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On the lim[sub(k)]p[sub(k)] (i;j)

1. We recall that Gk(x) is denoted by X'i = gi ) , Xt) = pk(i; 0) + Epk(i;jl, ,jt)x ..' Xt, jfo where limpk(i; 0) = limg)(0) x _i.

2. In the subcritical case, x^° = 1, we have 9)(1) -pk(i; O) = 1- pk(i; 0) = Zpk(i;jl,...,jt), jfo and hence limk ZjAo Pk(i; j) = 0.

3. In case 0 x^° 1, we have seen that 0 < x _i < 1 for all i. Hence, let m = min {x, . . ., x^} where 0 < C < 1. Then gi()()) - pkki;0)=) -= pk(i;j) (XI)L (Xt) j3o and hence limk 1Ej_oPk(i;) (x°)i1 .. (x °)j = 0. For every integer Uo> 1, we have* -Pk(i;j)(Xl°)^j l ... (X)j > E Pk(i; j)(X)^j . ..(XO)J > jo 1<aU(j)<a Z Pk(i;j)f^(j i⁾> 3 Pk(i;j)k since 0 < < 1 and r(j)< r l<a(j)<Oa l<a(j)<<a on the range of summation. It follows that linm E pk(i; j)W = 0 and hence 1<o(j)<a likm pk(i;j) = since e0. 1<a(j) a

4. In the only remaining case x^°= 0, we have Gk(0) = 0 and so gi()(0) = pk(i; O) = 0 and thus gk)(xi,...,xt) = EjoPk(i;j)x^l ..^j t^. We will show first the existence of a point x such that G(x) -< x 1. Since G(x) has for its component functions the gi indicated above, we have Ogi/dxi = Epi(i;j)jixl-'xlz² . ..xand so(gi/Oxi)o = pi(i; j0,. ..,0). Thus generally,(Ogi/0xj)o=pi(i; O,. . . , ).

Letting x = 1 in G(x), 1 = Ejo Pi (i;)>Ej=Pi (i; 0, . ., 1 , 0) = (9gi /lxj,)o Moreover, if 1 = (Ogi/Oxj)o, then all other pl(i;j) = 0 whence gi(x) = E t=Pi(i; O,..., ,..., 0)xj. But then (0² gi/&xjaxk) 0 for all x of It and all j, k, thus violating our assumption on second partials. Hence Ej(Qgi/Oxj)o < I for each i.

Define Si(x₁ ,...,Xt) = E(9gi/Oxj)x on It. Then Si(O) < 1, hence for each i there exists a 0i, 0 < ~i < 1, such that, for all x satisfying 0 <xj< (i we have Si(x) < 1 (continuity of S). * We use the notation a(j) = jl +... + jt, for j = (jl,...,jt).

Let U = min(f1,...,t}j , 0 < C < 1. Then for all x such that 0 <xj<< 1, we have Si(x) < 1 for all i. Now the Taylor form for gi(x) about x = 0 may be written gi(x) = 9i(O)+j(0gi/axj)txj = 0+E-(0gi/'xj),(i)xj, with 0 < x < x. Let x = (i,...,) where 5 is that above. Then gi(<,...,0) = ( /gi/)xj) (i) < since 0 < xj <~. Hence for x= (C,..., ) we have G(x) -< x ~ 1. It follows that 0 < ... < Gk(x)-Gk-1(x) - ...-G(x)-<x 1 so limGk(x) ~ 1 exists, and is a fixed point of G. Since there is only one fixed point (x^°= 0) other than 1, we have lim G(x) = 0.

For x= (,...,) then, lim g(k) () = limjoPk(i; j)(i) = Again fix a as any positive integer, and we have -Epk(i;j)e()^>pk(i;ij)>^(j)> pkP(i;j)". jO 1l<a(j)<al<o(j)<a So limk Zpk(i; j)l = 0 and thus limppk(i; j) = 0, where 1 <a(j)< a.

5. In summary, we have shown in this section that in all cases, for every a and every i lirm pk(i;j)= 0 k 1<a(j)<i and hence trivially, limkpk(i; jl... ,jt) = 0 for all i and all j 5 0.