If and are normed spaced and , then a mapping is called a linear operator if for every and (the underlying field),

If is a linear operator, then is injective if and only if implies .

**Proof**: Suppose is injective. Then , since is linear, it must map zero to zero, so . Now suppose that only for . If and , then and so (by linearity), and this means that , so .

Now, we say that is bounded if there exists a such that for every we have . If such a exists (i.e., if is bounded), then we say that the least such is the operator norm of , and write . We can then find as follows.

(This follows from rewriting the above earlier inequality as .) Note that by letting , we have that .

In a finite dimensional normed space, every linear operator is bounded. Suppose that and that is a basis for . If we let with , then we have the following.

Where and is the value such that .

Another important property is that a linear operator is bounded if and only if it is continuous.

**Proof**: Suppose is bounded and let be given. Let . Let be a fixed point in , and some other point in such that .

This shows that is continuous at . However, our was arbitrary, and didn’t depend on our choice of , so is uniformly continuous.

For the converse I’m going to copy the proof from Wikipedia, since I think it’s a bit clearer than Kreyszig’s proof.

Suppose that is continuous. Since is a subspace of , it contains the zero vector. Since is continuous, it’s continuous at the zero vector. This means there exists a such that for all satisfying we have that . Now let be any point in .

And so is bounded.

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Rudin’s approach is to define the operator norm (equivalently) as . Which turns out to be equivalent to just checking the sup over which is equivalent to what you did.

Note: when I post random stuff like this in the comments, it isn’t because I think you don’t know them, it is because my prelim is coming up, too, so it is a nice refresher for me.

Comment by hilbertthm90 — July 18, 2008 @ 5:00 pm |

[…] assume that is either or . Properties and theorems associated with “traditional” linear operators apply since and can be thought of as normed spaces with the “traditional” norms […]

Pingback by The Algebraic Dual Space « Mathematics Prelims — July 18, 2008 @ 10:27 pm |

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