- CHARACTERIZING ROUND SOLUTIONS 99
Indeed, suppose that rt C Bg(t) (k) for some t E (-oo, t~]. By (29.94) we have
Thus A+bt) < 2nk^2 , which contradicts (29.72) since t::; t~.
Recall from (29.66) that Lg(t) bt) ::; C for t E (-oo, tk]. By choosing k large
enough , we have that (29.95) implies
(29.96) rt n Bg<t)(3k/4) = 0.
Thus St(rt) contains the large almost Euclidean ball Bg(t)(3k/4). Let Xt E s:_(rt)
be such t hat (we drop the minus sign superscripts from our notation for Xt and Pt)
Pt~ d 9 (t)(Xt,/t) = sup d 9 (t)(Z, "ft)·
zES:_ (It)
Let Wt E rt be such t hat Pt= d 9 (t)(Xt,Wt)· Let
Note that Wt E at C s:_ (rt) and
(29.97)
Yt
' ' ' '
FIGURE 29.4.
Bg<t)(3k/4)
' ' '
' ' '
q ' '
From Lemmas 29. 26 and 29.29, we know that Lg(t) (at) ::; C and c ltl ::; Pt ::;
C itl. Let rt = dg(t)(q, Xt)· Then rt > Pt and q E Si~t)(rt)· By (29.90) and by
rt::; diam(g(t))::; CltJ from Lemma 29.29, we have that
Since q E sg~tJ(rt), this implies that sg~t)(rt) c Bg<tl(C) c B g(t)(k/2) fork large
enough.