a) Let (αi,βi) be n distinct pairs of real numbers ⩾0, distinct from (0,0), and such that infiαi=infiβi=0. Consider the equation
f(x,y)=i=1∑naixαiyβi(1+φi(x,y))=0 where the
ai are real numbers
=0 and the
φi are continuous functions on a square
0⩽x⩽a,0⩽y⩽a, tending to 0 as
(x,y) tends to
(0,0). Suppose that there exists a function
g which is positive and continuous on an interval
[0,b], tending to 0 with
x and such that
f(x,g(x))=0 for all
x∈[0,b]. For every real number
μ>0 show that
g(x)/xμ tends to a limit, finite or infinite, as
x tends to 0 (use V, p. 217, prop. 9, considering, for every number
t⩾0, the equation
f(x,txμ)=0).
b) For g(x)/xμ to tend to a finite limit =0 it is necessary that μ should be such that there exist at least two distinct pairs (αh,βh) and (αk,βk) such that αh+μβh=αk+μβk and such that, for every other pair (αi,βi), one has αi+μβi⩾αh+μβh. One thus obtains a finite number of possible values μj (1⩽j⩽r); the numbers −1/μj are the gradients of the affine lines in the plane R2 which contain at least two of the set of points (αi,βi) and such that all the other points of this set lie above the line considered ("Newton polygon").
c) Let μ1 be the smallest of the numbers μj. Show that g(x)/xμ1 tends to a finite limit (possibly zero). (Show first that one may always assume that if i and j are two distinct indices one cannot have both αi⩽αj and βi⩽βj; deduce that one can assume α1=0, αi>0 and βi<β1 for i=1; now, putting g(x)=t(x)xμ1, show that t(x) cannot tend to +∞ as x tends to 0.)
d) Deduce from c), by recursion on r, that there exists an index j such that g(x)/xμj tends to a finite limit =0 as x tends to 0.