By Pope C.N.

**Read or Download Geometry and topology in physics 2: Applications PDF**

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**Additional info for Geometry and topology in physics 2: Applications**

**Example text**

If y E domhp = {y > 0: hp(y) < +oo}, then (hp(y),y) E supp (p,L). Hence, ifdomhp = (O,+oo),thensupp(p,L) = {(a,y): y > O,a 'S hp(y)}. 11) holds with b = 0. Assume that there exists a pointy > 0 such that hp (y) = +oo. It means that (a, y) E supp (p, L ), for all a > 0. Then the normality of supp (p, L) implies that hp(y') = +oo, for all 0 < y'::::; y. Thus the set {y > 0: hp(y) = +oo }, if nonempty, is a segment. Upper semicontinuity of hp implies that this segment is closed (in IR++). 11) holds with b = sup{y: hp(y) = +oo}.

Then (y,hp 1 (y)) E supp (p1,L), (y,hp 2 (y)) E supp (p2,L), and (y, a) tj. 13) that hp(Y) = hp 1 (y) = min(hp 1 (y), hp 2 (y)). 3 and PI ~ P2· In the sequel we need the following simple assertion. A) > 0. ),y,\}for y > 0. ) 2) Xy ( Ay) = y).. A) as y --7 +0. Proof" The proof is straightforward. PH function defined on lR~. Then supp(1,y) y>O = suphp(y). y>O Proof" It follows from the definition of hp that supp (p, L) 0, 0 < 8 ~ hp(z)}. So, for y > 0, we have: p(1,y) : z > sup{ ((8, z), (1, y)) : (8, z) E supp (p, L)} sup z>0,6:Shp(z) Thus = {(8, z) min( 8, zy).

2 Let H be a set of continuous functions defined on a metric space Z and 2 E Z. Assume that each nonnegative continuous function f defined on Z is abstract convex at the point 2. Then, for each E E (0, 1) and 8 > 0, there exists a function h E H, which is a support to an Urysohn peak, corresponding to (2, E, 8). Proof To establish the result, we consider a 8-Urysohn peak fo, where 8 is an arbitrary positive number. Since fr; is abstract convex with respect to H at the point 2, it follows that for each c > 0 there exists a function h E: H, such that h fr; and h(z) > fo(2) - c = 1 - c.