Problem 5

Let

G

be a set of non-constant functions

f:\mathbb R\to\mathbb R

of the form

f(x)=ax+b

, with real

a,b

and

a\ne0

. Suppose: if

f,g\in G

then

g\circ f\in G

; each inverse

f^{-1}

belongs to

G

; and every

f\in G

has a fixed point. Prove that there is

k\in\mathbb R

fixed by every

f\in G

Step 4 of 4: Common fixed point

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In plain words

Equal fixed-point formulas identify one common point for every member.

\frac{b}{1-a}=\frac{d}{1-c}

Detailed analysis

Rearranging gives $b(1-c)=d(1-a)$ . If both maps are non-identity, their fixed points $b/(1-a)$ and $d/(1-c)$ are equal. If one map is the identity, choose the fixed point of the other; if both are identities, any $k$ works. Thus every pair has a common fixed point, and choosing a non-identity map when one exists gives a point fixed by all of $G$ .