VI—
On lim G[superscript(k)](x) = x[superscript(0)]
1. In case x° = 1, we have already seen that, for all x in It, lim Gk(x) = x°. We prove in this section that, in the remaining case, x° ¢ 1, lim Gk (x) = x° for all x / 1 of It. Indeed, we already know this when 0 - x -< x°.
2. We shall use the following Lemma. Let 0 <xi < 1 for all i, and p = max{x1 ,...,xt}, 0 < p < 1. Then for every e > 0 there exists a a > 0 such thatxil . . xt < e. 0-(j)>47
Analogies between Analogies
Proof. For arbitrary a > 0 we have (Margenau, Murphy p. 417) 00oo oo00 Xjl ... tjt _< E t a(j) -'o(j)= E tCm-lm o(j)>a a(j)>r m=a+l a(j)=m m=ar+ The series :=o Ct+m-l m converges by the ratio test: Ctmnm+l l/C't+m- Im = (t + m/m + 1) /-/< 1
3. Let x°0 1 and fix x in It, with 0 < xi< I for all i. Let l = max {ix,...,xt } and write g((x l...,Xt)-pk(i;0) = Ppk(i;j)x' ...xjt. Fix e > Oj30 Then by 2, there exists a a such that i x... xit < e/2. a(j)>a For this a, there exists a K such that, for all k>K, pk (i;j) < e/2. l1<a_(j)<a Hence, for all k > K, k) (Xl, . .,Xt) -pki)=Pk(i;j)x. ..x+pk(i;j)x .... t l<a(j)<ra a(j)>a < S pk(i;j) + E x' ....xtt < e/2 +e/2 e .1<oa(j)a a(o(j)>a Thus for every x such that 0 < xi < 1, all i, we have 0 = lim (gk)(x) - pk(i; 0)) since limpk(i; 0) = x i
4. Now suppose 0 - x 1. Then gi(x) < gi(1) = 1, for all i, so G(x) is a point of the type considered in 3. Hence x° = limGk(G(x)) = lim Gk+l(x) = lim Gk(x). This completes the proof that, in all cases, every point x7 1 of It has the lim Gk(x)=