Studia Scientiarium Mathematicarum Hungarica 32. (1996)

1-2. szám - Slezák B.: Implicit function theorem in uniform spaces

B. SLEZAK holds. Let (t', x') G H' x U'(:ro)- Then y' f(t', x') G IF (/(fo, £o)) and there exists an implicit function g: H x U(.x'o) belonging to y'. Let fi:H->V(x o), if 17^ í' if í = f'. It is clear that fi is an implicit function belonging to the point y' and passing through the point (t',x'). (iii) Let g: H -» V(xq) be a function such that g(h) = s, g is continuous at hand Vi Gif: f(t,g(t)) = f(h,s). Let W G By be arbitrary and let U G Bx be such that WteH: f(t,.)(U(s))^W(f(t,s)). Let the neighbourhood K of the point h be chosen so that the inclusions g(K) ^ U(s) and K ^ H hold. It will be shown that /(., s)(K) Q W(}{h, s)). Indeed, if t G K then for every (s,x) G U the set W contains the element (f(t,s),f(t,x)). On the other hand, if t G K then (s, g(t)) G {/, hence t € K (f(t,s), f (t, g{t)) G W. As the equality f(t, g(t)) = f(h, s) implies that (/(h, s), f(t, s)) is the element of W, it follows that for every t of the set K the point f{t,s) is in W(/(h, s)). It means that /(.,s) is continuous at h. Using Proposition 2 we get that / is continuous at {h,s). (iv) By Proposition 1 (ii) the function F is invertible. By Proposition 1 (iii) every implicit function is unique. As F is open at (to,xo) it follows that F~l is continuous at F(to, xq). Consequently, the function i?-1(., /(<0) ^0)) = = (idy, g) is continuous at to, where g is the implicit function passing through the point (to, xq). □ Theorem 2 (Implicit Function Theorem for metrizable topological groups). Let T be a topological space, G and Y be topological groups where the topology of G is defined by a translation invariant metric. Let us suppose that the topology of G is complete and the topology of Y is of Haus dorff-type. Let X B(x0; ro) QG be a closed ball, f: T X X —> T a function. Suppose that f(.,x0) is continuous at to and the mapping A: G —*Y is continuous, additive and open. Suppose further that there is a number k G]0,1[ so that for every B(x\ r)QX the inclusion (A — f(t, .))(B(x; r)) Q A(B(x\ kr)) — f(t, x) holds. Then (a) For every neighbourhood W of xq there is a neighbourhood H x V of the point (to,xo) that for every point (t,x) of H xV there is an implicit function g: H —>W such that f o (id^, g) = f(t, x) and g(t) = x. (b) The following three statements are equivalent: (i) Among the implicit functions passing through the point (t, x) there exists one which is continuous at t. (ii) The function f(.,x) is continuous at t. (iii) / is continuous at (t,x). (c) If the mapping A is injective then the implicit functions are unique. PROOF, (a) By Theorem 3 of [1] the family of functions T = {f(t,.) 11 G G T} is equiopen at every point x G X. By Lemma 2 in [1] IF is equicontinuous

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