13.13. BESICOVITCH COVERING THEOREM 379

≥ r∥x−y∥∥x−a∥

− r∥y−a∥|∥y−a∥−∥x−a∥|

∥x−a∥∥y−a∥

= r∥x−y∥∥x−a∥

− r∥x−a∥

|∥y−a∥−∥x−a∥| . (13.13.35)

There are two cases. First suppose that ∥y−a∥−∥x−a∥ ≥ 0. Then the above

= r∥x−y∥∥x−a∥

− r∥x−a∥

∥y−a∥+ r.

From the assumptions, ∥x−y∥ ≥ ry and also ∥y−a∥ ≤ r+ ry. Hence the above

≥ rry

∥x−a∥− r∥x−a∥

(r+ ry)+ r = r− rr

∥x−a∥

≥ r(

1− r∥x−a∥

)≥ r(

1− rrx

)≥ r(

1− 1α

)≥ r

α−1α +1

.

The other case is that ∥y−a∥−∥x−a∥ < 0 in 13.13.35. Then in this case 13.13.35equals

= r(∥x−y∥∥x−a∥

− 1∥x−a∥

(∥x−a∥−∥y−a∥))

=r

∥x−a∥(∥x−y∥− (∥x−a∥−∥y−a∥))

Then since ∥x−a∥≤ r+rx,∥x−y∥≥ ry,∥y−a∥≥ ry, and remembering that ry ≥ rx ≥αr,

≥ rrx + r

(ry− (r+ rx)+ ry)≥r

rx + r(ry− (r+ ry)+ ry)

≥ rrx + r

(ry− r)≥ rrx + r

(rx− r)≥ rrx +

rx

(rx−

rx

)=

r1+(1/α)

(1−1/α) =α−1α +1

r

Replacing r with something larger, 1α

rx is justified by the observation that x→ α−xα+x is

decreasing. This proves the estimate between Px and Py.Finally, in the case of the balls Bi having centers at xi, then as above, let Pxi = a+

r xi−a∥xi−a∥ . Then (Pxi −a)r−1 is on the unit sphere having center 0. Furthermore,

∥∥(Pxi −a)r−1− (Pyi −a)r−1∥∥= r−1 ∥Pxi −Pyi∥ ≥ r−1rα−1α +1

=α−1α +1

.

How many points on the unit sphere can be pairwise this far apart? The unit sphere iscompact and so there exists a 1

4

(α−1α+1

)net having L(p,α) points. Thus m cannot be any

larger than L(p,α) because if it were, then by the pigeon hole principal, two of the points(Pxi −a)r−1 would lie in a single ball B

(p, 1

4

(α−1α+1

))so they could not be α−1

α+1 apart.The above lemma has to do with balls which are relatively large intersecting a given

ball. Next is a lemma which has to do with relatively small balls intersecting a given ball.First is another lemma.

13.13. BESICOVITCH COVERING THEOREM 379x=yil lly =alllily—all = |x=all~ |x—all I|x—all |ly—allIIx—yl| r= - Illy —al| — ||x—all|. (13.13.35)|Ix—al] ||x—allThere are two cases. First suppose that ||y — al] — ||x —a|| > 0. Then the aboveIIx—yll r= - lly all +r.|Ix—al]— ||x—allFrom the assumptions, ||x — y|| > ry and also ||y —al| < r+ ry. Hence the abovery r r, (rtry)+r=rr _ — r-——_I|Ix—al]— |[x—all I|x—all|r r 1 a—1> rl l—-m]>r(1l—-—]>r(1—-—)]er .||x — al ry a a+lThe other case is that ||y —al| — ||x—al| < 0 in 13.13.35. Then in this case 13.13.35equalsIx—yl] 1r( = gy llx—all lly alI|x—all |xr~ x—al (\x—y|| —(|x—all —lly—all))Then since ||x—al| <r+r,, |x —y|| > 17, |ly—al] > ry, and remembering that ry > r, > ar,r2 pop Uy (tte) Fry) 2 (ty = (ry) +9)r r r 1> —r)> —r)> ——r/,— rae r) 2 x ieee G x")r a-—l= mm -(1-1/a) = —Ted /ay P/M =oaara-xa+x 1SReplacing r with something larger, ary is justified by the observation that x >decreasing. This proves the estimate between P, and Py.Finally, in the case of the balls B; having centers at x;, then as above, let Py, = a+Xi-2- Then (Px, —a)r! is on the unit sphere having center 0. Furthermore,r .IIx;—al_ _ _ _, a-1 a-tl\| (Pe —a)r '— (Py, —a)r ‘=r "Py Pull 7 ORT onl"How many points on the unit sphere can be pairwise this far apart? The unit sphere iscompact and so there exists a i (2) net having L(p,a@) points. Thus m cannot be anylarger than L(p, &) because if it were, then by the pigeon hole principal, two of the points(Px, —a)r~| would lie in a single ball B (p, + (45+)) so they could not be 2=} apart.The above lemma has to do with balls which are relatively large intersecting a givenball. Next is a lemma which has to do with relatively small balls intersecting a given ball.First is another lemma.