118 CHAPTER 4. LINEAR SPACES

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 +

1α

rx

(rx−

1α

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.

Lemma 4.5.6 Let Γ > 1 and B(a,Γr) be a ball and suppose {B(xi,ri)}mi=1 are balls

contained in B(a,Γr) such that r≤ ri and none of these balls contains the center of anotherball. Then there is a constant M (p,Γ) such that m≤M (p,Γ).

Proof: Let zi = xi−a. Then B(zi,ri) are balls contained in B(0,Γr) with no ballcontaining a center of another. Then B

(ziΓr ,

riΓr

)are balls in B(0,1) with no ball containing

the center of another. By compactness, there is a 18Γ

net for B(0,1), {yi}M(p,Γ)i=1 . Thus the

balls B(yi,

18Γ

)cover B(0,1). If m ≥M (p,Γ) , then by the pigeon hole principle, one of

these B(yi,

18Γ

)would contain some zi

Γr and z jΓr which requires

∥∥ ziΓr −

z jΓr

∥∥ ≤ 14Γ

<r j

4Γr soziΓr ∈ B

(z jΓr ,

r jΓr

). Thus m≤M (p,γ,Γ). ■

Intersections with small balls

Lemma 4.5.7 Let B be a ball having radius r and suppose B has nonempty intersectionwith the balls B1, · · · ,Bm having radii r1, · · · ,rm respectively, and as before, no Bi containsthe center of any other and the centers of the Bi are not contained in B. Suppose α > 1and r ≤ min(r1, · · · ,rm), each ri < αr. Then there exists a constant M (p,α) such thatm≤M (p,α).

Proof: Let B = B(a,r). Then each Bi is contained in B(a,2r+αr+αr) . This isbecause if y ∈ Bi ≡ B(xi,ri) ,

∥y−a∥ ≤ ∥y−xi∥+∥xi−a∥ ≤ ri + r+ ri < 2r+αr+αr