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

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

*Operator Norm*

We saw in the proof of the last theorem that given a bounded linear operator there are many choices for the bounding constant. Of course a natural question is to ask what is the minimum such constant? This leads us naturally to the notion of the operator norm which is defined to be the infimum of all such constants, and it turns out (as the name suggests) that this forms a norm on the vector space (it’s obvious that it is one under pointwise scalar multiplication and addition, by use of the triangle inequality, etc.) of all bounded linear operators denoted . To be more explicit if and are normed vector spaces we can define the function given by

We know that this function makes sense since the set on the right is non-empty (by assumption of boundedness) and bounded below by . We call the *operator norm *on . Before we actually show it’s a norm we find a few alternate definitions of this, as for now, function.

**Theorem: ***Let and be two normed vector spaces, then for any one has *

**Proof: **To prove the first equality we note by definition that for any which is an upper bound for (in the sense of the definition of boundendess) and any one has

since this is true for all upper bounds and all we may conclude that

That said, it’s evidently true that for every one has

and so evidently for every

and so

from where the first equality follows. To prove the second equality we note that evidently

since for one has that . That said

from where the reverse inequality readily follows and thus the entire second equality. Lastly, to prove the last equality we note that it’s evident that . That said, let and note that

and so the last equality holds.

**References:**

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