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.

  • $\begingroup$ 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 $\endgroup$ – reuns Apr 25 '19 at 20:05

Ok so note that since $|f(z)|<M$ 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.

  • 1
    $\begingroup$ So crispy! Also respect on the shoutout $\endgroup$ – jonan Nov 24 '19 at 16:10

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