- BUSEMANN FUNCTIONS 307
3.2. Constructing totally convex half spaces. An important use
of Busemann functions is to define arbitrarily large 'totally convex subsets'
in a complete noncompact manifold of nonnegative sectional curvature. As
a first step in their construction, we make the following definitions.
DEFINITION B.51. Given a ray 'Y : [O, oo) ---+ Mn emanating from a point
0 E Mn, the open right half space is the union
JIB 1 ~ LJ B ('Y ( s) , s) ,
sE(O,oo)where B ('Y ( s) , s) ~ { x E Mn : d ( x, 'Y ( s)) < s}. The closed left half space
IBLy is its complement:
JHI 1 ~ Mn\JIB'Y =Mn\ LJ B ('Y (s), s).
sE(O,oo)
The motivation for the term 'left half space' is the following result.LEMMA B.52. b'Y (x) :::;; 0 if and only if x E lHI'Y.
PROOF. By Lemma B.44, b'Y,s is monotone increasing ins. So b'Y (x) :::;; 0if and only if b'Y,s (x):::;; 0 for alls E (O,oo). But by definition
b'Y,s (x) :::;; 0 if and only if x tj. B ('Y (s), s).
Hence b'Y (x) :::;; 0 if and only if x tj. UsE(O,oo) B ('Y (s), s).^0DEFINITION B.53. A set X ~Mn is totally convex if for every x , y E
X and every minimizing geodesic a joining x and y, we have a~ X.Our interest in half spaces is explained by the following result.
PROPOSITION B.54. If 'Y is any ray in a complete noncompact Riemann-
ian manifold (Mn, g) of nonnegative sectional curvature, then the closed left
half-space lHI'Y is totally convex.
We shall give two proofs of the proposition, which are quite different in
character. The first uses the second variation formula, while the second uses
the Toponogov comparison theorem.
PROOF OF PROPOSITION B.54 IN THE CASE sect (g) > 0. If lHI'Y is not
totally convex, there are x, y E lHI'Y and a unit-speed geodesic a : [O, R] ---+ Mn
such that a (0) = x and a (R) = y, but a </:. lHI'Y" (We abuse notation by
writing a to denote its image o:([O,R]).) First observe that B('Y(t),t) ;2
B ('Y (s), s) for all t 2: s > 0, since by Lemma B.44,
p EB ('Y (s) 's) ::::} 0 < b')',S (p):::;; b')',t(p) ::::} p EB ('Y (t) 't).
Then note that by definition of lHI'Y, we have an B ('Y (so), so) =10 for some
so > 0, hence for all s 2: so. Fix any s 2: max {so, 1}. Since the image of a
is compact, there is Ts E [O, R] and a minimal geodesic f3s : [O, Rs] ---+Mn such
that f3s (0) = a(Ts) and f3s (Rs)= 'Y(s), where Rs~ L(f3s) = d(a,"((s)) < s.