## Bounded Linear Transformations

In the context of normed vector spaces we have a topology on our spaces and so it makes sense to ask that maps between them be continuous. In the finite-dimensional case, all linear functions are continuous, so this hasn’t really come up before in our study of linear algebra. But for functional analysis, it becomes much more important.

Now, really we only need to require continuity at one point — the origin, to be specific — because if it’s continuous there then it’ll be continuous everywhere. Indeed, continuity at means that for any there is a so that implies . In particular, if , then this means implies . Clearly if this holds, then the general version also holds.

But it turns out that there’s another equivalent condition. We say that a linear transformation is “bounded” if there is some such that for all . That is, the factor by which stretches the length of a vector is bounded. By linearity, we only really need to check this on the unit sphere , but it’s often just as easy to test it everywhere.

Anyway, I say that a linear transformation is continuous if and only if it’s bounded. Indeed, if is bounded, then we find

so as we let approach — as approaches — the difference between and approaches zero as well. And so is continuous.

Conversely, if is continuous, then it is bounded. Since it’s continuous, we let and find a so that for all vectors with . Thus for all nonzero we find

Thus we can use and conclude that is bounded.

The least such that works in the condition for to be bounded is called the “operator norm” of , which we write as . It’s straightforward to verify that , and that if and only if is the zero operator. It remains to verify the triangle identity.

Let’s say that we have bounded linear transformations and with operator norms and , respectively. We will show that works as a bound for , and thus conclude that . Indeed, we check that

and our assertion follows. In particular, when our base field is itself a normed linear space (like or itself) we can conclude that the “continuous dual space” consisting of bounded linear functionals is a normed linear space using the operator norm on .

[…] (but not its Hölder conjugate ) is infinite. And I say that when is -finite, the space of bounded linear functionals on is isomorphic to […]

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