244 Trigonometric functions in calculus Ch. 12
COMMENT. This equation forμis referred to asDarboux’s functional equa-
tion.
There is a number p> 0 such that for 0 ≤x≤ 360 ,μ(x)=px.
Proof. We start by writing out a well known proof that any functionμwhich
satisfies Darboux’s functional equation on[ 0 , 360 ]satisfiesμ(r)=rμ( 1 )for all ra-
tional numbersrin[ 0 , 360 ]. On first takingy=xwe haveμ( 2 x)= 2 μ(x);thentaking
y= 2 x,wegetμ( 3 x)= 3 μ(x);nexttakingy= 3 x,wegetμ( 4 x)= 4 μ(x); and pro-
ceeding this way we evidently get
μ(nx)=nμ(x)for alln∈Nand 0≤x≤360
n.
This can be confirmed by induction onn.
On applying this withn=m,x= 1 /mfor anym∈Nwe deduce that
μ( 1 )=mμ(
1
m)
so that
μ(
1
m)
=
1
mμ( 1 ).Now on takingnandx= 1 /mwe see that
μ(n
m)
=μ(
n.1
m)
=nμ(
1
m)
=n.1
mμ( 1 )=n
mμ( 1 ).Thusμ(r)=rμ( 1 )for every rational numberrwith 0<r≤360, and this extends
to the caser=0 also as puttingx=y=0 in Darboux’s functional equation gives
μ( 0 )=0.
It remains to show thatμ(y)=yμ( 1 )for each irrational numberywith 0<y<
- Choose a decreasing sequence of rational numberstn(y)which converge toy.
By the case for rational numbers, we haveμ(tn)=tnμ( 1 ).Nowtn=(tn−y)+yso
by Darboux’s functional equation
μ(tn)=μ(tn−y)+μ(y),and so
μ(y)=tnμ( 1 )−μ(tn−y).
We wish to maken→∞in this and take limits.
Now as definedμ(z)>0for0<z<360. Then if 0<x<y<360 asμ(x)+μ(y−
x)=μ(y)andμ(y−x)>0, we must haveμ(x)<μ(y). Thusμis an increasing
function. Hence aszdecreases to 0,μ(z)decreases and is positive and so the one-
sided limit
l=z→lim 0 +μ(z)=z→lim 0 ,z> 0 μ(z)