bounded operator (original) (raw)

Let X and Y be as above, and let T:X→Y be a bounded operator. Then the norm of T is defined as the real number

∥T∥:=sup⁢{∥T⁢x∥Y∥x∥X\|x∈X∖{0}}.
Thus the operator norm is the smallest constant C∈𝐑 such that
∥T⁢x∥Y
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Now for any x∈X∖{0}, if we let y=x/∥x∥, then linearity implies that
∥T⁢y∥Y=∥T⁢(x∥x∥X)∥Y=∥T⁢x∥Y∥x∥X
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and thus it easily follows that
∥T∥
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=
In the special case when X={𝟎} is the zero vector space, any linear map T:X→Y is the zero map since T⁢(𝟎)=T⁢(𝟎𝟎)=0⁢T⁢(𝟎)=0. In this case, we define ∥T∥:=0.

1. To avoid cumbersome notational stuff usually one can simplify the symbols like ||x||X and ||T⁢x||Yby writing only ||x||, ||T⁢x||since there is a little danger in confusing which is space about calculating norms.

0.0.1 TO DO:

Lemma Any bounded operator with a finite dimensional kernel and cokernel has a closed image.

Proof By Banach’s isomorphism theorem.