22.10. IMPLICIT FUNCTION THEOREM 729

Proof: First of all, why does the series make sense?∥∥∥∥∥ q

∑k=p

Ak

∥∥∥∥∥≤ q

∑k=p

∥∥∥Ak∥∥∥≤ q

∑k=p∥A∥k ≤

∞

∑k=p

rk ≤ rp

1− r

and so the partial sums are Cauchy in L (X ,X) . Therefore, the series converges to some-thing in L (X ,X) by completeness of this normed linear space. Now why is it the inverse?

∞

∑k=0

Ak (I−A) = limn→∞

n

∑k=0

Ak (I−A) = limn→∞

(n

∑k=0

Ak−n+1

∑k=1

Ak

)= lim

n→∞

(I−An+1)= I

because∥∥An+1

∥∥≤ ∥A∥n+1 ≤ rn+1. Similarly,

(I−A)∞

∑k=0

Ak = limn→∞

(I−An+1)= I

and so this shows that this series is indeed the desired inverse.Next suppose A ∈ O so A−1 ∈ L (X ,X) . Then suppose ∥A−B∥ < r

1+∥A−1∥ ,r < 1.

Does it follow that B is also invertible?

B = A− (A−B) = A[I−A−1 (A−B)

]Then

∥∥A−1 (A−B)∥∥≤ ∥∥A−1

∥∥∥A−B∥< r and so[I−A−1 (A−B)

]−1 exists. Hence

B−1 =[I−A−1 (A−B)

]−1A−1

Thus O is open as claimed. As to continuity, let A,B be as just described. Then using theNeuman series,

∥IA−IB∥=∥∥∥A−1−

[I−A−1 (A−B)

]−1A−1

∥∥∥=

∥∥∥∥∥A−1−∞

∑k=0

(A−1 (A−B)

)kA−1

∥∥∥∥∥=∥∥∥∥∥ ∞

∑k=1

(A−1 (A−B)

)kA−1

∥∥∥∥∥≤

∞

∑k=1

∥∥A−1∥∥k+1 ∥A−B∥k = ∥A−B∥∥∥A−1∥∥2

∞

∑k=0

∥∥A−1∥∥k(

r1+∥A−1∥

)k

≤ ∥B−A∥∥∥A−1∥∥2 1

1− r.

Thus I is continuous at A ∈ O .

Lemma 22.10.4 LetO ≡ {A ∈L (X ,Y ) : A−1 ∈L (Y,X)}

and letI : O →L (Y,X) , IA≡ A−1.