## Linear Operators and the Operator Norm (Pt. III)

**Point of Post: **This post is a continuation of this one.

With this we can now prove that the operator norm is, in fact, a norm:

**Theorem: ***Let and be two normed vector spaces, then the operator norm is a vector norm on . In particular for every and (where is either or ) the following holds*

**Proof: **

**: **We merely note that for every we have

Since was arbitrary we have that

: This is clear by definition since

: For this we merely note that if then

and so evidently for all and so for all . The conclusion follows.

One of the most important qualities about the operator norm is that it enables one to bound the actual value of a linear operator. More precisely:

**Theorem: ***Let and be normed vector spaces and . Then, for every one has that *

**Proof: **If the result is clear, if not we note that by definition

from where the result follows immediately.

From this we can easily see that the operator norm is submultiplicative in the sense that:

**Theorem: ***Let be normed vector spaces for and let be bounded linear operators. Then, .*

**Proof: **To prove this we merely note that for any then

and so

**References:**

1. Kreyszig, Erwin. *Introductory Functional Analysis with Applications*. New York: Wiley, 1978. Print.

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