Projections on Banach and Hilbert spaces

[jostpuur]

I've now encountered two different definitions for a projection.

Let X be a Banach space. An operator P on it is a projection if P^2=P.

Let H be a Hilbert space. An operator P on it is a projection if P^2=P and if P is self-adjoint.

But the Hilbert space is also a Banach space, and there's two different definitions for projections then. Are these common definitions anyway?

 

[mathwonk]

in hilbert space angles make sense whilst in banach space they do not. so in hilbert space it is reasonable to require that the projections be "orthogonal" in the sense that the kernel be orthogonal to the image.

this is guaranted by making the projections self adjoint. say are your projections also required to be bounded? or is that automatic?

 

[jostpuur]

say are your projections also required to be bounded? or is that automatic?

The text mentions "bounded linear operator" once in the beginning, and from then on talks only about "linear operators". I think that linear operator here always means a bounded linear operator, so... I would interpret this so that the projections are required to be bounded, because in the definition "linear operators" are defined to be projections with P^2=P and P^*=P conditions.

Is it common to call also non-bounded linear mappings "linear operators"? hmh... oh that is common, because derivative operators are linear operators

I somehow got an impression that ||P||=1 always, but now when we started talking about this, I'm not sure how to prove this. Is this true only in inner product spaces, or in Hilbert spaces?

 

[jostpuur]

Argh! I think I've just lived with a misunderstanding for weeks I thought that $\mathcal{L}(X,Y)$ is the set of bounded linear mappings, but in reality that is the set of all linear mappings?

 

[morphism]

L(X,Y) usually refers to the set of all linear maps from X to Y; B(X,Y) (or C(X,Y)) are the bounded ones.

The zero map is a projection (both in the Banach space and Hilbert space sense), and ||0|| = 0. On the other hand if P is a nonzero projection, then
(1) ||P|| = ||P^2|| <= ||P||^2, which implies that ||P|| >= 1; and
(2) ||x||^2 = ||Px + (1-P)x||^2 = ||Px||^2 + ||(1-P)x||^2, which implies that ||Px|| <= ||x||, and thus ||P|| <= 1.

(1) & (2) combined give us that ||P||=1. This proof is only valid on a Hilbert space, but not necessarily on a general inner product space. Do you see why?

 

[jostpuur]

L(X,Y) usually refers to the set of all linear maps from X to Y; B(X,Y) (or C(X,Y)) are the bounded ones.

The zero map is a projection (both in the Banach space and Hilbert space sense), and ||0|| = 0. On the other hand if P is a nonzero projection, then
(1) ||P|| = ||P^2|| <= ||P||^2, which implies that ||P|| >= 1; and
(2) ||x||^2 = ||Px + (1-P)x||^2 = ||Px||^2 + ||(1-P)x||^2, which implies that ||Px|| <= ||x||, and thus ||P|| <= 1.

(1) & (2) combined give us that ||P||=1. This proof is only valid on a Hilbert space, but not necessarily on a general inner product space. Do you see why?

I suppose my first explanation was wrong. Since nobody has been quoting it yet, I'm changing it.

The proof assumes (Px|x-Px)=0. If P is self-adjoint, then (Px|Px)=(P^2x|x)=(Px|x) -> (Px|x-Px)=0. I suppose that doesn't come any other way.