $$f(z)=(\varphi^{-1}\circ g\circ \varphi)(z)$$ Thus $$f'(a)=(\varphi^{-1})'(0)g'(0)\varphi'(a)$$ As $(\varphi^{-1}\circ \varphi)(z)=z$ we get $$(\varphi^{-1})'(0)\varphi'(a)=1$$ hence $$f'(a)=g'(0)$$ Thus $|f'(a)|\le 1.$