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Fall 2017 - Problem 9

entire functions

Consider a map $\func{F}{\C \times \C}{\C}$ with the following properties:

For each fixed $z \in \C$ the map $w \mapsto F\p{z, w}$ is injective.
For each fixed $w \in \C$ the map $z \mapsto F\p{z, w}$ is holomorphic.
$F\p{0, w} = w$ for $w \in \C$ .

Show that then

F\p{z, w} = a\p{z}w + b\p{z}

for $z, w \in \C$ , where $a$ and $b$ are entire functions with $a\p{0} = 1$ , $b\p{0} = 0$ , and $a\p{z} \neq 0$ for $z \in \C$ .

Hint: Consider $\frac{F\p{z, w} - F\p{z, 0}}{F\p{z, 1} - F\p{z, 0}}$ .

Solution.

Let $G\p{z, w} = \frac{F\p{z, w} - F\p{z, 0}}{F\p{z, 1} - F\p{z, 0}}$ . By injectivity, $F\p{z, 1} - F\p{z, 0} \neq 0$ for any $z \in \C$ , so $z \mapsto G\p{z, w}$ is entire.

Notice that $G\p{z, 0} = 0$ and $G\p{z, 1} = 1$ . If $w \neq 1, 2$ , then by injectivity, $F\p{z, w} \neq F\p{z, 0}$ and $F\p{z, w} \neq F\p{z, 1}$ , so $G\p{z, w}$ misses $0$ and $1$ . By Picard's little theorem, $z \mapsto G\p{z, w}$ is constant. In all cases, $z \mapsto G\p{z, w}$ is constant, so

G\p{z, w} = G\p{0, w} = \frac{F\p{0, w} - F\p{0, 0}}{F\p{0, 1} - F\p{0, 0}} = w.

Thus,

w = \frac{F\p{z, w} - F\p{z, 0}}{F\p{z, 1} - F\p{z, 0}} \implies F\p{z, w} = \p{F\p{z, 1} - F\p{z, 0}}w + F\p{z, 0}.

If we set $a\p{z} = F\p{z, 1} - F\p{z, 0}$ and $b\p{z} = F\p{z, 0}$ , then $a\p{0} = 0$ , $b\p{0} = 0$ , and by injectivity, $a\p{z} \neq 0$ , which completes the proof.