= Solution
The <Schwarz-Pick lemma> says that every <holomorphic> self-map $F$ of the <unit disc> contracts <pseudohyperbolic distance> and therefore <hyperbolic distance>:
$$
\left|\frac{F(z)-F(w)}{1-\overline{F(w)}F(z)}\right|
\le\left|\frac{z-w}{1-\overline wz}\right|,\qquad
\rho(F(z),F(w))\le\rho(z,w).
$$
Its infinitesimal form is
$$
\boxed{\frac{|F'(w)|}{1-|F(w)|^2}\le\frac1{1-|w|^2}.}
$$
Equality for distinct points in the distance inequality, or at one point in the <derivative> inequality, holds exactly when $F$ is a disc <Möbius transformation>.
For completeness, first prove the <Schwarz lemma>. If $H:\mathbb D\to\mathbb D$ is <holomorphic> and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The <maximum modulus principle>, followed by $r\uparrow1$, gives $|H(z)|\le|z|$ and $|H'(0)|\le1$. Equality at a nonzero point or in the <derivative> forces that quotient to be a constant of modulus one, again by the <maximum modulus principle>; hence $H(z)=e^{i\theta}z$.
Now let $\phi_a(z)=(z-a)/(1-\overline az)$ and apply this result to $H=\phi_{F(w)}\circ F\circ\phi_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho=\log((1+\delta)/(1-\delta))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|\phi_a'(a)|=(1-|a|^2)^{-1}$ and $|({\phi_w^{-1}})'(0)|=1-|w|^2$, so the <chain rule> proves the <derivative> inequality. The equality statement follows from the rotation case of the <Schwarz lemma> after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.
There is a maximum of the <derivative> modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and <Cauchy estimates> bound their <derivatives> at zero. Take a sequence whose <derivative> moduli approach the finite supremum. By <Montel theorem> it has a subsequence converging locally uniformly to a <holomorphic function> $f$, with $f(0)=1$ and $e^{-1}\le|f|\le e$. The upper equality cannot occur at an interior point, because the <maximum modulus principle> would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is <holomorphic> because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. <Locally uniform convergence> gives convergence of <derivatives> by the <Cauchy integral formula>, so this limit attains the supremum.
To compute it and all extremizers, the nonvanishing function $f$ on the <simply connected> disc has a <holomorphic logarithm> $h$ with $h(0)=0$. One can construct it by integrating $f'/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\operatorname{Re}h<1$. The map
$$
T(h)=\tan\frac{\pi h}{4}
$$
is a <biholomorphism> from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi/2,\pi/2)$. Therefore
$$
s(z)=\frac4\pi\arctan z=\frac{2}{\pi i}\log\frac{1+iz}{1-iz}
$$
has real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the <tangent>, so these are inverse maps.
Applying the <Schwarz lemma> to $T\circ h$ gives $(\pi/4)|h'(0)|\le1$. Since $f'(0)=h'(0)$, the exact maximum is
$$
\boxed{\max_{f\in\mathcal F}|f'(0)|=\frac4\pi.}
$$
It is attained by $f(z)=\exp((4/\pi)\arctan z)$. The equality case requires $T(h(z))=e^{i\theta}z$, and hence all maximizing functions are
$$
\boxed{f_\theta(z)=\exp\left(\frac4\pi\arctan(e^{i\theta}z)\right),\qquad\theta\in\mathbb R.}
$$
Their <derivatives> are $(4/\pi)e^{i\theta}$, so \b[the extremizer is not unique]: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the <extremal derivative of a disc map into a symmetric annulus> for $L=1$.
Back to article page
