728 CHAPTER 22. THE DERIVATIVE

follows X ,Y will be Banach spaces, complete normed linear spaces. Thus these are com-plete normed linear space and L (X ,Y ) is the space of bounded linear maps. I will alsocease trying to write the vectors in bold face partly to emphasize that these are not in Rn.

Definition 22.10.1 Let (X ,∥·∥X ) and (Y,∥·∥Y ) be two normed linear spaces. L (X ,Y )denotes the set of linear maps from X to Y which also satisfy the following condition. ForL ∈L (X ,Y ) ,

lim∥x∥X≤1

∥Lx∥Y ≡ ∥L∥< ∞

Recall that this operator norm is less than infinity is always the case where X is fi-nite dimensional. However, if you wish to consider infinite dimensional situations, youassume the operator norm is finite as a qualification for being in L (X ,Y ). Then here is animportant theorem.

Theorem 22.10.2 If Y is a Banach space, then L (X ,Y ) is also a Banach space.

Proof: Let {Ln} be a Cauchy sequence in L (X ,Y ) and let x ∈ X .

||Lnx−Lmx|| ≤ ||x|| ||Ln−Lm||.

Thus {Lnx} is a Cauchy sequence. Let

Lx = limn→∞

Lnx.

Then, clearly, L is linear because if x1,x2 are in X , and a,b are scalars, then

L(ax1 +bx2) = limn→∞

Ln (ax1 +bx2)

= limn→∞

(aLnx1 +bLnx2)

= aLx1 +bLx2.

Also L is bounded. To see this, note that {||Ln||} is a Cauchy sequence of real numbersbecause |||Ln||− ||Lm||| ≤ ||Ln−Lm||. Hence there exists K > sup{||Ln|| : n ∈ N}. Thus, ifx ∈ X ,

∥Lx∥= limn→∞||Lnx|| ≤ K ∥x∥ .

The following theorem is really nice. The series in this theorem is called the Neumanseries.

Lemma 22.10.3 Let (X ,∥·∥) is a Banach space, and if A ∈ L (X ,X) and ∥A∥ = r < 1,then

(I−A)−1 =∞

∑k=0

Ak ∈L (X ,X)

where the series converges in the Banach space L (X ,X). If O consists of the invertiblemaps in L (X ,X) , then O is open and if I is the mapping which takes A to A−1, then I iscontinuous.