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Schwarz-Pick Lemma


Let f be analytic on the unit disk, and assume that

1. |f(z)|<=1 for all z, and

2. f(a)=b for some a,b in D(0,1), the unit disk.

Then

 |f^'(a)|<=(1-|b|^2)/(1-|a|^2).
(1)

Furthermore, if f(a_1)=b_1 and f(a_2)=b_2, then

 |(b_2-b_1)/(1-b^__1b_2)|<=|(a_2-a_1)/(1-a^__1a_2)|,
(2)

where z^_ is the complex conjugate (Krantz 1999, p. 78). As a consequence, if either

 |f^'(a)|=(1-|b|^2)/(1-|a|^2)
(3)

or

 |(b_2-b_1)/(1-b^__1b_2)|=|(a_2-a_1)/(1-a^__1a_2)|
(4)

for a_1!=a_2, then f is a conformal self-map of D(0,1) to itself.

Stated succinctly, the Schwarz-Pick lemma guarantees that if f is an analytic map of the disk D into D and f preserves the hyperbolic distance between any two points, then f is a disk map and preserves all distances.


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References

Busemann, H. The Geometry of Geodesics. New York: Academic Press, p. 41, 1955.Krantz, S. G. "The Schwarz-Pick Lemma." §5.5.2 in Handbook of Complex Variables. Boston, MA: Birkhäuser, p. 78, 1999.

Referenced on Wolfram|Alpha

Schwarz-Pick Lemma

Cite this as:

Weisstein, Eric W. "Schwarz-Pick Lemma." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Schwarz-PickLemma.html

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