# Schwarz lemma in Stein book

Show that if $$f\colon D(0,R)→\mathbb{C}$$ is holomorphic with $$|f(z)|\leq M$$ for some $$M>0$$ then $$\bigg| \frac{f(z)-f(0)}{M^2-\bar{f(0)}f(z)}\bigg|\leq \frac{|z|}{MR}$$ This is an exercise of Stein's book Schwarz lemma section. But I do not have any idea how can I solve it, first at all, the Schwarz lemma needs that the origin is fixed by the function, but I do not have any value, only that it is bounded. So have I to do a conformal map first?

I would be grateful if you give any idea.

• Take $M=R=1$ assume $|f(0)| < M$ the LHS is $\phi(f(z))$ for some biholomorphism of the unit disk to which you can apply en.wikipedia.org/wiki/Schwarz_lemma – reuns Apr 25 '19 at 20:05

Ok so note that since $$|f(z)| you can actually re-write: $$f:D_0(R)\to \overline{D_0(M)}$$. However by the maximum modulus principle we know that f doesn't achieve its supremum on the open set $$D_0(R)$$ and thus we actually have: $$f:D_0(R)\to D_0(M)$$ Now by the Riemann Mapping Theorem (although honestly it's not really necessary to see what i'm about to do) we know that every simply connected open subset of $$\mathbb{C}$$ can be mapped conformally to the unit disc $$\mathbb{D}$$. That gives us the existence of the following two maps: $$\varphi : D_0(R)\to\mathbb{D} \quad \text{and} \quad \pi : D_0(M)\to\mathbb{D}$$ Both of which are conformal maps defined by $$z\mapsto z/R$$ and $$z\mapsto z/M$$ repectively. Now we define the function: $$F = \pi\circ f\circ \varphi^{-1} : \mathbb{D}\to\mathbb{D}$$ Note that by the chain rule we have that $$F$$ is holomorphic and so we have the following inequality (following from the Schwarz-Pick theorem letting $$z_1 = z$$ and $$z_2 = 0$$) [Shoutout to my boi @Dominic Petti for pointing this out]: $$\left|\frac{F(z) - F(0)}{1 -\overline{F(0)}F(z)}\right|\leq |z| \implies \left|\frac{\pi\circ f(zR) - \pi\circ f(0)} {1 -\overline{\pi\circ f(0)}\pi\circ f(zR)}\right|\leq |z|$$ Since we define $$\pi(z) = z/M$$, $$\implies \left|\frac{\frac{1}{M}(f(zR) - f(0))} {1 -\frac{1}{M^2}\overline{f(0)}f(zR)}\right|\leq |z| \implies M \left|\frac{f(zR) - f(0)}{M^2 -\overline{f(0)}f(zR)}\right|\leq |z|$$ Now letting $$\zeta = zR$$ and thus $$|z| = |\zeta|/R$$ we obtain the desired inequality: $$\left|\frac{f(\zeta) - f(0)}{M^2 -\overline{f(0)}f(\zeta)}\right|\leq \frac{|\zeta|}{MR}$$ A crispy result.