1456 CHAPTER 43. INTERPOLATION IN BANACH SPACE

Then let ui ≡ a0,i− a0,i−1 = a1,i−1− a1,i. The reason these are equal is a = a0,i + a1,i =a0,i−1 +a1,i−1. Then

n

∑i=−m

ui = a0,n−a0,−(m+1) = a1,−(m+1)−a1,n.

It follows a−∑ni=−m ui = a−

(a0,n−a0,−(m+1)

)= a0,−(m+1)+a1,n, and both terms converge

to zero as m and n converge to ∞ by 43.7.42. Therefore,

K

(1,a−

n

∑i=−m

ui

)≤∣∣∣∣a0,−(m+1)

∣∣∣∣+ ||a1,n||

and so this shows a = ∑∞i=−∞ ui which is one of the claims of the lemma. Also

J(2i,ui

)≡max

(||ui||A0

,2i ||ui||A1

)≤ ||ui||A0

+2i ||ui||A1

≤ ||a0,i||A0+2i ||a1,i||A1

+

≤2(||a0,i−1||A0

+2i−1||a1,i−1||A1

)︷ ︸︸ ︷||a0,i−1||A0

+2i ||a1,i−1||A1

≤ (1+ ε)K(2i,a

)+2(1+ ε)K

(2i−1,a

)≤ 3(1+ ε)K

(2i,a

)because t→ K (t,a) is nondecreasing. This proves the lemma.

Lemma 43.7.4 If a ∈ A0∩A1, then K (t,a)≤min(1, t

s

)J (s,a) .

Proof: If s≥ t, then min(1, t

s

)= t

s and so

min(

1,ts

)J (s,a) =

ts

max(||a||A0

,s ||a||A1

)≥( t

s

)s ||a||A1

= t ||a||A1≥ K (t,a) .

Now in case s < t, then min(1, t

s

)= 1 and so

min(

1,ts

)J (s,a) = max

(||a||A0

,s ||a||A1

)≥ ||a||A0

≥ K (t,a) .

This proves the lemma.

Theorem 43.7.5 Let A0,A1,K and J be as described above. Then for all q ≥ 1 and θ ∈(0,1) ,

(A0,A1)θ ,q = (A0,A1)θ ,q,J

and furthermore, the norms are equivalent.