328 CHAPTER 13. NORMS

Corollary 13.0.8 Suppose X is a finite dimensional linear space with the field of scalarseither C or R and ||·|| and |||·||| are two norms on X. Then there exist positive constants, δand ∆, independent of x ∈ X such that

δ |||x||| ≤ ||x|| ≤ ∆ |||x||| .

Thus any two norms are equivalent.

This is very important because it shows that all questions of convergence can be consid-ered relative to any norm with the same outcome.

Proof: Let {v1, · · · ,vn} be a basis for X and let |·| be the norm taken with respect tothis basis which was described earlier. Then by Theorem 13.0.4, there are positive constantsδ1,∆1, δ2,∆2, all independent of x ∈X such that

δ2 |||x||| ≤ |x| ≤ ∆2 |||x||| , δ1 ||x|| ≤ |x| ≤ ∆1 ||x|| .

Then

δ2 |||x||| ≤ |x| ≤ ∆1 ||x|| ≤∆1

δ1|x| ≤ ∆1∆2

δ1|||x|||

and soδ2∆1

|||x||| ≤ ||x|| ≤ ∆2

δ1|||x||| ■

Definition 13.0.9 Let X and Y be normed linear spaces with norms ||·||X and ||·||Y re-spectively. Then L (X,Y ) denotes the space of linear transformations, called bounded lineartransformations, mapping X to Y which have the property that

||A|| ≡ sup {||Ax||Y : ||x||X ≤ 1} <∞.

Then ||A|| is referred to as the operator norm of the bounded linear transformation A.

It is an easy exercise to verify that ||·|| is a norm on L (X,Y ) and it is always the casethat

||Ax||Y ≤ ||A|| ||x||X .

Furthermore, you should verify that you can replace ≤ 1 with = 1 in the definition. Thus

||A|| ≡ sup {||Ax||Y : ||x||X = 1} .

Theorem 13.0.10 Let X and Y be finite dimensional normed linear spaces of dimensionn and m respectively and denote by ||·|| the norm on either X or Y . Then if A is any linearfunction mapping X to Y, then A ∈ L (X,Y ) and (L (X,Y ) , ||·||) is a complete normedlinear space of dimension nm with

||Ax|| ≤ ||A|| ||x|| .

Also if A ∈ L (X,Y ) and B ∈ L (Y, Z) where X,Y, Z are normed linear spaces,

∥BA∥ ≤ ∥B∥ ∥A∥

Proof: It is necessary to show the norm defined on linear transformations really is anorm. Again the first and third properties listed above for norms are obvious. It remains to