Find and verify an inclusion radius

Suppose every function in one Ma–Minda class is dilated by \(f_r(z)=f(rz)/r\). How large can \(r\) be before the dilated function leaves a target class? That largest value is the inclusion radius.

This page starts with a case where the first contact occurs on the negative real axis, then shows why that shortcut does not work in general.

Run it

python examples/research_workflows/sharp_radius.py \
  --source sine \
  --s 1/2 \
  --output /tmp/gfa-sharp-radius

What you will see

The directory contains a JSON result and three SVG figures. Together they show the source generator, the target generator, the point where their boundaries first meet, and how the radius changes with the target parameter \(s\).

The target family is

\[ L_s(z)=(1+sz)^2. \]

Rows and formulas always mean the direction

\[ \mathcal S^*(\phi)\longrightarrow\mathcal S^*(L_s). \]

Reversing the arrow is a different question.

The sine calculation

For \(\phi(z)=1+\sin z\), take the principal square root

\[ H(z)=\sqrt{\phi(z)}=\cos(z/2)+\sin(z/2). \]

Membership in the target is equivalent to \(|H(z)-1|<s\). For \(|z|\le r\), the maximum occurs at \(z=-r\) and equals

\[ 1-\cos(r/2)+\sin(r/2). \]

Setting this equal to \(s\) gives

\[ R(s)=\min\{1,\arcsin(2s-s^2)\}. \]

At \(s=1/2\) the answer is therefore

\[ R=\arcsin(3/4). \]

The Schwarz lemma carries this generator inclusion to every member of the source class. The canonical member reaches the same boundary, so a larger radius cannot work. This proves sharpness.

The detailed maximum argument is in the sine/limaçon proof.

Why checking only the real axis can fail

Run the second example:

python examples/research_workflows/sharp_radius.py \
  --source off_axis \
  --s 1/4 \
  --output /tmp/gfa-off-axis-radius

Here

\[ h(z)=\frac z4-\frac{z^3}{32}, \qquad \phi(z)=(1+h(z))^2. \]

Every point on the real diameter satisfies \(|h(x)|<1/4\), but on a circle \(|z|=r\) the maximum is

\[ \max |h(z)|=\frac r4+\frac{r^3}{32}, \]

attained at \(z=\pm ir\). The signs cancel on the real axis and reinforce on the imaginary axis. For \(s=1/4\), the sharp radius is the unique root in \((0,1)\) of

\[ r^3+8r-8=0, \]

approximately \(0.9067953030\).

The lesson is simple: real-axis contact is a theorem to prove for a particular generator, not a general rule for radius problems.

Sources: Masih–Kanas, DOI 10.3390/sym12060942, for the target family; Cho et al., DOI 10.1007/s41980-018-0127-5, for the sine class.