Schwarz-Pick lemma on the unit disc

Put $b=f(a)$ and $F={\phi }_b\circ f\circ {\phi }_a$. By [F1], the two Blaschke factors are disc automorphisms, so $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and satisfies $F(0)={\phi }_b(f(a))=0$.

Applying [F2] to $F$ at the point ${\phi }_a(z)\in \mathbb{D}$ gives $\mathrm{\mid }F({\phi }_a(z))\mathrm{\mid }\le \mathrm{\mid }{\phi }_a(z)\mathrm{\mid }$, that is, $\mathrm{\mid }{\phi }_b(f(z))\mathrm{\mid }\le \mathrm{\mid }{\phi }_a(z)\mathrm{\mid }$. This is exactly the displayed pseudohyperbolic inequality.

Since ${\phi }_a^{\mathrm{\prime }}(0)=\mathrm{\mid }a{\mathrm{\mid }}^2-1$ and ${\phi }_b^{\mathrm{\prime }}(b)=-1\mathrm{/}(1-\mathrm{\mid }b{\mathrm{\mid }}^2)$, the chain rule [F3] gives $\mathrm{\mid }{F}^{\mathrm{\prime }}(0)\mathrm{\mid }=\mathrm{\mid }{\phi }_b^{\mathrm{\prime }}(b)\mathrm{\mid }\mathrm{\mid }{f}^{\mathrm{\prime }}(a)\mathrm{\mid }\mathrm{\mid }{\phi }_a^{\mathrm{\prime }}(0)\mathrm{\mid }=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{\mid }{f}^{\mathrm{\prime }}(a)\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }b{\mathrm{\mid }}^2)$. Applying the derivative part of [F2] to $F$ yields the stated bound for $\mathrm{\mid }{f}^{\mathrm{\prime }}(a)\mathrm{\mid }$.

If equality holds in the pseudohyperbolic inequality for some $z\mathrm{\ne }a$, then equality holds in Schwarz's lemma for $F$ at the nonzero point ${\phi }_a(z)$; if equality holds in the derivative inequality, then $\mathrm{\mid }{F}^{\mathrm{\prime }}(0)\mathrm{\mid }=1$. In either case [F2] makes $F$ a rotation, so $f={\phi }_b\circ F\circ {\phi }_a$ is an automorphism by [F1].